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## Big Ideas Math Book Algebra 1 Answer Key Chapter 8 Graphing Quadratic Functions

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- Graphing Quadratic Functions Maintaining Mathematical Proficiency – Page 417
- Graphing Quadratic Functions Mathematical Practices – Page 418
- Lesson 8.1 Graphing f(x) = ax2 – Page (419-424)
- Graphing f(x) = ax2 8.1 Exercises – Page (423-424)
- Lesson 8.2 Graphing f(x) = ax2 + c – Page (425-430)
- Graphing f(x) = ax2 + c 8.2 Exercises – Page (429-430)
- Lesson 8.3 Graphing f(x) = ax2 + bx + c – Page (431-438)
- Graphing f(x) = ax2 + bx + c 8.3 Exercises – Page (436-438)
- Graphing Quadratic Functions Study Skills: Learning Visually – Page 439
- Graphing Quadratic Functions 8.1 – 8.3 Quiz – Page 440
- Lesson 8.4 Graphing f(x) = a(x – h)2 + k – Page (441-448)
- Graphing f(x) = a(x – h)2 + k 8.4 Exercises – Page (446-448)
- Lesson 8.5 Using Intercept Form – Page (449-458)
- Using Intercept Form 8.5 Exercises – Page (455-458)
- Lesson 8.6 Comparing Linear, Exponential, and Quadratic Functions – Page (459-468)
- Comparing Linear, Exponential, and Quadratic Functions 8.6 Exercises – Page (465-468)
- Graphing Quadratic Functions Performance Task: Asteroid Aim – Page 469
- Graphing Quadratic Functions Chapter Review – Page (470-472)
- Graphing Quadratic Functions Chapter Test – Page 473
- Graphing Quadratic Functions Cumulative Assessment – Page (474-475)

### Graphing Quadratic Functions Maintaining Mathematical Proficiency

**Graph the linear equation.**

Question 1.

y = 2x – 3

Answer:

y = 2x – 3

If x = 0 → 2(0) – 3 = -3

If x = 1 → 2(1) – 3 = -1

If x = 2 → 2(2) – 3 = 1

If x = 3 → 2(3) – 3 = 3

If x = 4 → 2(4) – 3 = 5

Question 2.

y = -3x + 4

Answer:

Given,

y = -3x + 4

If x = 0 → -3(0) + 4 = 4

If x = 1 → -3(1) + 4 = 1

If x = 2 → -3(2) + 4 = -2

If x = 3 → -3(3) + 4 = -5

If x = 4 → -3(4) + 4 = -8

Question 3.

y = – \(\frac{1}{2}\)x – 2

Answer:

Given,

y = – \(\frac{1}{2}\)x – 2

If x = 0 → – \(\frac{1}{2}\)(0) – 2 = -2

If x = 1 → – \(\frac{1}{2}\)(1) – 2 = -2\(\frac{1}{2}\)

If x = 2 → –\(\frac{1}{2}\)(2) – 2 = -3

If x = 3 → – \(\frac{1}{2}\)(3) – 2 = –\(\frac{7}{2}\)

If x = 4 → – \(\frac{1}{2}\)(4) – 2 = -4

Question 4.

y = x + 5

Answer:

Given,

y = x + 5

If x = 0 → 0 + 5 = 5

If x = 1 → 1 + 5 = 6

If x = 2 → 2 + 5 = 7

If x = 3 → 3 + 5 = 8

If x = 4 → 4 + 5 = 9

**Evaluate the expression when x = −2.**

Question 5.

5x^{2} – 9

Answer:

Given,

5x^{2} – 9

Now we have to substitute x = -2 in the above expression

5(-2)^{2} – 9

= 5(4) – 9

= 20 – 9

= 11

Question 6.

3x^{2} + x – 2

Answer:

Given,

3x^{2} + x – 2

Now we have to substitute x = -2 in the above expression

3(-2)^{2} + (-2) – 2

= 3(4) – 2 – 2

= 12 – 4

= 8

Question 7.

-x^{2} + 4x + 1

Answer:

Given,

-x^{2} + 4x + 1

Now we have to substitute x = -2 in the above expression

-(-2)^{2} + 4(-2) + 1

= -4 – 8 + 1

= -12 + 1

= -11

Question 8.

x^{2} + 8x + 5

Answer:

Given,

x^{2} + 8x + 5

Now we have to substitute x = -2 in the above expression

(-2)^{2} + 8(-2) + 5

= 4 – 16 + 5

= -7

Question 9.

-2x^{2} – 4x + 3

Answer:

Given,

-2x^{2} – 4x + 3

Now we have to substitute x = -2 in the above expression

= -2(-2)^{2} – 4(-2) + 3

= -2(4) + 8 + 3

= -8 + 8 + 3

= 3

Question 10.

-4x^{2} + 2x – 6

Answer:

Given,

-4x^{2} + 2x – 6

Now we have to substitute x = -2 in the above expression

-4(-2)^{2} + 2(-2) – 6

= -16 – 4 – 6

= -26

Question 11.**ABSTRACT REASONING**

Complete the table. Find a pattern in the differences of consecutive y-values. Use the pattern to write an expression for y when x = 6.

Answer:

### Graphing Quadratic Functions Mathematical Practices

Mathematically proficient students try special cases of the original problem to gain insight into its solution.

**Monitoring Progress**

**Graph the quadratic function. Then describe its graph.**

Question 1.

y = -x^{2}

Answer:

Question 2.

y = 2x^{2}

Answer:

Question 3.

f(x) = 2x^{2} + 1

Answer:

Question 4.

f(x) = 2x^{2} – 1

Answer:

Question 5.

f(x) = \(\frac{1}{2}\)x^{2} + 4x + 3

Answer:

Question 6.

f(x) = \(\frac{1}{2}\) x^{2} – 4x + 3

Answer:

Question 7.

y = -2(x + 1)^{2} + 1

Answer:

Question 8.

y = -2(x – 1)^{2} + 1

Answer:

Question 9.

How are the graphs in Monitoring Progress Questions 1-8 similar? How are they different?

Answer: All the graphs from 1-8 are quadratic. The quadrants of the graphs are different and they are not equal to zero.

### Lesson 8.1 Graphing f(x) = ax^{2}

**Essential Question** What are some of the characteristics of the graph of a quadratic function of the form f(x) = ax^{2}?

**EXPLORATION 1**

Graphing Quadratic Functions

Work with a partner. Graph each quadratic function. Compare each graph to the graph of f(x) = x^{2}.

Answer:

a.

b.

c.

d.

**Communicate Your Answer**

Question 2.

What are some of the characteristics of the graph of a quadratic function of the form f(x) = ax^{2}?

Answer:

The graph of a quadratic function is U-shaped and known as a parabola.

Parabolas have several recognizable features that characterize their shape and placement on the Cartesian plane.

Question 3.

How does the value of a affect the graph of f(x) = ax^{2}? Consider 0 < a < 1, a> 1, -1 < a < 0, and a < -1. Use a graphing calculator to verify your answers.

Answer:

0 < a < 1

a> 1

-1 < a < 0

a < -1

Question 4.

The figure shows the graph of a quadratic function of the form y = ax^{2}. Which of the intervals in Question 3 describes the value of a? Explain your reasoning.

Answer:

**Monitoring Progress**

**Identify characteristics of the quadratic function and its graph.**

Question 1.

Answer:

Vertex is (2, -3)

The axis of symmetry is 2

The domain is all real numbers

the Range is all real numbers greater than or equal to -3

Question 2.

Answer:

Vertex is (-3, 7)

The axis of symmetry is -3

The domain is all real numbers

the Range is all real numbers greater than or equal to 7

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**Question 3.

g(x) = 5x

^{2}

Answer:

Question 4.

h(x) = \(\frac{1}{3}\)x^{2}

Answer:

Question 5.

n(x) = \(\frac{3}{2}\)x^{2}

Answer:

Question 6.

p(x) = -3x^{2}

Answer:

Question 7.

q(x) = -0.1x^{2}

Answer:

Question 8.

g(x) = –\(\frac{1}{4}\)x^{2}

Answer:

Question 9.

The cross section of a spotlight can be modeled by the graph of y = 0.5x^{2}, where x and y are measured in inches and -2 ≤ x ≤ 2. Find the width and depth of the spotlight.

Answer:

### Graphing f(x) = ax^{2} 8.1 Exercises

**Vocabulary and Core Concept Check**

Question 1.**VOCABULARY**

What is the U-shaped graph of a quadratic function called?

Answer:

The U-shaped graph of a quadratic function is called a parabola.

Question 2.**WRITING**

When does the graph of a quadratic function open up? open down?

Answer:

When a < 0 it opens down

When a > 0 it opens up

**Monitoring Progress and Modeling with Mathematics**

**In Exercises 3 and 4, identify characteristics of the quadratic function and its graph.**

Question 3.

Answer:

Question 4.

Answer:

The vertex is (-2, 4)

The axis of symmetry is x = -2

The domain is all real numbers.

The range is y ≥ 4

When x < -2, y increases as x decreases.

When x > -2, y increases as x decreases.

**In Exercises 5–12, graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 5.

g(x) = 6x

^{2}

Answer:

Question 6.

b(x) = 2.5x^{2}

Answer:

b(x) = 2.5x^{2}

Axis of symmetry is x = 0

Graph of 2.5x^{2 }is narrower than x^{2 } since it is stretched vertically by factor 2.5

Question 7.

h(x) = \(\frac{1}{4}\)x^{2}

Answer:

Question 8.

j(x) = 0.75x^{2}

Answer:

We observe that the two graphs have the same domain, range, axis of symmetry and are both parabolas. The main difference is that the graph of the given function opens wider than the graph of f(x) = x^{2}

Question 9.

m(x) = -2x^{2}

Answer:

Question 10.

q(x) = –\(\frac{9}{2}\)x^{2}

Answer:

We observe that the two graphs have the same domain, range, axis of symmetry and are both parabolas. The main difference is that the graph of the given function opens wider than the graph of f(x) = x^{2 }has been reflected about the x-axis.

Question 11.

k(x) = -0.2x^{2}

Answer:

Question 12.

p(x) = –\(\frac{2}{3}\)x^{2}

Answer:

We observe that the two graphs have the same vertex, range, axis of symmetry and are both parabolas. The main difference is that the graph of the given function opens wider than the graph of f(x) = x^{2 }has been reflected about the x-axis and vertically stretched by a factor 2/3

**In Exercises 13–16, use a graphing calculator to graph the function. Compare the graph to the graph of y = −4x ^{2}.**

Question 13.

y = 4x

^{2}

Answer:

Question 14.

y = -0.4x^{2}

Answer:

We observe that the two graphs have the same domain, range, axis of symmetry and are both parabolas. The main difference is that the graph of the given function opens wider than the graph of f(x) = x^{2}

Question 15.

y = -0.04x^{2}

Answer:

Question 16.

y = -0.004x^{2}

Answer:

We observe that the two graphs have the same domain, range, axis of symmetry and are both parabolas. The main difference is that the graph of the given function opens wider than the graph of f(x) = -4x^{2}

Question 17.**ERROR ANALYSIS**

Describe and correct the error in graphing and comparing y = x^{2} and y = 0.5x^{2}.

Answer:

Question 18.**MODELING WITH MATHEMATICS**

The arch support of a bridge can be modeled by y = -0.0012x^{2}, where x and y are measured in feet. Find the height and width of the arch.

Answer:

Width of arch = 500 – (-500)

= 1000 feet

Height of arch = 0 – (-300)

= 300 feet

Question 19.**PROBLEM SOLVING**

The breaking strength z (in pounds) of a manila rope can be modeled by z = 8900d^{2}, where d is the diameter (in inches) of the rope.

a. Describe the domain and range of the function.

b. Graph the function using the domain in part (a).

c. A manila rope has four times the breaking strength of another manila rope. Does the stronger rope have four times the diameter? Explain.

Answer:

Question 20.**HOW DO YOU SEE IT?**

Describe the possible values of a.

Answer:

a. The graph of g(x) = ax² will be narrower than the graph of f(x) = x² if a > 1

b. The graph of g(x) = ax² will be wider than the graph of f(x) = x² if 0 < |a| < 1. However we also note than the graph has been reflected about the x-axis and thus a has to be negative. Then we know that -1 < a < 0.

**ANALYZING GRAPHS** In Exercises 21–23, use the graph.

Question 21.

When is each function increasing?

Answer:

Question 22.

When is each function decreasing?

Answer: f is decreasing when g < 0. g is decreasing when x > 0

Question 23.

Which function could include the point (-2, 3)? Find the value of a when the graph passes through (-2, 3).

Answer:

Question 24.**REASONING**

Is the x-intercept of the graph of y = x^{2} always 0? Justify your answer.

Answer:

The x-intercept is the intersection of the graph of the function with the x-axis and thus y = 0.

Thus let us replace y with 0 in the given function

0 = ax^{2}

0 = x^{2}

0 = x

Then we note that 0 is the only x-intercept.

Question 25.**REASONING**

A parabola opens up and passes through (-4, 2) and (6, -3). How do you know that (-4, 2) is not the vertex?

Answer:

**ABSTRACT REASONING** In Exercises 26–29, determine whether the statement is always, sometimes, or never true. Explain your reasoning.

Question 26.

The graph of f(x) = x^{2} is narrower than the graph of g(x) = x^{2} when a > 0.

Answer:

The given statement is sometimes true, because the graph f(x) = x^{2} is narrower than the graph of g(x) = x^{2} if a > 1, wider if a < 1 and equaly wide of a = 1.

Question 27.

The graph of f(x) = x^{2} is narrower than the graph of g(x) = x^{2} when |a| > 1.

Answer:

Question 28.

The graph of f(x) = x^{2} is wider than the graph of g(x) = x^{2} when 0 < |a| < 1.

Answer:

The given statement is sometimes true, because the graph f(x) = x^{2} is narrower than the graph of g(x) = x^{2} if a > 1, wider if -1< a < 0 or 0 < a < 1.

Question 29.

The graph of f(x) = x^{2} is wider than the graph of g(x) = dx^{2} when |a | > |d| .

Answer:

Question 30.**THOUGHT PROVOKING**

Draw the isosceles triangle shown. Divide each leg into eight congruent segments. Connect the highest point of one leg with the lowest point of the other leg. Then connect the second highest point of one leg to the second lowest point of the other leg. Continue this process. Write a quadratic equation whose graph models the shape that appears.

Answer:

From the parent function y = x², the transformation is reflection about the x-axis so we have y = -ax²

To find a, substitute either (-6, -4) or (6, -4)

-4 = -a(6)²

-4 = -36a

a = 1/9

y = -1/9x²

Question 31.**MAKING AN ARGUMENT**

The diagram shows the parabolic cross section of a swirling glass of water, where x and y are measured in centimeters.

a. About how wide is the mouth of the glass?

b. Your friend claims that the rotational speed of the water would have to increase for the cross section to be modeled by y = 0.1x^{2}. Is your friend correct? Explain your reasoning.

Answer:

**Maintaining Mathematical Proficiency**

**Evaluate the expression when n = 3 and x = −2.**

Question 32.

n^{2} + 5

Answer:

n^{2} + 5 =3^{2} + 5

= 9 + 5

= 14

Question 33.

3x^{2} – 9

Answer:

Question 34.

-4n^{2} + 11

Answer:

-4n^{2} + 11 = -4(3)^{2} + 11

= -4(9) + 11

= -36 + 11

= -25

Question 35.

n + 2x^{2}

Answer:

### Lesson 8.2 Graphing f(x) = ax^{2} + c

**Essential Question** How does the value of c affect the graph of f(x) = -ax^{2} + c?

**EXPLORATION 1**

Graphing y = ax^{2} + c

Work with a partner. Sketch the graphs of the functions in the same coordinate plane. What do you notice?

Answer:

a.

b.

**EXPLORATION 2**

Finding x-Intercepts of Graphs

Work with a partner. Graph each function. Find the x-intercepts of the graph. Explain how you found the x-intercepts.

Answer:

a.

b.

**Communicate Your Answer**

Question 3.

How does the value of c affect the graph of f(x) = ax^{2} + c?

Answer:

Given equation

f(x) = ax^{2} + c

‘c’ tell us the intercept of graph.

Question 4.

Use a graphing calculator to verify your answers to Question 3.

Answer:

Question 5.

The figure shows the graph of a quadratic function of the form y = ax^{2} + c. Describe possible values of a and c. Explain your reasoning.

Answer:

**Monitoring Progress**

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 1.

g(x) = x

^{2}– 5

Answer:

Question 2.

h(x) = x^{2} + 3

Answer:

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 3.

g(x) = 2x

^{2}– 5

Answer:

Question 4.

h(x) = – \(\frac{1}{4}\)x^{2} + 4

Answer:

Question 5.

Let f(x) = 3x^{2} – 1 and g(x) = f (x) + 3.

a. Describe the transformation from the graph of f to the graph of g. Then graph f and g in the same coordinate plane.

b. Write an equation that represents g in terms of x.

Answer:

g(x) = f(x) + 3

g(x) = (3x^{2} – 1 )+ 3

= 3x^{2} + 2

Question 6.

Explain why only nonnegative values of t are used in Example 4.

Answer:

Question 7.**WHAT IF?**

The egg is dropped from a height of 100 feet. After how many seconds does the egg hit the ground?

Answer:

h = 0 and t = 2.5 sec

### Graphing f(x) = ax^{2} + c 8.2 Exercises

**Vocabulary and Core Concept Check**

Question 1.**VOCABULARY**

State the vertex and axis of symmetry of the graph of y = ax^{2} + c.

Answer:

The graph of y = ax^{2} + c has a vertex of (0, c) and an axis of symmetry of x = 0

Question 2.**WRITING**

How does the graph of y = ax^{2} + c compare to the graph of y = ax^{2}?

Answer:

y = ax^{2}

y = ax^{2} + c

The graph of the function y = ax^{2} + c is the graph of the function y = ax^{2 }translated up with c units if c > 0 and down with |c| units if c < 0

**0Monitoring Progress and Modeling with Mathematics**

**In Exercises 3–6, graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 3.

g(x) = x

^{2}+ 6

Answer:

Question 4.

h(x) = x^{2} + 8

Answer:

We note that both graphs are parabolas and have the same range and axis of symmetry. The main difference between the two graphs is that the graph of the given function is the graph of f(x) = x^{2} translated up by 8 units.

Question 5.

p(x) = x^{2} – 3

Answer:

Question 6.

q(x) = x^{2} – 1

Answer:

We note that both graphs are parabolas and have the same range and axis of symmetry. The main difference between the two graphs is that the graph of the given function is the graph of f(x) = x^{2} translated up by 1 unit.

**In Exercises 7–12, graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 7.

g(x) = -x

^{2}+ 3

Answer:

Question 8.

h(x) = -x^{2} – 7

Answer:

We note that both graphs are parabolas and have the same range and axis of symmetry. The main difference between the two graphs is that the graph of the given function is the graph of f(x) = x^{2} translated down by 7 units.

Question 9.

s(x) = 2x^{2} – 4

Answer:

Question 10.

t(x) = -3x^{2} + 1

Answer:

We note that both graphs are parabolas and have the same range and axis of symmetry. The main difference between the two graphs is that the graph of the given function is the graph of f(x) = x^{2} translated up by 1 unit.

Question 11.

p(x) = – \(\frac{1}{3}\)x^{2} – 2

Answer:

Question 12.

q(x) = \(\frac{1}{2}\)x^{2} + 6

Answer:

We note that both graphs are parabolas and have the same range and axis of symmetry. The main difference between the two graphs is that the graph of the given function is the graph of f(x) = x^{2} translated up by 6 units.

In Exercises 13–16, describe the transformation from the graph of f to the graph of g. Then graph f and g in the same coordinate plane. Write an equation that represents g in terms of x.

Question 13.

f(x) = 3x^{2} + 4

g(x) = f(x) + 2

Answer:

Question 14.

f(x) = \(\frac{1}{2}\)x^{2} + 1

g(x) = f(x) – 4

Answer:

We note that the function g is the function f decreased by 4 and thus its graph will be translated down 4 units.

Question 15.

f(x) = – \(\frac{1}{4}\)x^{2} – 6

g(x) = f(x) – 3

Answer:

Question 16.

f(x) = 4x^{2} – 5

g(x) = f(x) + 7

Answer:

We note that the function g is the function f increased by 7 and thus its graph will be translated up by 7 units.

Question 17.**ERROR ANALYSIS**

Describe and correct the error in comparing the graphs.

Answer:

Question 18.**ERROR ANALYSIS**

Describe and correct the error in graphing and comparing f(x) = x^{2} and g(x) = x^{2} – 10.

Answer:

The error is that the graph g(x) = x^{2} – 10 is the graph of f(x) = x^{2} translated down by 10 units instead of up.

**In Exercises 19–26, find the zeros of the function.**

Question 19.

y = x^{2} – 1

Answer:

Question 20.

y = x^{2} – 36

Answer:

y = x^{2} – 36

x^{2} – 36 = 0

x^{2} = 36

x = ±6

Question 21.

f(x) = -x^{2} + 25

Answer:

Question 22.

f(x) = -x^{2} + 49

Answer:

f(x) = -x^{2} + 49

-x^{2} + 49 = 0

-x^{2} = -49

x = ±7

Question 23.

f(x) = 4x^{2} – 16

Answer:

Question 24.

f(x) = 3x^{2} – 27

Answer:

f(x) = 3x^{2} – 27

3x^{2} – 27 = 0

3x^{2} = 27

x^{2} = 9

x = ±3

Question 25.

f(x) = -12x^{2} + 3

Answer:

Question 26.

f(x) = -8x^{2} + 98

Answer:

f(x) = -8x^{2} + 98

-8x^{2} + 98 = 0

-8x^{2} = -98

x = ±7/2

Question 27.**MODELING WITH /MATHEMATICS**

A water balloon is dropped from a height of 144 feet.

a. After how many seconds does the water balloon hit the ground?

b. Suppose the initial height is adjusted by k feet. How does this affect part (a)?

Answer:

Question 28.**MODELING WITH MATHEMATICS**

The function y = -16x^{2} + 36 represents the height y (in feet) of an apple x seconds after falling from a tree. Find and interpret the x- and y-intercepts.

Answer:

y = -16x^{2} + 36

-16x^{2} + 36 = 0

x^{2} = 36/16

x = ±√9/4

x = ±3/2 = ±1.5

So, the x-intercepts are (-1.5, 0) and (1.5, 0)

x = -1.5 does not make sense in the context of the problem

x = 1.5 represents 1.5s, the time the apple will hit the ground.

To find the y-intercept set x = 0 and solve for y:

y = -16(0) + 36

y = 36

The y-intercept is (0, 36)

y = 36 represents 36 ft, the initial height the apple was dropped.

**In Exercises 29–32, sketch a parabola with the given characteristics.**

Question 29.

The parabola opens up, and the vertex is (0, 3).

Answer:

Question 30.

The vertex is (0, 4), and one of the x-intercepts is 2.

Answer:

Question 31.

The related function is increasing when x < 0, and the zeros are -1 and 1.

Answer:

Question 32.

The highest point on the parabola is (0, -5).

Answer:

Question 33.**DRAWING CONCLUSIONS**

You and your friend both drop a ball at the same time. The function h(x) = -16x^{2} + 256 represents the height (in feet) of your ball after x seconds. The function g(x) = -16x^{2} + 300 represents the height (in feet) of your friend’s ball after x seconds.

a. Write the function T(x) = h(x) – g(x). What does T(x) represent?

b. When your ball hits the ground, what is the height of your friend’s ball? Use a graph to justify your answer.

Answer:

Question 34.**MAKING AN ARGUMENT**

Your friend claims that in the equation y = ax^{2} + c, the vertex changes when the value of a changes. Is your friend correct? Explain your reasoning.

Answer:

No, a only determines how narrow the function is and if it is reflected about the x-axis (if a < 0). If c changes however, then the vertex will also change.

Question 35.**MATHEMATICAL CONNECTIONS**

The area A (in square feet) of a square patio is represented by A = x^{2}, where x is the length of one side of the patio. You add 48 square feet to the patio, resulting in a total area of 192 square feet. What are the dimensions of the original patio? Use a graph to justify your answer.

Answer:

Question 36.**HOW DO YOU SEE IT?**

The graph of f(x) = ax^{2} + c is shown. Points A and B are the same distance from the vertex of the graph of f. Which point is closer to the vertex of the graph of f as c increases?

Answer:

As c increases, the graph will be translated up and thus the vertex will go closer to the upper point A.

Question 37.**REASONING**

Describe two algebraic methods you can use to find the zeros of the function f(t) = -16t^{2} + 400. Check your answer by graphing.

Answer:

Question 38.**PROBLEM SOLVING**

The paths of water from three different garden waterfalls are given below. Each function gives the height h (in feet) and the horizontal distance d (in feet) of the water.

Waterfall 1 h = -3.1d^{2} + 4.8

Waterfall 2 h = -3.5d^{2} + 1.9

Waterfall 3 h = -1.1d^{2} + 1.6

a. Which waterfall drops water from the highest point?

Answer: The constant term represents the initial height so from the three waterfall 1 drops water from the highest point.

b. Which waterfall follows the narrowest path?

Answer: Waterfall 3 follows the narrowest path

c. Which waterfall sends water the farthest?

Answer: Waterfall 1 sends water the farthest.

Question 39.**WRITING EQUATIONS**

Two acorns fall to the ground from an oak tree. One falls 45 feet, while the other falls 32 feet.

a. For each acorn, write an equation that represents the height h (in feet) as a function of the time t (in seconds).

b. Describe how the graphs of the two equations are related.

Answer:

Question 40.**THOUGHT PROVOKING**

One of two classic problems in calculus is to find the area under a curve. Approximate the area of the region bounded by the parabola and the x-axis. Show your work.

Answer:

One way to approximate the area is by redrawing the parabola into a graph with grids then count the number of squares. Note that the graph has an equation of y = -x² + 4 because of reflection in the x-axis and vertical translation.

In this case, each square is 1 square unit so the area is approximately 11 square units.

Question 41.**CRITICAL THINKING**

A cross section of the parabolic surface of the antenna shown can be modeled by y = 0.012x^{2}, where x and y are measured in feet. The antenna is moved up so that the outer edges of the dish are 25 feet above x-axis. Where is the vertex of the cross section located? Explain.

Answer:

**Maintaining Mathematical Proficiency**

**Evaluate the expression when a = 4 and b = −3.**

Question 42.

\(\frac{a}{4b}\)

Answer:

Given,

\(\frac{a}{4b}\)

a = 4 and b = -3

\(\frac{4}{4(-3)}\)

\(\frac{4}{-12}\) = \(\frac{1}{-3}\)

Question 43.

–\(\frac{b}{2a}\)

Answer:

Question 44.

\(\frac{a-b}{3 a+b}\)

Answer:

Given,

\(\frac{a-b}{3 a+b}\)

a = 4 and b = -3

\(\frac{4-(-3)}{3 4 – 3}\)

\(\frac{7}{3}\)

Question 45.

–\(\frac{b+2 a}{a b}\)

Answer:

### Lesson 8.3 Graphing f(x) = ax^{2} + bx + c

**Essential Question** How can you find the vertex of the graph of f(x) = ax^{2} + bx + c?

**EXPLORATION 1**

Comparing x-Intercepts with the Vertex

Work with a partner.

a. Sketch the graphs of y = 2x^{2} – 8x and y = 2x^{2} – 8x + 6.

b. What do you notice about the x-coordinate of the vertex of each graph?

c. Use the graph of y = 2x^{2} – 8x to find its x-intercepts. Verify your answer by solving 0 = 2x^{2} – 8x.

d. Compare the value of the x-coordinate of the vertex with the values of the x-intercepts.

**EXPLORATION 2**

Finding x-Intercepts

Work with a partner.

a. Solve 0 = ax^{2} + bx for x by factoring.

b. What are the x-intercepts of the graph of y = ax^{2} + bx?

c. Copy and complete the table to verify your answer.

**EXPLORATION 3**

Deductive Reasoning

Work with a partner. Complete the following logical argument.

**Communicate Your Answer**

Question 4.

How can you find the vertex of the graph of f(x) = ax^{2} + bx + c?

Answer:

Question 5.

Without graphing, find the vertex of the graph of f(x) = x^{2} – 4x + 3. Check your result by graphing.

Answer:

f(x) = x^{2} – 4x + 3

x² – 4x + 3 = 0

x² – 1x – 3x + 3 = 0

x(x – 1) – 3(x – 1) = 0

(x – 1)(x – 3) = 0

x = 1, 3

**Monitoring Progress**

**Find (a) the axis of symmetry and (b) the vertex of the graph of the function.**

Question 1.

f(x) = 3x^{2} – 2x

Answer:

Given,

f(x) = 3x^{2} – 2x

y = 3x^{2} – 2x

Differentiate

y’ = 6x – 2

6x – 2 = 0

6x = 2

x = 2/6

x = 1/3

y = 3(1/3)^{2} – 2(1/3)

y = 1/3 – 2/3

y = -1/3

Hence the vertex is (1/3, -1/3)

Question 2.

g(x) = x^{2} + 6x + 5

Answer:

g(x) = x^{2} + 6x + 5

y = x^{2} + 6x + 5

Differentiate

y’ = 2x + 6

2x + 6 = 0

2x = -6

x = -6/2

x = -3

y = x^{2} + 6x + 5

y = (-3)^{2} + 6(-3) + 5

y = 9 – 18 + 5

y = -4

Hence the vertex is (-3, -4)

Question 3.

h(x) = – \(\frac{1}{2}\)x^{2} + 7x – 4

Answer:

Given,

h(x) = – \(\frac{1}{2}\)x^{2} + 7x – 4

y = – \(\frac{1}{2}\)x^{2} + 7x – 4

Differentiate

y’ = -x + 7

0 = -x + 7

x = 7

y = – \(\frac{1}{2}\)x^{2} + 7x – 4

y = – \(\frac{1}{2}\)(7)^{2} + 7(2) – 4

y = –\(\frac{49}{2}\) + 14 – 4

y = -24.5 + 10

y = -14.5

Hence the vertex is (7, -14.5)

**Graph the function. Describe the domain and range.**

Question 4.

h(x) = 2x^{2} + 4x + 1

Answer:

x from -2.3 to 0.3

Question 5.

k(x) = x^{2} – 8x + 7

Answer:

Range: {y∈R:y≥-9}

(x from 0.4 to 7.6)

Question 6.

p(x) = -5x^{2} – 10x – 2

Answer:

Range:{y∈R:y≤3}

**Tell whether the function has a minimum value or a maximum value. Then find the value.**

Question 7.

g(x) = 8x^{2} – 8x + 6

Answer:

Graph the function above and graph y = x^{2}

g(x) = 8x^{2} – 8x + 6

x = -b/2a

x = 8/2(8) = 1/2 = 0.5

Substitute in 0.5 for x and you get 4 for y

your vertex is 0.5, 4

axis of symmetry equation is x – 0.5

Question 8.

h(x) = – \(\frac{1}{4}\)x^{2} + 3x + 1

Answer: maximum value; 10

Question 9.

The cables between the two towers of the Tacoma Narrows Bridge in Washington form a parabola that can be modeled by y = 0.00016x^{2} – 0.46x + 507, where x and y are measured in feet. What is the height of the cable above the water at its lowest point?

Answer:

Given,

The cables between the two towers of the Tacoma Narrows Bridge in Washington form a parabola that can be modeled by y = 0.00016x^{2} – 0.46x + 507, where x and y are measured in feet.

x = -b/2a

= -(-0.37)/2(0.000098) = 1888

Now substitute 1888 for x in the equation to find the y-coordinate of the vertex.

y = 0.00016(1888)^{2} – 0.46(1888) + 507 = 203

The cable is about 203 feet above the water at its lower point.

Question 10.

Which balloon is in the air longer? Explain your reasoning.

Answer:

Given one air balloon follows a parabolic path f(t) = -16t² + 80t + 5 and the graph of the second water balloon.

f(t) = -16t² + 80t + 5

125 > 105, hence the second balloon stays in the air for a longer time than the 1st balloon.

Question 11.

Which balloon reaches its maximum height faster? Explain your reasoning.

Answer:

### Graphing f(x) = ax^{2} + bx + c 8.3 Exercises

**Vocabulary and Core Concept Check**

Question 1.**VOCABULARY**

Explain how you can tell whether a quadratic function has a maximum value or a minimum value without graphing the function.

Answer:

The equation of a quadratic function is of the form ax2 + bx + c and if a < 0 then the function will have a maximum value of if a > 0 then the function will have a minimum value.

Question 2.**DIFFERENT WORDS, SAME QUESTION**

Consider the quadratic function f(x) = -2x^{2} + 8x + 24. Which is different? Find “both” answers.

Answer:

f(x) = -2x^{2} + 8x + 24

a = -2 and b = 8

The axis of symmetry is at x = -b/2a

x = -8/2(-2)

x = 2

The other equation asks for the y-value of the vertex. Since the vertex is located at the axis of symmetry, we can find the y-coordinate by substituting x = 2

y = -2(2)^{2} + 8(2) + 24

y = 32

**Monitoring Progress and Modeling with Mathematics**

**In Exercises 3–6, find the vertex, the axis of symmetry, and the y-intercept of the graph.**

Question 3.

Answer:

The vertex is (2, -1). The axis of symmetry is x = 2.

The y-intercept of the graph is 1.

Question 4.

Answer:

The vertex is (-3, 2). The axis of symmetry is x = -3.

The y-intercept of the graph is -1.

Question 5.

Answer:

The vertex is (-2, 0). The axis of symmetry is x = -2.

The y-intercept of the graph is -3.

Question 6.

Answer:

The vertex is (-1, 1). The axis of symmetry is x = -1.

The y-intercept of the graph is 5.

**In Exercises 7–12, find (a) the axis of symmetry and (b) the vertex of the graph of the function.**

Question 7.

f(x) = 2x^{2} – 4x

Answer:

Question 8.

y = 3x^{2} + 2x

Answer:

y = 3x^{2} + 2x

The axis of symmetry when a = 3 and b = 2

x = -b/2a

x = -1/3

The axis of symmetry is -1/3

Now use the function to find the y-coordinate of the vertex

y = 3x^{2} + 2x

y = 3(-1/3)^{2} + 2(-1/3)

y = -1/3

Thus the vertex is (-1/3, -1/3)

Question 9.

y = -9x^{2} – 18x – 1

Answer:

Question 10.

f(x) = -6x^{2} + 24x – 20

Answer:

Given,

f(x) = -6x^{2} + 24x – 20

a = -6 and b = 24

x = -b/2a

= -24/2(-6) = 2

The axis of symmetry is x = 2, so the x-coordinate of the vertex is 2.

Use the function to find the y-coordinate of the vertex.

f(x) = -6x^{2} + 24x – 20

= -6(2)^{2} + 24(2) – 20

= -24 + 48 – 20

= 4

Vertex is (2, 4)

Question 11.

f(x) = \(\frac{2}{5}\)x^{2} – 4x + 14

Answer:

Question 12.

y = – \(\frac{3}{4}\) x^{2} + 9x – 18

Answer:

y = – \(\frac{3}{4}\) x^{2} + 9x – 18

a = – \(\frac{3}{4}\) and b = 9

x = -b/2a

Substitute – \(\frac{3}{4}\) for a and 9 for b

x = – 9/2(-\(\frac{3}{4}\))

x = 6

Thus the axis of symmetry is x = 6

y = – \(\frac{3}{4}\) (6)^{2} + 9(6) – 18 = 9

The vertex is (6, 9)

**In Exercises 13–18, graph the function. Describe the domain and range.**

Question 13.

f(x) = 2x^{2} + 12x + 4

Answer:

Question 14.

y = 4x^{2} + 24x + 13

Answer:

a = 4 and b = 24

x = -b/2a

x = – 24/2(4) = -3

The axis of symmetry is x = -3.

The y-coordinate of vertex is y(-3)

y(-3) = 4(-3)^{2} + 24(-3) + 13

= 36 – 72 + 13

= -23

Vertex is (-3, -23)

Finding 2 more points on graph using y intercept

y intercept is c = 13.

Thus the parabola passes through (0, 13).

Using the fact that the axis of symmetry is x = -3, the parabola passes through (-6, 13)

Question 15.

y = -8x^{2} – 16x – 9

Answer:

Question 16.

f(x) = -5x^{2} + 20x – 7

Answer:

The domain contains all possible x-values that the function can take on and thus in this case the domain is all real values R.

The range contains all possible y values that the function can take on and we note in this case that the function only takes on values less or equal to 13.

Question 17.

y = \(\frac{2}{3}\)x^{2} – 6x + 5

Answer:

Question 18.

f(x) = – \(\frac{1}{2}\)x^{2} – 3x – 4

Answer:

The domain contains all possible x-values that the function can take on and thus in this case the domain is all real values R.

The range contains all possible y values that the function can take on and we note in this case that the function only takes on values less or equal to 0.5

.

Question 19.**ERROR ANALYSIS**

Describe and correct the error in finding the axis of symmetry of the graph of y = 3x^{2} – 12x + 11.

Answer:

Question 20.**ERROR ANALYSIS**

Describe and correct the error in graphing the function f(x) = -x^{2}+ 4x + 3.

Answer:

The formula for axis of symmetry is incorrect.

The formula should be -b/2a

As a result, the entire solution is incorrect.

The axis of symmetry is

x = -b/2a

= – 4/2(1) = -2

f(-2) = (-2)^{2}+ 4(-2) + 3 = -1

So, the vertex is (-2, -1)

The y-intercept is 3.

So, the points (0, 3) and (-4, 3) lie on the graph.

**In Exercises 21–26, tell whether the function has a minimum value or a maximum value. Then find the value.**

Question 21.

y = 3x^{2} – 18x + 15

Answer:

Question 22.

f(x) = -5x^{2} + 10x + 7

Answer:

Since the leading coefficient is negative, the function has a maximum value.

f(x) = -5x^{2} + 10x + 7

Axis of symmetry:

x = -10/-10 = 1

f(x) = -5(1)^{2} + 10(1) + 7

= 12

Maximum value is 12.

Question 23.

f(x) = -4x^{2} + 4x – 2

Answer:

Question 24.

y = 2x^{2} – 10x + 13

Answer:

Since the leading coefficient is positive, the function has a minimum value.

y = 2x^{2} – 10x + 13

Axis of symmetry is x = 5/2

y = 2(5/2)^{2} – 10(5/2) + 13 = 1/2

Minimum value: 1/2

Question 25.

y = – \(\frac{1}{2}\)x^{2} – 11x + 6

Answer:

Question 26.

f(x) = \(\frac{1}{5}\)x^{2} – 5x + 27

Answer:

f(x) = \(\frac{1}{5}\)x^{2} – 5x + 27

Since a>0, the minimum value for function exists.

The minimum value is attained at the vertex of parabola.

x = –\(\frac{b}{2a}\)

x = \(\frac{25}{2}\)

y0 = f(\(\frac{25}{2}\))

= \(\frac{1}{5}\)(\(\frac{25}{2}\))^{2} – 5(\(\frac{25}{2}\)) + 27

= -4.25

The minimum value is -4.25

Question 27.**MODELING WITH MATHEMATICS**

The function shown represents the height h (in feet) of a firework t seconds after it is launched. The firework explodes at its highest point.

a. When does the firework explode?

b. At what height does the firework explode?

Answer:

Question 28.**MODELING WITH MATHEMATICS**

The function h(t) = -16t^{2} + 16t represents the height (in feet) of a horse t seconds after it jumps during a steeplechase.

a. When does the horse reach its maximum height?

b. Can the horse clear a fence that is 3.5 feet tall? If so, by how much?

c. How long is the horse in the air?

Answer:

a. h(t) = -16t^{2} + 16t

t0 = –\(\frac{16}{2(-16)}\)

= 1/2

Horse reaches its maximum height at t = 0.5s

b.

y0 = h(1/2)

= -16(1/2)^{2} + 16(1/2)

= -4 + 8

= 4

Since, the maximum value is 4 feet, the fence would be cleared by 0.5 feet.

c.

Let t1 be the time at which horse hits the ground.

Then h(t1) = 0 and t1 > 0

h(t1) = 0

-16t1^{2} + 16t1 = 0

-16t(t1 – 1) = 0

t1 = 0 or 1

Since t1 > 0, we have

t1 = 1s

Question 29.**MODELING WITH MATHEMATICS**

The cable between two towers of a suspension bridge can be modeled by the function shown, where x and y are measured in feet. The cable is at road level midway between the towers.

a. How far from each tower shown is the lowest point of the cable?

b. How high is the road above the water?

c. Describe the domain and range of the function shown.

Answer:

Question 30.**REASONING**

Find the axis of symmetry of the graph of the equation y = ax^{2} + bx + c when b = 0. Can you find the axis of symmetry when a = 0? Explain.

Answer:

x = –\(\frac{b}{2a}\)

when b = 0

x = 0

when a = 0

x = undefined because of division by 0.

Infact, there is no axis of symmetry because the equation will become linear.

y = bx + c

Question 31.**ATTENDING TO PRECISION**

The vertex of a parabola is (3, -1). One point on the parabola is (6, 8). Find another point on the parabola. Justify your answer.

Answer:

Question 32.**MAKING AN ARGUMENT**

Your friend claims that it is possible to draw a parabola through any two points with different x-coordinates. Is your friend correct? Explain.

Answer:

f(x) = ax^{2} + bx + c

A parabola passing through 2 points give two linear equations with 3 variables that need to satisfied.

Let (x0, y0) and (x1, y1) be two points through which parabola passes.

ax0^{2} + bx0 + c = y0 – c

ax1^{2} + b1 + c = y1 – c

Thus there would be the infinite number of parabolas passing through the two points.

**USING TOOLS** In Exercises 33–36, use the minimum or maximum feature of a graphing calculator to approximate the vertex of the graph of the function.

Question 33.

y = 0.5x^{2} + \(\sqrt{2x}\) x – 3

Answer:

Question 34.

y = -6.2x^{2} + 4.8x – 1

Answer:

y = -6.2x^{2} + 4.8x – 1

Maximum feature gives vertex as (0.387, -0.071)

Question 35.

y = -πx^{2}+ 3x

Answer:

Question 36.

y = 0.25x^{2} – 5^{2/3}x + 2

Answer:

Question 37.**MODELING WITH MATHEMATICS**

The opening of one aircraft hangar is a parabolic arch that can be modeled by the equation y = -0.006x^{2}+ 1.5x, where x and y are measured in feet. The opening of a second aircraft hangar is shown in the graph.

a. Which aircraft hangar is taller?

b. Which aircraft hangar is wider?

Answer:

Question 38.**MODELING WITH MATHEMATICS**

An office supply store sells about 80 graphing calculators per month for $120 each. For each $6 decrease in price, the store expects to sell eight more calculators. The revenue from calculator sales is given by the function R(n) = (unit price)(units sold), or R(n) = (120 – 6n)(80 + 8n), where n is the number of $6 price decreases.

a. How much should the store charge to maximize monthly revenue?

Answer:

(120 – 6n) represents the unit price so we need to find the value of n where R is maximum which is located at the vertex.

R(n) = (120 – 6n)(80 + 8n)

R(n) = -48n² + 480n + 9600

x = -b/2a

a = -48 and b = 480

n = – 480/2(-48) = 5

R(n) = (120 – 6(5))(80 + 8(5)) = 10800

The vertex is at (5, 10800)

p = 120 – 6(5)

p = 120 – 30 = 90

b. Using a different revenue model, the store expects to sell five more calculators for each $4 decrease in price. Which revenue model results in a greater maximum monthly revenue? Explain.

Answer:

R(n) = (120 – 4n)(80 + 5n)

R(n) = -20n² + 280n + 9600

n = -b/2a

a = -20 and b = 280

n = – 280/2(-20) = 7

R(n) = (120 – 4(7))(80 + 5(7)) = 10580

The vertex is at (7, 10580)

This means that the maximum revenue is $10580 which is less than the original revenue at $10800.

Therefore, the original revenue results in a greater maximum monthly revenue.

**MATHEMATICAL CONNECTIONS** In Exercises 39 and 40, (a) find the value of x that maximizes the area of the figure and (b) find the maximum area.

Question 39.

Answer:

Question 40

Answer:

Area of trapezoid = \(\frac{1}{2}\)(b1 + b2)h

A = \(\frac{1}{2}\)(12 – 4x + 12)(x + 2)

A = \(\frac{1}{2}\)(24 – 4x)(x + 2)

A = \(\frac{1}{2}\)(-4x² + 16x + 48)

A = -2x² + 8x + 24

The maximum area occurs at the vertex and the x-coordinate that gives this is x = –\(\frac{b}{2a}\)

a = -2, b = 8

x = –\(\frac{8}{2(-2)}\)

x = 2

A = -2(2)² + 8(2) + 24

A = 32 sq. ft

Question 41.**WRITING**

Compare the graph of g(x) = x^{2} + 4x + 1 with the graph of h(x) = x^{2} – 4x + 1.

Answer:

Question 42.**HOW DO YOU SEE IT?**

During an archery competition, an archer shoots an arrow. The arrow follows the parabolic path shown, where x and y are measured in meters.

a. What is the initial height of the arrow?

Answer:

The initial height is the y-intercept which is y = 1.5

b. Estimate the maximum height of the arrow.

Answer: Approximately 1.6 meters

c. How far does the arrow travel?

Answer:

Find the x-coordinate where y = 0 which is at x = 90

x = 90 meters

Question 43.**USING TOOLS**

The graph of a quadratic function passes through (3, 2), (4, 7), and (9, 2). Does the graph open up or down? Explain your reasoning.

Answer:

Question 44.**REASONING**

For a quadratic function f, what does f(-\(\frac{b}{2a}\)) represent? Explain your reasoning.

Answer:

Since x = –\(\frac{b}{2a}\) represents the x-coordinate of the vertex, the f(-\(\frac{b}{2a}\)) is the y-coodinate of the vertex.

Question 45.**PROBLEM SOLVING**

Write a function of the form y = ax^{2} + bx whose graph contains the points (1, 6) and (3, 6).

Answer:

Question 46.**CRITICAL THINKING**

Parabolas A and B contain the points shown. Identify characteristics of each parabola, if possible. Explain your reasoning.

Answer:

Parabola A may be opening upward or downward as shown because there are only 2 points.

Parabola B can only be opening up and the vertex is at (3, -4) due to symmetry:

Question 47.**MODELING WITH MATHEMATICS**

At a basketball game, an air cannon launches T-shirts into the crowd. The function y = – \(\frac{1}{8}\)x^{2} + 4x represents the path of a T-shirt. The function 3y = 2x – 14 represents the height of the bleachers. In both functions, y represents vertical height (in feet) and x represents horizontal distance (in feet). At what height does the T-shirt land in the bleachers?

Answer:

Question 48.**THOUGHT PROVOKING**

One of two classic problems in calculus is finding the slope of a tangent line to a curve. An example of a tangent line, which just touches the parabola at one point, is shown.

Approximate the slope of the tangent line to the graph of y = x^{2} at the point (1, 1). Explain your reasoning.

Answer:

By using the grid lines as guide, we can find an exact point by going 1 unit up and 1 unit to the right which means the slope is m = rise/run = 1

m = 1

Question 49.**PROBLEM SOLVING**

The owners of a dog shelter want to enclose a rectangular play area on the side of their building. They have k feet of fencing. What is the maximum area of the outside enclosure in terms of k? (Hint: Find the y-coordinate of the vertex of the graph of the area function.)

Answer:

**Maintaining Mathematical Proficiency**

**Describe the transformation(s) from the graph of f(x) = |x| to the graph of the given function.**

Question 50.

q(x) = |x + 6|

Answer:

The graph is shifted to the left by 6.

Question 51.

h(x) = -0.5|x|

Answer:

Question 52.

g(x) = |x – 2| + 5

Answer:

The graph is shifted to the right by 2 and up by 5.

Question 53.

p(x) = 3|x + 1|

Answer:

### Graphing Quadratic Functions Study Skills: Learning Visually

**8.1– 8.3 What Did You Learn?**

**Core Vocabulary**

**Core Concepts**

Section 8.1

Characteristics of Quadratic Functions, p. 420

Graphing f(x) = ax^{2} When a > 0, p. 421

Graphing f (x) = ax^{2}When a < 0, p. 421

Section 8.2

Graphing f(x) = ax^{2} + c, p. 426

Section 8.3

Graphing f(x) = ax^{2} + bx + c, p. 432

Maximum and Minimum Values, p. 433

**Mathematical Practices**

Question 1.

Explain your plan for solving Exercise 18 on page 423.

Answer:

Question 2.

How does graphing a function in Exercise 27 on page 429 help you answer the questions?

Answer:

Question 3.

What definition and characteristics of the graph of a quadratic function did you use to answer Exercise 44 on page 438?

Answer:

**Study Skills: Learning Visually**

- Draw a picture of a word problem before writing a verbal model. You do not have to be an artist.
- When making a review card for a word problem, include a picture. This will help you recall the information while taking a test.
- Make sure your notes are visually neat for easy recall

### Graphing Quadratic Functions 8.1 – 8.3 Quiz

**Identify characteristics of the quadratic function and its graph.**

Question 1.

Answer:

The parabola opens downward with the following characteristics

Vertex is (1, 4)

Axis of symmetry is x = 1

Domain is (-∞, +∞)

Range: (-∞, 4]

Question 2.

Answer:

The parabola opens downward with the following characteristics

Vertex is (-2, 5)

Axis of symmetry is x = -2

Domain is (-∞, +∞)

Range: [5, -∞)

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 3.

h(x) = -x

^{2}

Answer:

The given functions’s graph involves a reflection about the x-axis of the parent function f(x) = x

^{2}

Question 4.

p(x) = 2x^{2} + 2

Answer:

Question 5.

r(x) = 4x^{2} – 16

Answer:

Question 6.

b(x) = 8x^{2}

Answer:

Question 7.

g(x) = \(\frac{2}{5}\)x^{2}

Answer:

Question 8.

m(x) = – \(\frac{1}{2}\)x^{2} – 4

Answer:

The vertical compression by a factor 1/2

a reflection about the x-axis

a 4 unit shift downwards

Describe the transformation from the graph of f to the graph of g. Then graph f and g in the same coordinate plane. Write an equation that represents g in terms of x.

Question 9.

f(x) = 2x^{2} + 1; g(x) = f(x) + 2

Answer:

The function g(x) = f(x) + 2 involes additiion of 2 to f(x) therefore g(x) involes a 2 unit shift upward of the graph of f(x).

g(x) = f(x) + 2

g(x) = 2x^{2} + 1 + 2

g(x) = 2x^{2} + 3

Question 10.

f(x) = -3x^{2} + 12; g(x) = f(x) – 9

Answer:

The function g(x) = f(x) – 9 involves subtraction of 9 to f(x) therefore g(x) involves a 9 unit shift downward of the graph of f(x)

Question 11.

f(x) = \(\frac{1}{2}\)x^{2} – 2; g(x) = f(x) – 6

Answer:

The function g(x) = f(x) – 6 involves subtraction of 6 to f(x) therefore g(x) involves a 6 unit shift downward of the graph of f(x)

g(x) = f(x) – 6

g(x) = \(\frac{1}{2}\)x^{2} – 2- 6 = \(\frac{1}{2}\)x^{2} – 8

Question 12.

f(x) = 5x^{2} – 3; g(x) = f(x) + 1

Answer:

The function g(x) = f(x) + 1 involves addition of 1 to f(x) therefore g(x) involves a 1 unit shift upward of the graph of f(x)

g(x) = f(x) + 1

g(x) = 5x^{2} – 3 + 1

g(x) = 5x^{2} – 2

**Graph the function. Describe the domain and range.**

Question 13.

f(x) = -4x^{2} – 4x + 7

Answer:

x = -b/2a

= -(-4)/2(-4) = -1/2

f(-1/2) = -4(-1/2)^{2} – 4(-1/2) + 7 = 8

Therefore the vertex of the parabola is at (-1/2, 8)

The domain is the set of real numbers.

The range is (-∞, 8]

Question 14.

f(x) = 2x^{2} + 12x + 5

Answer:

x = -b/2a

= -12/2(5) = -3

f(-3) = 2. (-3)^{2} + 12(-3) + 5 = -13

Therefore the vertex of the parabola is at (-3, -13)

The domain is the set of real numbers.

The range is [-13, +∞)

Question 15.

y = x^{2} + 4x – 5

Answer:

x = -b/2a

= -4/2(1) = -2

f(-2) = (-2)^{2} + 4(-2) – 5 = -9

Therefore the vertex of the parabola is at (-2, -9)

The domain is the set of real numbers.

The range is [-9, +∞)

Question 16.

y = -3x^{2} + 6x + 9

Answer:

y = -3x^{2} + 6x + 9

x = -b/2a

= -6/2(-3) = 1

f(1) = -3(1)^{2} + 6(1) + 9 = 12

Therefore the vertex of the parabola is at (1, 12)

The domain is the set of real numbers.

The range is (-∞, 12]

**Tell whether the function has a minimum value or a maximum value. Then find the value.**

Question 17.

f(x) = 5x^{2} + 10x – 3

Answer:

f(x) = 5x^{2} + 10x – 3

x = -10/10 = -1

f(x) = 5(-1)^{2} + 10(-1) – 3

= -8

Minimum value: -8

Question 18.

f(x) = – \(\frac{1}{2}\)x^{2} + 2x + 16

Answer:

Since the leading coefficient is negative, the parabola contains a maximum value

f(x) = – \(\frac{1}{2}\)x^{2} + 2x + 16

x = -2/-1 = 2

f(x) = – \(\frac{1}{2}\)(2)^{2} + 2(2) + 16 = 18

Maximum value: 18

Question 19.

y = -x^{2} + 4x + 12

Answer:

y = -x^{2} + 4x + 12

Since a < 0, maximum value exists.

The maximum value is attained at vertex.

x0 = -b/2a

= – 4/2(-1)

= 2

y0 = y(2)

= -(2)^{2} + 4(2) + 12

= -4 + 8 + 12

= 16

Question 20.

y = 2x^{2} + 8x + 3

Answer:

y = 2x^{2} + 8x + 3

x = -8/4

x = 2

y = 2(-2)^{2} + 8(-2) + 3

y = -5

Minimum value is -5

Question 21.

The distance y (in feet) that a coconut falls after t seconds is given by the function y = 16t^{2}. Use a graph to determine how many seconds it takes for the coconut to fall 64 feet.

Answer:

y = 16t^{2}

16t^{2 }= 64

t^{2 }= 4

t = 2

Question 22.

The function y = -16t^{2} + 25 represents the height y (in feet) of a pinecone t seconds after falling from a tree.

a. After how many seconds does the pinecone hit the ground?

Answer:

The pinecone hits the ground at y = 0

0 = -16t^{2} + 25

16t^{2} = 25

t^{2} = 25/16

t = 5/4 = 1.25 seconds

b. A second pinecone falls from a height of 36 feet. Which pinecone hits the ground in the least amount of time? Explain.

Answer:

The constant 25 represents the initial height of 25 feet. Therefore, the first pinecone will hit the ground in the least amount of time.

Question 23.

The function shown models the height (in feet) of a softball t seconds after it is pitched in an underhand motion. Describe the domain and range. Find the maximum height of the softball.

Answer:

h(t) = = -16t^{2} + 32t + 2

x0 = -b/2a

= – 32/2(-16)

= 1

y0 = h(x0)

= -16(1)^{2} + 32(1) + 2

= -16+ 32 + 2

= 18

For the given problem, the domain is from the point softball is thrown, till it hits the ground.

h(t) = 0

-16t^{2} + 32t + 2 = 0

8t^{2} – 16t – 1 = 0

t = (4 + 3√2)/4

Thus the domain is [0, (4 + 3√2)/4]

The softball attains the maximum height at the vertex. Thus maximum height is 18 feet and after achieving it the softball hits the ground.

Thus the range is [0, 18]

The domain is set of all real numbers R.

The maximum height is 18 feet.

Thus the range is (-∞, 18]

### Lesson 8.4 Graphing f(x) = a(x – h)^{2} + k

**Essential Question** How can you describe the graph of f(x) = a(x – h)^{2}?

**EXPLORATION 1**

Graphing y = a(x − h)^{2} When h > 0

Work with a partner. Sketch the graphs of the functions in the same coordinate plane. How does the value of h affect the graph of y = a(x – h)^{2}?

Answer:

a.

b.

**EXPLORATION 2**

Graphing y = a(x − h)^{2} When h < 0

Work with a partner. Sketch the graphs of the functions in the same coordinate plane. How does the value of h affect the graph of y = a(x – h)^{2}?

Answer:

a.

b.

**Communicate Your Answer**

Question 3.

How can you describe the graph of f(x) = a(x – h)^{2}?

Answer:

Question 4.

Without graphing, describe the graph of each function. Use a graphing calculator to check your answer.

a. y = (x – 3)^{2}

b. y = (x + 3)^{2}

c. y = -(x – 3)^{2}

Answer:

a. y = (x – 3)^{2}

b.

c.

**Monitoring Progress**

**Determine whether the function is even, odd, or neither.**

Question 1.

f(x) = 5x

Answer:

Given function

f(x) = 5x

f(-x) = 5(-x)

f(-x) = -5x

Thus the function is odd

Question 2.

g(x) = 2x

Answer:

Given function

g(x) = 2x

g(-x) = 2(-x) = -2x

Thus the function is odd

Question 3.

h(x) = 2x^{2} + 3

Answer:

Given function

h(x) = 2x^{2} + 3

h(-x) = 2(-x)^{2} + 3

= 2x^{2} + 3

Thus the function is even.

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 4.

g(x) = 2(x + 5)

^{2}

Answer:

Question 5.

h(x) = -(x – 2)^{2}

Answer:

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 6.

g(x) = 3(x – 1)

^{2}+ 6

Answer:

Question 7.

h(x) = \(\frac{1}{2}\)(x + 4)^{2} – 2

Answer:

Question 8.

Consider function g in Example 3. Graph f(x) = g(x) – 3

Answer:

Question 9.**WHAT IF?**

The vertex is (3, 6). Write and graph a quadratic function that models the path.

Answer:

### Graphing f(x) = a(x – h)^{2} + k 8.4 Exercises

**Vocabulary and Core Concept Check**

Question 1.**VOCABULARY**

Compare the graph of an even function with the graph of an odd function.

Answer:

The graph of an even function is symmetric about the y-axis. The graph of an odd function is symmetric about the origin.

Question 2.**OPEN-ENDED**

Write a quadratic function whose graph has a vertex of (1, 2).

Answer:

f(x) = a(x – 1)^{2} + 2

Equation must have a vertex at (1, 2)

Question 3.**WRITING**

Describe the transformation from the graph of f(x) = ax^{2} to the graph of g(x) = a(x – h)^{2} + k.

Answer:

The graph of g is a horizontal translation h units right if h is positive or |h| units left if h is negative and a vertical translation k units up if k is positive or |k| units down if k is negative of the graph of f.

Question 4.**WHICH ONE DOESN’T BELONG?**

Which function does not belong with the other three? Explain your reasoning.

Answer:

f(x) = a(x – 2)^{2} + 4 does not belong because it is the only graph that will not have a vertical stretch or shrink.

**Monitoring Progress and Modeling with Mathematics**

**In Exercises 5–12, determine whether the function is even, odd, or neither.**

Question 5.

f(x) = 4x + 3

Answer:

Question 6.

g(x) = 3x^{2}

Answer:

Given function

g(x) = 3x^{2}

g(-x) = 3(-x)^{2}

g(-x) = 3x^{2}

Even

Question 7.

h(x) = 5^{x} + 2

Answer:

Question 8.

m(x) = 2x^{2} – 7x

Answer:

m(x) = 2x^{2} – 7x

m(-x) = 2(-x)^{2} – 7(-x)

m(-x) = 2x^{2} + 7x

Neither

Question 9.

p(x) = -x^{2} + 8

Answer:

Question 10.

f(x) = – \(\frac{1}{2}\)x

Answer:

f(x) = – \(\frac{1}{2}\)x

f(-x) = – \(\frac{1}{2}\)(-x)

f(-x) = \(\frac{1}{2}\)x

Odd

Question 11.

n(x) = 2x^{2} – 7x + 3

Answer:

Question 12.

r(x) = -6x^{2} + 5

Answer:

To determine is function odd, even, neither, you should replace x with -x.

1. If r(-x) = -r(x) function is odd

2. If r(-x) = r(x) function is even

r(x) = -6x^{2} + 5

r(-x) = -6(-x)^{2} + 5

r(-x) = -6x^{2} + 5

Thus r(-x) = r(x)

Even

**In Exercises 13–18, determine whether the function represented by the graph is even, odd, or neither.**

Question 13.

Answer:

The graph is symmetric about the y-axis. So, the function is even.

Question 14.

Answer: The graph is neither even nor odd because it is not symmetric about the y-axis or the origin.

Question 15.

Answer:

The graph is neither even nor odd because it is not symmetric about the y-axis or the origin.

Question 16.

Answer: The graph is even because it is symmetric about the y-axis.

Question 17.

Answer:

The graph is symmetric about the origin. So, the function is odd.

Question 18.

Answer:

The graph is neither even nor odd because it is not symmetric about the y-axis or the origin.

**In Exercises 19–22, find the vertex and the axis of symmetry of the graph of the function.**

Question 19.

f(x) = 3(x + 1)^{2}

Answer:

Question 20.

f(x) = \(\frac{1}{4}\)(x – 6)^{2}

Answer:

f(x) = \(\frac{1}{4}\)(x – 6)^{2}

Find the axis of symmetry and vertex:

Since h = 6, the axis of symmetry is x = 6 and the vertex is (6, 0)

Question 21.

y = – \(\frac{1}{8}\)(x – 4)^{2}

Answer:

Question 22.

y = -5(x + 9)^{2}

Answer:

y = -5(x + 9)^{2}

Find the axis of symmetry and vertex:

Since h = -9, the axis of symmetry is x = -9 and the vertex is (-9, 0)

**In Exercises 23–28, graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 23.

g(x) = 2(x + 3)

^{2}

Answer:

Question 24.

p(x) = 3(x – 1)^{2}

Answer:

Given function,

p(x) = 3(x – 1)^{2}

Axis of symmetry: x = 1

Vertex: (1, 0)

Point 1: (0, 3)

point 2: (2, 3)

The graph is a translation right 1 unit and a vertical stretch by a factor of 3 of the parent function.

Question 25.

r(x) = \(\frac{1}{4}\)(x + 10)^{2}

Answer:

Question 26.

n(x) = \(\frac{1}{4}\)(x – 6)^{2}

Answer:

Given function,

n(x) = \(\frac{1}{4}\)(x – 6)^{2}

Axis of symmetry: x = 6

Vertex: (6, 0)

Point 1: (3, 3)

point 2: (9, 3)

The graph is a translation right 6 units and a vertical shrink by a factor of 1/3 of the parent function.

Question 27.

d(x) = \(\frac{1}{5}\)(x – 5)^{2}

Answer:

Question 28.

q(x) = 6(x + 2)^{2}

Answer:

Given function,

q(x) = 6(x + 2)^{2}

Axis of symmetry: x = -2

Vertex: (-2, 0)

Point 1: (-3, 6)

point 2: (-1, 6)

The graph is a translation left 2 units and a vertical stretch by a factor of 6 of the parent function.

Question 29.**ERROR ANALYSIS**

Describe and correct the error in determining whether the function f(x) = x^{2} + 3 is even, odd, or neither.

Answer:

Question 30.**ERROR ANALYSIS**

Describe and correct the error in finding the vertex of the graph of the function.

Answer:

Since h = -8, the parent function will be shifted to the left 8 units, so the vertex is (-8, 0) not (0, -8)

**In Exercises 31–34, find the vertex and the axis of symmetry of the graph of the function.**

Question 31.

y = -6(x + 4)^{2} – 3

Answer:

Question 32.

f(x) = 3(x – 3)^{2} + 6

Answer:

Given,

f(x) = 3(x – 3)^{2} + 6

Vertex: (3, 6)

Axis of symmetry: x = 3

Question 33.

f(x) = -4(x + 3)^{2} + 1

Answer:

Question 34.

y = -(x – 6)^{2} – 5

Answer:

y = -(x – 6)^{2} – 5

Vertex: (6, -5)

Axis of symmetry: x = 6

**In Exercises 35–38, match the function with its graph.**

Question 35.

y = -(x + 1)^{2} – 3

Answer:

Question 36.

y = – \(\frac{1}{2}\)(x – 1)^{2} + 3

Answer:

y = – \(\frac{1}{2}\)(x – 1)^{2} + 3

This equation belongs to Graph A because it is the only graph with a vertex of (1, 3) and a reflection over the x-axis.

Question 37.

y = \(\frac{1}{3}\)(x – 1)^{2} + 3

Answer:

Question 38.

y = 2(x + 1)^{2} – 3

Answer:

Given,

y = 2(x + 1)^{2} – 3

This equation belongs to Graph B because it is the only graph with a vertex of (-1, -3) and a vertical stretch by a factor of 2.

**In Exercises 39–44, graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 39.

h(x) = (x – 2)

^{2}+ 4

Answer:

Question 40.

g(x) = (x + 1)^{2} – 7

Answer:

g(x) = (x + 1)^{2} – 7

Axis of symmetry: x = -1

Vertex: (-1, -7)

Point 1: (0, -6)

point 2: (-2, -6)

Question 41.

r(x) = 4(x – 1)^{2} – 5

Answer:

Question 42.

n(x) = -(x + 4)^{2} + 2

Answer:

Given,

n(x) = -(x + 4)^{2} + 2

Axis of symmetry: x = -4

Vertex: (-4, 2)

Point 1: (-5, 1)

point 2: (-3, 1)

Question 43.

g(x) = – \(\frac{1}{3}\)(x + 3)^{2} – 2

Answer:

Question 44.

r(x) = \(\frac{1}{2}\)(x – 2)^{2} – 4

Answer:

r(x) = \(\frac{1}{2}\)(x – 2)^{2} – 4

Axis of symmetry: x = 2

Vertex: (2, -4)

Point 1: (0, -2)

point 2: (4, -2)

**In Exercises 45–48, let f(x) = (x − 2) ^{2} + 1. Match the function with its graph.**

Question 45.

g(x) = f(x – 1)

Answer:

Question 46.

r(x) = f(x + 2)

Answer:

Given,

f(x) = (x – 2)² + 1

r(x) = f(x + 2)

Rewrite r(x):

r(x) = x² + 1

Match with a graph:

The function r(x) belongs to Graph C because it is the only graph with a vertex at (0, 1)

Question 47.

h(x) = f(x) + 2

Answer:

Question 48.

p(x) = f(x) – 3

Answer:

f(x) = (x – 2)² + 1

p(x) = f(x) – 3

Rewrite p(x):

p(x) = (x – 2)² – 2

Match with a graph:

The function p(x) belongs to Graph D because it is the only graph with a vertex at (2, -2)

**In Exercises 49–54, graph g.**

Question 49.

f(x) = 2(x – 1)^{2} + 1; g(x) = f(x + 3)

Answer:

Question 50.

f(x) = -(x + 1)^{2} + 2; g(x) = \(\frac{1}{2}\)f(x)

Answer:

g(x) = \(\frac{1}{2}\)f(x)

\(\frac{1}{2}\)(-(x + 1)^{2} + 2) = –\(\frac{1}{2}\)(x + 1)^{2} + 1.

In given function g(x) = –\(\frac{1}{2}\)(x + 1)^{2} + 1

a. a vertical contraction by a factor of \(\frac{1}{2}\)

b. a vertical shift of 1 unit downward

c. reflection about the x-axis

d. horizontal shift of 1 unit to the right of the parent function y = ax^{2}

Question 51.

f(x) = -3(x + 5)^{2} – 6; g(x) = 2f(x)

Answer:

Question 52.

f(x) = 5(x – 3)^{2} – 1; g(x) = (x) – 6

Answer:

Given,

f(x) = 5(x – 3)^{2} – 1

g(x) = f(x) – 6

The graph of g is vertical translation 6 units down of graph of f.

Question 53.

f(x) = (x + 3)^{2} + 5; g(x) = f(x – 4)

Answer:

Question 54.

f(x) = -2(x – 4)^{2} – 8; g(x) = -f(x)

Answer:

f(x) = -2(x – 4)^{2} – 8

g(x) = -f(x)

Question 55.**MODELING WITH MATHEMATICS**

The height (in meters) of a bird diving to catch a fish is represented by h(t) = 5(t – 2.5)^{2}, where t is the number of seconds after beginning the dive.

a. Graph h.

b. Another bird’s dive is represented by r(t) = 2h(t). Graph r.

c. Compare the graphs. Which bird starts its dive from a greater height? Explain.

Answer:

Question 56.**MODELING WITH MATHEMATICS**

A kicker punts a football. The height (in yards) of the football is represented by f(x) = – \(\frac{1}{9}\)(x – 30)^{2} + 25, where x is the horizontal distance (in yards) from the kicker’s goal line.

a. Graph f. Describe the domain and range.

Answer:

f(x) = – \(\frac{1}{9}\)(x – 30)^{2} + 25

Domain: 15 ≤ x ≤ 45

Range: 0 ≤ f(x) ≤ 25

b. On the next possession, the kicker punts the football. The height of the football is represented by g(x) = f (x + 5). Graph g. Describe the domain and range.

Answer:

g(x) = f (x + 5)

Domain: 10 ≤ x ≤ 40

Range: 0 ≤ f(x) ≤ 25

c. Compare the graphs. On which possession does the kicker punt closer to his goal line? Explain.

Answer:

The graph of g(x) is a translation left 5 units of f(x), The kicker kicks closer to his goal line the second time.

**In Exercises 57–62, write a quadratic function in vertex form whose graph has the given vertex and passes through the given point.**

Question 57.

vertex: (1, 2); passes through (3, 10)

Answer:

Question 58.

vertex: (-3, 5); passes through (0, -14)

Answer:

Vertex form is

f(x) = a(x – h)^{2} + k

f(x) = a(x + 3)^{2} + 5

-14 = a(0 + 3)^{2} + 5

-14 = a(3)^{2} + 5

9a + 5 = -14

9a = -14 – 5

9a = -19

a = -19/9

f(x) = -19/9(x + 3)^{2} + 5

Question 59.

vertex: (-2, -4); passes through (-1, -6)

Answer:

Question 60.

vertex: (1, 8); passes through (3, 12)

Answer:

Vertex form is

f(x) = a(x – h)^{2} + k

f(x) = a(x – 1)^{2} + 8

12 = a(3 – 1)^{2} + 8

12 = a(2)^{2} + 8

4a + 8 = 12

4a = 12 – 8

4a = 4

a = 1

f(x) = 1(x – 1)^{2} + 8

Question 61.

vertex: (5, -2); passes through (7, 0)

Answer:

Question 62.

vertex: (-5, -1); passes through (-2, 2)

Answer:

Vertex form is

f(x) = a(x – h)^{2} + k

f(x) = a(x + 5)^{2} – 1

2 = a(-2 + 5)^{2} – 1

2 = a(3)^{2} – 1

9a – 1 = 2

9a = 3

a = 1/3

f(x) = 1/3(x + 5)^{2} – 1

Question 63.**MODELING WITH MATHEMATICS**

A portion of a roller coaster track is in the shape of a parabola. Write and graph a quadratic function that models this portion of the roller coaster with a maximum height of 90 feet, represented by a vertex of (25, 90), passing through the point (50, 0).

Answer:

Question 64.**MODELING WITH MATHEMATICS**

A flare is launched from a boat and travels in a parabolic path until reaching the water. Write and graph a quadratic function that models the path of the are with a maximum height of 300 meters, represented by a vertex of (59, 300), landing in the water at the point (119, 0).

Answer:

f(x) = a(x – h)^{2} + k

h = 59 and k = 300

f(x) = a(x – 59)^{2} + 300

As (119, 0) lies on the function, putting the values in function we get

0 = a(119 – 59)^{2} + 300

0 = 3600a + 300

a = -300/3600 = -1/12

f(x) = -1/12(x – 59)^{2} + 300

Step 1: Graph the axis of symmetry: h = 59, axis of symmetry is x = 59.

Step 2: Plot the vertex: Vertex is (h, k) = (59, 300)

Step 3: Plot 2 more coordinates. We are given that (119, 0) lies on the graph.

Finding y-coordinate for x = 29, we get that (29, 225) lies on the graph.

Step 4: Draw smooth curve through the points.

**In Exercises 65–68, rewrite the quadratic function in vertex form.**

Question 65.

y = 2x)^{2} – 8x + 4

Answer:

Question 66.

y = 3x^{2} + 6x – 1

Answer:

Given,

y = 3x^{2} + 6x – 1

x = -b/2a

x = -6/6

x = -1

y = 3(-1)^{2} + 6(-1) – 1

y = 3 – 6 – 1

y = -4

Vertex: (-1, -4)

f(x) = a(x – h)^{2} + k

y = a(x + 1)^{2} – 4

-1 = a(0 + 1)^{2} – 4

-1 = a – 4

a = -1 + 4

a = 3

f(x) = 3(x + 1)^{2} – 4

Question 67.

f(x) = -5x)^{2} + 10x + 36

Answer:

Question 68.

f(x) = -x^{2} + 4x + 2

Answer:

Given,

f(x) = -x^{2} + 4x + 2

x = -b/2a

x = -4/2(-1)

x = -2

y = -(-2)^{2} – 4(-2) + 2

y = -4+8+2

y = 6

Vertex: (-2, 6)

f(x) = a(x – h)^{2} + k

y = a(x + 2)^{2} + k

2 = a(0 + 2)^{2} + 6

2 = 4a + 6

4a + 6 = 2

a = -1

f(x) = -(x + 2)^{2} + 6

Question 69.**REASONING**

Can a function be symmetric about the x-axis? Explain.

Answer:

A function cannot be symmetric about the x-axis because it would not pass the vertical line test.

Question 70.**HOW DO YOU SEE IT?**

The graph of a quadratic function is shown. Determine which symbols to use to complete the vertex form of the quadratic function. Explain your reasoning.

Answer:

Since the graph has a vertex of (-2, -3), the equation must be y = a(x + 2)^{2} – 3

since h = -2 and k = -3

In Exercises 71–74, describe the transformation from the graph of f to the graph of h. Write an equation that represents h in terms of x.

Question 71.

f(x) = -(x + 1))^{2} – 2

h(x) = f(x) + 4

Answer:

Question 72.

f(x) = 2(x – 1))^{2} + 1

h(x) = f(x – 5)

Answer:

given,

f(x) = 2(x – 1))^{2} + 1

h(x) = f(x – 5)

The graph of h(x) is a horizontal translation right 5 units of the graph of f(x)

y = 2(x – 6)^{2} + 1

Question 73.

f(x) = 4(x – 2))^{2} + 3

h(x) = 2f(x)

Answer:

Question 74.

f(x) = -(x + 5))^{2} – 6

h(x) = \(\frac{1}{3}\)f(x)

Answer:

Given,

f(x) = -(x + 5))^{2} – 6

h(x) = \(\frac{1}{3}\)f(x)

The graph of h(x) is a vertical shrink by a factor of 1/3 of the graph of f(x)

y = –\(\frac{1}{3}\)(x + 5)^{2} – 2

Question 75.**REASONING**

The graph of y = x^{2} is translated 2 units right and 5 units down. Write an equation for the function in vertex form and in standard form. Describe advantages of writing the function in each form.

Answer:

Question 76.**THOUGHT PROVOKING**

Which of the following are true? Justify your answers.

a. Any constant multiple of an even function is even.

Answer:

Let f(x) be an even function.

Let g(x) = af(x)

g(-x) = af(-x)

Since f is even function, we get

g(-x) = af(x)

= g(x)

Thus, g is an even function.

b. Any constant multiple of an odd function is odd.

Answer:

Let f(x) be an odd function.

Let g(x) = af(x)

g(-x) = af(-x)

Since f is even function, we get

g(-x) = a(-f(x))

= -g(x)

Thus, g is an odd function.

c. The sum or difference of two even functions is even.

Answer:

Let f, h be even function.

Let g(x) = f(x) + ah(x)

g(-x) = f(-x) + ah(-x)

Since f, h are even functions.

Let g(x) = f(x) + ah(x)

g(-x) = f(-x) + ah(-x)

g(x) = f(x) + ah(x) = g(x)

Thus g is an even function

d. The sum or difference of two odd functions is odd.

Answer:

Let f, h be odd function.

Let g(x) = f(x) + ah(x)

g(-x) = f(-x) + ah(-x)

Since f, h are even functions.

Let g(x) = f(x) + ah(x)

g(-x) = -f(x) + a(-h(x)) = -g(x)

Thus g is an odd function

e. The sum or difference of an even function and an odd function is odd.

Answer:

f(x) = x²

h(x) = x

g(x) = f(x) + h(x)

f(1) = h(1) = 1,

g(1) = 2 ≠ -g(-1) = 0

Thus, g is not odd function

Question 77.**COMPARING FUNCTIONS**

A cross section of a birdbath can be modeled by y = \(\frac{1}{81}\)(x – 18)^{2} – 4, where x and y are measured in inches. The graph shows the cross section of another birdbath.

a. Which birdbath is deeper? Explain.

b. Which birdbath is wider? Explain.

Answer:

Question 78.**REASONING**

Compare the graphs of y = 2x^{2} + 8x +8 and y = x^{2} without graphing the functions. How can factoring help you compare the parabolas? Explain.

Answer:

y = 2x^{2} + 8x +8

y = 2(x^{2} + 4x +4)

y = 2(x+2)(x + 2)

y = 2(x+2)²

The graph is a translation left 2 units and a vertical stretch by a factor of 2 of the graph y = x^{2}

Question 79.**MAKING AN ARGUMENT**

Your friend says all absolute value functions are even because of their symmetry. Is your friend correct? Explain.

Answer:

**Maintaining Mathematical Proficiency**

**Solve the equation.**

Question 80.

x(x – 1) = 0

Answer:

Given,

x(x – 1) = 0

x = 0 or x – 1 = 0

x = 0 or x = 1

Question 81.

(x + 3)(x – 8) = 0

Answer:

Question 82.

(3x – 9)(4x + 12) = 0

Answer:

Given,

(3x – 9)(4x + 12) = 0

3x – 9 = 0 or 4x + 12 = 0

3x = 9 or 4x = -12

x = 3 or x = -3

### Lesson 8.5 Using Intercept Form

**Essential Question** What are some of the characteristics of the graph of f(x) = a(x – p)(x – q)?

**EXPLORATION 1**

Using Zeros to Write Functions

Work with a partner. Each graph represents a function of the form f(x) = (x – p)(x – q) or f(x) = -(x – p)(x – q). Write the function represented by each graph. Explain your reasoning.

**Communicate Your Answer**

Question 2.

What are some of the characteristics of the graph of f(x) = a(x – p)(x – q)?

Answer:

Question 3.

Consider the graph of f(x) = a(x – p)(x – q).

a. Does changing the sign of a change the x-intercepts? Does changing the sign of a change the y-intercept? Explain your reasoning.

b. Does changing the value of p change the x-intercepts? Does changing the value of p change the y-intercept? Explain your reasoning.

Answer:

**Monitoring Progress**

**Graph the quadratic function. Label the vertex, axis of symmetry, and x-intercepts. Describe the domain and range of the function.**

Question 1.

f(x) = (x + 2)(x – 3)

Answer:

Range: {y ∈ R: y≥ -25/4}

x from -5.5 to 6.5

Question 2.

g(x) = -2(x – 4)(x + 1)

Answer:

x-intercept:(-1, 0) and (4, 0)

Range: {y ∈ R : y ≤ 25/2}

Question 3.

h(x) = 4x^{2} – 36

Answer:

x-intercept: (-3, 0) and (3, 0)

Range: {y ∈ R : y ≥ -36}

**Find the zero(s) of the function.**

Question 4.

f(x) = (x – 6)(x – 1)

Answer:

f(x) = (x – 6)(x – 1)

(x – 6)(x – 1) = 0

x² – 6x -x + 6 = 0

x – 6 = 0 or x – 1 = 0

x = 6 or x = 1

Question 5.

g(x) = 3x^{2} – 12x + 12

Answer:

Given,

g(x) = 3x^{2} – 12x + 12

3x^{2} – 12x + 12 = 0

3x^{2} – 12x + 12 = 0

3( x^{2} – 4x + 4) = 0

x^{2} – 4x + 4 = 0

Question 6.

h(x) = x(x^{2} – 1)

Answer:

Given,

h(x) = x(x^{2} – 1)

x(x^{2} – 1) = 0

x = 0 or x^{2} – 1 = 0

x = 0 or x = 1

**Use zeros to graph the function.**

Question 7.

f(x) = (x – 1)(x – 4)

Answer:

f(x) = (x – 1)(x – 4)

(x – 1)(x – 4) = 0

x – 1 = 0 or x – 4 = 0

x = 1, 4

Question 8.

g(x) = x^{2} + x – 12

Answer:

x^{2} + x – 12 = 0

x^{2} + 4x – 3x – 12

x(x + 4) – 3(x + 4) = 0

(x + 4)(x – 3) = 0

x = -3, 4

**Write a quadratic function in standard form whose graph satisfies the given condition(s).**

Question 9.

x-intercepts: -1 and 1

Answer:

Question 10.

vertex: (8, 8)

Answer:

Let a = 1

f(x) = a(x – h)² + k

f(x) = 1(x – 8)² + 8

f(x) = (x – 8)² + 8

f(x) = x² – 16x + 64 + 8

f(x) = x² – 16x + 72

11. passes through (0, 0), (10, 0), and (4, 12)

Answer:

Question 12.

passes through (-5, 0), (4, 0), and (3, -16)

Answer:

**Use zeros to graph the function.**

Question 13.

g(x) = (x – 1)(x – 3)(x + 3)

Answer:

g(x) = (x – 1)(x – 3)(x + 3)

(x – 1)(x – 3)(x + 3) = 0

x – 1 = 0 or x – 3 = 0 or x + 3 = 0

x = 1, 3, -3

Question 14.

h(x) = x^{3} – 6x^{2} + 5x

Answer:

Question 15.

The zeros of a cubic function are -3, -1, and 1. The graph of the function passes through the point (0, -3). Write the function.

Answer:

### Using Intercept Form 8.5 Exercises

**Vocabulary and Core Concept Check**

Question 1.**COMPLETE THE SENTENCE**

The values p and q are __________ of the graph of the function f(x) = a(x – p)(x – q).

Answer:

The values p and q are x-intercepts of the graph of the function f(x) = a(x – p)(x – q).

Question 2.**WRITING**

Explain how to find the maximum value or minimum value of a quadratic function when the function is given in intercept form.

Answer:

The intercept form is:

f(x) = a(x – p)(x – q)

x = (p+q)/2

Substitute the x-coordinate to the function to find the y-coordinate which is the maximum or minimum value.

**Monitoring Progress and Modeling with Mathematics**

**In Exercises 3–6, find the x-intercepts and axis of symmetry of the graph of the function.**

Question 3.

Answer:

Question 4.

Answer:

y = -2(x – 2)(x – 5)

x – intercepts of parabola in intercept form are given p = 2 and q = 5

Axis of symmetry is

x = (p + q)/2

= (2 + 5)/2 = 7/2 = 3.5

Question 5.

f(x) = -5(x + 7)(x – 5)

Answer:

Question 6.

g(x) = \(\frac{2}{3}\) x(x + 8)

Answer:

g(x) = \(\frac{2}{3}\) x(x + 8)

x – intercepts of parabola in intercept form are given p = 0 and q = -8

Axis of symmetry is

x = (p + q)/2

= (0 – 8)/2 = -4

**In Exercises 7–12, graph the quadratic function. Label the vertex, axis of symmetry, and x-intercepts. Describe the domain and range of the function.**

Question 7.

f(x) = (x + 4)(x + 1)

Answer:

Question 8.

y = (x – 2)(x + 2)

Answer:

y = (x – 2)(x + 2)

Step 1: The intercepts of the parabola are p = 2 and q = -2.

Thus the points (2, 0) and (-2, 0) lie on the graph.

Step 2:

The axis of symmetry is

x = (p + q)/2

x = 2 + (-2)/2 = 0

Plot axis of symmetry, x = 0

Step 3: Find and plot the vertex. The y-coordinate of vertex is y(0)

y(0) = (0 – 2)(0 + 2) = -4

So, the vertex is (0, -4)

Step 4: Draw a parabola through vertex and x-intercept points.

We can see in graph that domain is set of all real numbers, R and range is [-4, ∞)

Question 9.

y = -(x + 6)(x – 4)

Answer:

Question 10.

h(x) = -4(x – 7)(x – 3)

Answer:

In the function f(x) = a(x – p)(x – q)

Step 1: The x-intercepts are 7 and 3. Plot the points (7, 0) and (3, 0)

Step 2: Graph the axis of symmetry:

x = (7 + 3)/2 = 5

Step 3:

The x-coordinate of the vertex is 5. Find the y-value of the vertex by substituting 5 to x

h(5) = -4(5 – 7)(5 – 3)

= -4(-2)(2) = 16

So, the vertex is (5, 16)

We can see in graph that domain is set of all real numbers, R and range is (-∞, 16]

Question 11.

g(x) = 5(x + 1)(x + 2)

Answer:

Question 12.

y = -2(x – 3)(x + 4)

Answer:

In the function f(x) = a(x – p)(x – q)

Step 1: The x-intercepts are 3 and -4.

Plot the points (3, 0) and (-4, 0)

Step 2:

Graph the axis of symmetry x = 3 + (-4)/2 = -0.5

Step 3:

The x-coordinate of the vertex is -0.5.

f(-0.5) = -2(-0.5 – 3)(-0.5 + 4) = 24.5

So, the vertex is (-0.5, 24.5)

**In Exercises 13–20, graph the quadratic function. Label the vertex, axis of symmetry, and x-intercepts. Describe the domain and range of the function.**

Question 13.

y = x^{2} – 9

Answer:

Question 14.

f(x) = x^{2} – 8x

Answer:

In the function f(x) = a(x – p)(x – q)

Step 1: The x-intercepts are 0 and 8.

Plot the points (0, 0) and (8, 0)

Step 2:

Graph the axis of symmetry x = 0 + 8/2 = 4

Step 3:

The x-coordinate of the vertex is 4

f(4) = (4)^{2} – 8(4) = -5

So, the vertex is (4, -16)

We can see in graph that domain is set of all real numbers, R and range is [-16, ∞)

Question 15.

h(x) = -5x^{2} + 5x

Answer:

Question 16.

y = 3x^{2} – 48

Answer:

In the function f(x) = a(x – p)(x – q)

Step 1: The x-intercepts are 4 and -4.

Plot the points (4, 0) and (-4, 0)

Step 2:

Graph the axis of symmetry x = 4 + (-4)/2 = 0

Step 3:

The x-coordinate of the vertex is 0

f(0) = 3(0)^{2} – 48 = -48

So, the vertex is (0, -48)

We can see in graph that domain is set of all real numbers, R and range is [-48, ∞)

Question 17.

q(x) = x^{2} + 9x + 14

Answer:

Question 18.

p(x) = x^{2} + 6x – 27

Answer:

In the function f(x) = a(x – p)(x – q)

Step 1: The x-intercepts are -9 and 3.

Plot the points (-9, 0) and (3, 0)

Step 2:

Graph the axis of symmetry x = -9 + 3/2 = -3

Step 3:

The x-coordinate of the vertex is -3

f(-3) = (-3)^{2} + 6(-3) – 27 = -36

So, the vertex is (-3, -36)

We can see in graph that domain is set of all real numbers, R and range is [-36, ∞)

Question 19.

y = 4x^{2} – 36x + 32

Answer:

Question 20.

y = -2x^{2} – 4x + 30

Answer:

In the function f(x) = a(x – p)(x – q)

Step 1: The x-intercepts are 3 and -5.

Plot the points (3, 0) and (-5, 0)

Step 2:

Graph the axis of symmetry x = 3 + (-5)/2 = -1

Step 3:

The x-coordinate of the vertex is -1

f(-1) = -2(-1)^{2} – 4(-1) + 30 = 32

So, the vertex is (-1, 32)

We can see in graph that domain is set of all real numbers, R and range is (-∞, 32]

**In Exercises 21–30, find the zero(s) of the function.**

Question 21.

y = -2(x – 2)(x – 10)

Answer:

Question 22.

f(x) = \(\frac{1}{3}\)(x + 5)(x – 1)

Answer:

f(x) = \(\frac{1}{3}\)(x + 5)(x – 1)

\(\frac{1}{3}\)(x + 5)(x – 1) = 0

x + 5 = 0 or x – 1 = 0

x = -5 or x = 1

Question 23.

g(x) = x^{2} + 5x – 24

Answer:

Question 24.

y = x^{2} – 17x + 52

Answer:

Given,

y = x^{2} – 17x + 52

x^{2} – 17x + 52 = 0

x^{2} – 4x – 13x + 52 = 0

x(x – 4) – 13(x – 4) = 0

(x – 4)(x – 13) = 0

x – 4 = 0 or x – 13 = 0

x = 4 or x = 13

Question 25.

y = 3x^{2} – 15x – 42

Answer:

Question 26.

g(x) = -4x^{2} – 8x – 4

Answer:

g(x) = -4x^{2} – 8x – 4

-4x^{2} – 8x – 4 = 0

(x + 1)^{2} = 0

x + 1 = 0

x = -1

Question 27.

f(x) = (x + 5)(x^{2} – 4)

Answer:

Question 28.

h(x) = (x^{2} – 36)(x – 11)

Answer:

h(x) = (x^{2} – 36)(x – 11)

(x^{2} – 36)(x – 11) = 0

(x + 6)(x – 6)(x – 11) = 0

x + 6 = 0 or x – 6 = 0 or x – 11 = 0

x = -6 or 6 or 11

Question 29.

y = x^{3} – 49x

Answer:

Question 30.

y = x^{3} – x^{2} – 9x + 9

Answer:

y = x^{3} – x^{2} – 9x + 9

x^{3} – x^{2} – 9x + 9 = 0

x^{2} (x – 1) -9(x – 1) = 0

(x – 1)(x² – 9) = 0

(x – 1)(x – 3)(x + 3) = 0

x – 1 = 0 or x – 3 = 0 or x + 3 = 0

x = 1 or x = 3 or x = -3

**In Exercises 31–36, match the function with its graph.**

Question 31.

y = (x + 5)(x + 3)

Answer:

Question 32.

y = (x + 5)(x – 3)

Answer:

y = (x + 5)(x – 3)

(x + 5)(x – 3) = 0

x + 5 = 0 or x – 3 = 0

x = -5 or x = 3

Question 33.

y = (x – 5)(x + 3)

Answer:

Question 34.

y = (x – 5)(x – 3)

Answer:

y = (x – 5)(x – 3)

(x – 5)(x – 3) = 0

x – 5 = 0 or x – 3 = 0

x = 5 or x = 3

Question 35.

y = (x + 5)(x – 5)

Answer:

Question 36.

y = (x + 3)(x – 3)

Answer:

y = (x + 3)(x – 3)

(x + 3)(x – 3) = 0

x + 3 = 0 or x – 3 = 0

x = -3 or x = 3

**In Exercises 37–42, use zeros to graph the function.**

Question 37.

f(x) = (x + 2)(x – 6)

Answer:

Question 38.

g(x) = -3(x + 1)(x + 7)

Answer:

g(x) = -3(x + 1)(x + 7)

x = 0

y = -3(0 + 1)(0 + 7) = -21

x = -1 and x = -7

We plot the points (0, -21), (-1, 0), (-7, 0) and join them to sketch the parabola.

Question 39.

y = x^{2} – 11x + 18

Answer:

Question 40.

y = x^{2} – x – 30

Answer:

Given,

y = x^{2} – x – 30

x = 0

y = 0^{2} – 0 – 30 = -30

x^{2} – x – 30 = 30

x^{2} + 5x – 6x – 30 = 0

(x + 5)(x – 6) = 0

x + 5 = 0 or x – 6 = 0

x = -5 or x = 6

Question 41.

y = -5x^{2} – 10x + 40

Answer:

Question 42.

h(x) = 8x^{2} – 8

Answer:

h(x) = 8x^{2} – 8

x = 0

y = 8(0)^{2} – 8 = -8

8x^{2} – 8 = 0

8(x^{2} – 1) = 0

8(x + 1)(x – 1) = 0

x + 1 = 0 or x – 1 = 0

x = -1 or x = 1

**ERROR ANALYSIS** In Exercises 43 and 44, describe and correct the error in finding the zeros of the function.

Question 43.

Answer:

Question 44.

Answer:

y = (x + 4)(x^{2} – 9)

(x + 4)(x + 3)(x – 3) = 0

x + 4 = 0 or x + 3 = 0 or x – 3 = 0

x = -4 or x = -3 or x = 3

The error was in solving the equation x^{2} – 9 = 0 in a wrong way. It was treated as a linear equation, but it is a quadratic equation that can be solved either directly with the square root method or by factoring the expression and applying the zero product property.

**In Exercises 45–56, write a quadratic function in standard form whose graph satisfies the given condition(s).**

Question 45.

vertex: (7, -3)

Answer:

Question 46.

vertex: (4, 8)

Answer:

f(x) = a(x – h)² + k

V(h, k) = (4, 8)

h = 4, k = 8

f(x) = a(x – 4)² + 8

a = 1

f(x) = 1(x – 4)² + 8

f(x) = (x – 4)² + 8 = x² – 8x + 24

Question 47.

x-intercepts: 1 and 9

Answer:

Question 48.

x-intercepts: -2 and -5

Answer:

Given,

x-intercepts: -2 and -5

f(x) = a[x – (-2)][x – (-5)]

f(x) = a(x + 2)(x + 5)

a = 1

f(x) = 1(x + 2)(x + 5)

f(x) = (x + 2)(x + 5)

= x² + 5x + 2x + 10

= x² + 7x + 10

Question 49.

passes through (-4, 0), (3, 0), and (2, -18)

Answer:

Question 50.

passes through (-5, 0), (-1, 0), and (-4, 3)

Answer:

f(x) = a[x – (-5)][x – (-1)]

f(x) = a(x + 5)(x + 1)

a(-4 + 5)(-4 + 1) = 3

-3a = 3

a = -1

f(x) = -1(x + 5)(x + 1)

f(x) = -(x² + x + 5x + 5)

– x² – 6x – 5

Question 51.

passes through (7, 0)

Answer:

Question 52.

passes through (0, 0) and (6, 0)

Answer:

The points of parabola (0, 0) and (6, 0)

f(x) = a[x – 0][x – 6]

f(x) = ax(x – 6)

a=1

f(x) = x(x – 6)

f(x) = x² – 6x

Question 53.

axis of symmetry: x = -5

Answer:

Question 54.

y increases as x increases when x < 4; y decreases as x increases when x > 4.

Answer:

The axis of symmetry of a quadratic function with equation of the form

f(x) = a(x – p)(x – q)

x = (p + q)/2

The behavior of the graph changes at x = 4. This means that the axis of symmetry x = 4/

x = (p + q)/2 = 4

f(x) = -2(x)(x – 8)

f(x) = -2x² + 16x

Question 55.

range: y ≥ -3

Answer:

Question 56.

range: y ≤ 10

Answer:

The function f(x) = ax² + x will have a range of

y ≥ c if a > 0 or

y ≤ c if if a < 0

This means that in the given function a > 0 and c = 10

f(x) = -2x² + 10

**In Exercises 57–60, write the quadratic function represented by the graph.**

Question 57.

Answer:

Question 58.

Answer:

For the given graph the x-intercepts are 1 and 7.

p = and q = 7

f(x) = a(x – 1)(x – 7)

To find the value of a, substitute the x and y values of the point on the parabola (6, 5)

5 = a(6 – 1)(6 – 7)

5 = a . 5 . -1

5 = -5a

a = -1

Question 59.

Answer:

Question 60.

Answer:

For the given graph the x-intercepts are 6 and 10.

p = 6 and q = 10

f(x) = a(x – 6)(x – 10)

To find the value of a, substitute the x and y values of the point on the parabola (6, 5)

-2 = a(8 – 6)(8 – 10)

-2 = a . 2 . -2

-2 = -4a

a = 1/2

**In Exercises 61–68, use zeros to graph the function.**

Question 61.

y = 5x(x + 2)(x – 6)

Answer:

Question 62.

f(x) = -x(x + 9)(x + 3)

Answer:

The function completely and the zeros are: 0, -9, -3.

Plot the corresponding points: (-9, 0), (-3, 0), (0, 0)

To help determine the shape of the graph, find the points between the zeros and plot them.

Question 63.

h(x) = (x – 2)(x + 2)(x + 7)

Answer:

Question 64.

y = (x + 1)(x – 5)(x – 4)

Answer:

The function completely and the zeros are: -1, 5, 4

Plot the corresponding points: (-1, 0), (4, 0), (5, 0)

To help determine the shape of the graph, find the points between the zeros and plot them.

Question 65.

f(x) = 3x^{3} – 48x

Answer:

Question 66.

y = -2x^{3} + 20x^{2} – 50x

Answer:

y = -2x^{3} + 20x^{2} – 50x

y = -2x(x^{2} – 10x + 25)

y = -2x(x – 5)(x – 5)

Plot the corresponding points: (0, 0), (5, 0)

To help determine the shape of the graph, find the points between the zeros and plot them

Question 67.

y = -x^{3} – 16x^{2} – 28x

Answer:

Question 68.

g(x) = 6x^{3} + 30x^{2} – 36x

Answer:

Given,

g(x) = 6x^{3} + 30x^{2} – 36x

g(x) = 6x(x^{2} + 5x – 6)

g(x) = 6x(x – 1)(x + 6)

The function completely and the zeros are: 0, 1, -6

Plot the corresponding points: (-6, 0), (0, 0), (1, 0)

To help determine the shape of the graph, find the points between the zeros and plot them

**In Exercises 69–72, write the cubic function represented by the graph.**

Question 69.

Answer:

Question 70.

Answer:

From the graph, the x-intercepts are -3, 0, 2

f(x) = a(x – p)(x – q)(x – r)

f(x) = a(x – (-3))(x – 0)(x – 2)

f(x) = a(x + 3)(x)(x – 2)

Use the other point (-1, -36) to find the value of a

-36 = a(-1 + 3)(-1)(-1 – 2)

-36 = 6a

a = -6

f(x) = -6(x + 3)(x)(x – 2)

= -6x^{3} -6x^{2} + 36x

Question 71.

Answer:

Question 72.

Answer:

From the graph, the x-intercepts are 1, 3, 6.

f(x) = a(x – p)(x – q)(x – r)

f(x) = a(x – 1)(x – 3)(x – 6)

Use the other point (5, -40) to find the value of a

-40 = a(5 – 1)(5 – 3)(5 – 6)

-40 = -8a

a = 5

f(x) = 5(x – 1)(x – 3)(x – 6)

f(x) = 5x³ + -50x² + 135x – 90

**In Exercises 73–76, write a cubic function whose graph satisfies the given condition(s).**

Question 73.

x-intercepts: -2, 3, and 8

Answer:

Question 74.

x-intercepts: -7, -5, and 0

Answer:

Given the x-intercepts use the intercept form:

f(x) = a(x – p)(x – q)(x – r)

a = 1

f(x) = 1(x – (-7))(x – (-5))(x – 0)

f(x) = (x + 7)(x + 5)(x)

f(x) = x² + 12x² + 35x

Question 75.

passes through (1, 0) and (7, 0)

Answer:

Question 76.

passes through (0, 6)

Answer:

Note that the given is not x-intercept

Since we need a cubic function, we need 3 intercepts.

For simplicity’s sake we will use (-1, 0), (1, 0) and (2, 0)

Given the x-intercepts use the intercept form:

f(x) = a(x – p)(x – q)(x – r)

f(x) = a(x – (-1))(x – 1)(x – 2)

Use the other point (0, 6) to find the value of a:

6 = a(0 + 1)(0 – 1)(0 – 2)

6 = 2a

a = 3

f(x) = 3(x + 1)(x – 1)(x – 2)

f(x) = 3x³ – 6x² – 3x + 6

**In Exercises 77–80, all the zeros of a function are given. Use the zeros and the other point given to write a quadratic or cubic function represented by the table.**

Question 77.

Answer:

Question 78.

Answer:

Given the x-intercepts use the intercept form:

f(x) = a(x – p)(x – q)(x – r)

The x-intercepts are -3 and 4.

p = -3 and q = 4

f(x) = a(x +3)(x-4)

-72 = a(1 +3)(1-4)

-72 = -12a

a = 6

Therefore the equation of the quadratic function whose graph is given

f(x) = 6(x + 3)(x – 4)

Question 79.

Answer:

Question 80.

Answer:

Given the x-intercepts (-8, -3, 0) use the intercept form:

f(x) = a(x – p)(x – q)(x – r)

f(x) = a(x – (-8))(x – (-3))(x – 0)

f(x) = a(x +8)(x + 3)(x)

-36 = a(-6 + 8)(-6 + 3)(-6)

-36 = 36a

a = 1

f(x) = -1(x +8)(x + 3)(x)

f(x) = -x³ – 11x² + 24x

**In Exercises 81–84, sketch a parabola that satisfies the given conditions.**

Question 81.

x-intercepts: -4 and 2; range: y ≥ -3

Answer:

Question 82.

axis of symmetry: x = 6; passes through (4, 15)

Answer:

Let the equation of parabola be y = a(x – h)² + k

x = 6, h = 6

The parabola passes through (4, 15).

15 = a(4 – 6)² + k

15 = 4a + k

a = 1

15 = 4 + k

k = 11

Question 83.

range: y ≤ 5; passes through (0, 2)

Answer:

Question 84.

x-intercept: 6; y-intercept: 1; range: y ≥ -4

Answer:

Question 85.**MODELING WITH MATHEMATICS**

Satellite dishes are shaped like parabolas to optimally receive signals. The cross section of a satellite dish can be modeled by the function shown, where x and y are measured in feet. The x-axis represents the top of the opening of the dish.

a. How wide is the satellite dish?

b. How deep is the satellite dish?

c. Write a quadratic function in standard form that models the cross section of a satellite dish that is 6 feet wide and 1.5 feet deep.

Answer:

Question 86.**MODELING WITH MATHEMATICS**

A professional basketball player’s shot is modeled by the function shown, where x and y are measured in feet.

a. Does the player make the shot? Explain.

Answer:

The player will make the shot if the point (3, 0) is on the parabola.

0 = –\(\frac{1}{20}\)(3² – 19(3) + 48)

0 = –\(\frac{1}{20}\)(9 – 57 + 48)

0 = 0

Yes, he will make the shot

b. The basketball player releases another shot from the point (13, 0) and makes the shot. The shot also passes through the point (10, 1.4). Write a quadratic function in standard form that models the path of the shot.

Answer:

Two intercepts are given (3, 0) and (13, 0)

y = a(x – p)(x – q)

y = a(x – 3)(x – 13)

Use the other point (10, 1.4) to find the value of a

1.4 = a(10 – 3)(10 – 13)

1.4 = -21a

a = –\(\frac{1}{15}\)

f(x) = –\(\frac{1}{15}\) (x – 3)(x – 13)

f(x) = –\(\frac{1}{15}\) (x² – 16x + 39)

**USING STRUCTURE** In Exercises 87–90, match the function with its graph.

Question 87.

y = -x^{2} + 5x

Answer:

Question 88.

y = x^{2} – x – 12

Answer:

The given is a quadratic function, so it is either A or D.

Substituting x = 0, the y-intercept is

y = (0)^{2} – 0 – 12 = -12

This corresponds to Graph A.

Question 89.

y = x^{3} – 2x^{2} – 8x

Answer:

Question 90.

y = x^{3} – 4x^{2} – 11x + 30

Answer:

The given is a cubic function. So it is either B or C.

Substituting x = 0, the y-intercept is

y = x^{3} – 4x^{2} – 11x + 30

y = (0)^{3} – 4(0)^{2} – 11(0) + 30 = 30

This corresponds to Graph B.

Question 91.**CRITICAL THINKING**

Write a quadratic function represented by the table, if possible. If not, explain why.

Answer:

It is not possible to write a quadratic function represented by the table. Because -5 and 1 are the x-intercepts, the axis of symmetry is x = -2. The points (-3, 12) and (-1, 4) are the same horizontal distance from the axis of symmetry, so for both of them to lie on the parabola, they would have the same y-coordinate.

Question 92.**HOW DO YOU SEE IT?**

The graph shows the parabolic arch that supports the roof of a convention center, where x and y are measured in feet.

a. The arch can be represented by a function of the form f(x) = a(x – p)(x – q). Estimate the values of p and q.

Answer:

From the graph, the x-intercepts are 75 and 425.

p = 75 and q = 425

b. Estimate the width and height of the arch. Explain how you can use your height estimate to calculate a.

Answer:

The width is the distance between the 2 x-intercepts

width = |75 – 425|

width = 350 feet

The height is approximately 60 feet

Since the height corresponds to the location of the vertex, we can use it as an additional point to find the value of a.

**ANALYZING EQUATIONS** In Exercises 93 and 94,

(a) rewrite the quadratic function in intercept form and

(b) graph the function using any method. Explain the method you used.

Question 93.

f(x) = -3(x + 1)^{2} + 27

Answer:

Question 94.

g(x) = 2(x – 1)^{2} – 2

Answer:

The vertex is at (h, k) and

g(x) = 2(x – 1)^{2} – 2

g(x) = 2(x^{2} – 2x + 1)– 2

g(x) = 2x^{2} – 4x + 2 – 2

g(x) = 2x^{2} – 4x

g(x) = 2(x)(x – 2)

The x-intercepts of the given function are 0 and 2.

Plot the points (0, 0) and (2, 0)

Question 95.**WRITING**

Can a quadratic function with exactly one real zero be written in intercept form? Explain.

Answer:

Question 96.**MAKING AN ARGUMENT**

Your friend claims that any quadratic function can be written in standard form and in vertex form. Is your friend correct? Explain.

Answer: My friend is correct

Any quadratic function can be written in vertex form.

y = a(x² – 2hx + h²) + k

b = -2ahx

c = h² + k

Any quadratic function can be written in standard form. When a quadratic function is in standard form, completing the square can be applied so that it can be transformed to vertex form.

Question 97.**PROBLEM SOLVING**

Write the function represented by the graph in intercept form.

Answer:

Question 98.**THOUGHT PROVOKING**

Sketch the graph of each function. Explain your procedure.

a. f(x) = (x^{2} – 1)(x^{2} – 4)

b. g(x) = x(x^{2} – 1)(x^{2} – 4)

Answer:

Question 99.**REASONING**

Let k be a constant. Find the zeros of the function f(x) = kx^{2} – k^{2}x – 2k^{3} in terms of k.

Answer:

**PROBLEM SOLVING** In Exercises 100 and 101, write a system of two quadratic equations whose graphs intersect at the given points. Explain your reasoning.

Question 100.

(-4, 0) and (2, 0)

Answer:

Given the x-intercepts (-4, 2) use the intercept form:

y = a(x – p)(x – q)

y = a(x – (-4))(x – 2)

y = a(x + 4)(x – 2)

y = a(x² + 2x – 8)

To find 2 equations let a be any constant value other than 0

y = x² + 2x – 8

y = -x² – 2x + 8

Question 101.

(3, 6) and (7, 6)

Answer:

**Maintaining Mathematical Proficiency**

The scatter plot shows the amounts x (in grams) of fat and the numbers y of calories in 12 burgers at a fast-food restaurant.

Question 102.

How many calories are in the burger that contains 12 grams of fat?

Answer:

From the above graph, we observe that there is approximately 300 calories in a burger containing 12 grams of fat.

Question 103.

How many grams of fat are in the burger that contains 600 calories?

Answer:

Question 104.

What tends to happen to the number of calories as the number of grams of fat increases?

Answer: As the number of grams of fat increases, the number of calories also increases.

Determine whether the sequence is arithmetic, geometric, or neither. Explain your reasoning.

Question 105.

3, 11, 21, 33, 47, . . .

Answer:

Question 106.

-2, -6, -18, -54, . . .

Answer:

-6/-2 = -18/-6….. geometric sequence

Question 107.

26, 18, 10, 2, -6, . . .

Answer:

Question 108.

4, 5, 9, 14, 23, . . .

Answer:

5 – 4 ≠ 9 – 5 not an arithmetic sequence

5/4 ≠ 9/5 not geometric sequence

### Lesson 8.6 Comparing Linear, Exponential, and Quadratic Functions

**Essential Question** How can you compare the growth rates of linear, exponential, and quadratic functions?

**EXPLORATION 1**

Comparing Speeds

Work with a partner. Three cars start traveling at the same time. The distance traveled in t minutes is y miles. Complete each table and sketch all three graphs in the same coordinate plane. Compare the speeds of the three cars. Which car has a constant speed? Which car is accelerating the most? Explain your reasoning.

**EXPLORATION 2**

Comparing Speeds

Work with a partner. Analyze the speeds of the three cars over the given time periods. The distance traveled in t minutes is y miles. Which car eventually overtakes the others?

**Communicate Your Answer**

Question 3.

How can you compare the growth rates of linear, exponential, and quadratic functions?

Answer:

Question 4.

Which function has a growth rate that is eventually much greater than the growth rates of the other two functions? Explain your reasoning.

Answer:

**Monitoring Progress**

**Plot the points. Tell whether the points appear to represent a linear, an exponential, or a quadratic function.**

Question 1.

(-1, 5), (2, -1), (0, -1), (3, 5), (1, -3)

Answer:

Question 2.

(-1, 2), (-2, 8), (-3, 32), (0, \(\frac{1}{2}\)), (1, \(\frac{1}{8}\))

Answer:

Question 3.

(-3, 5), (0, -1), (2, -5), (-4, 7), (1, -3)

Answer:

Question 4.

Tell whether the table of values represents a linear, an exponential, or a quadratic function.

Answer:

Question 5.

Tell whether the table of values represents a linear, an exponential, or a quadratic function. Then write the function.

Answer:

Question 6.

Compare the websites in Example 4 by calculating and interpreting the average rates of change from Day 0 to Day 10.

Answer:

Question 7.**WHAT IF?**

Tinyville’s population increased by 8% each year. In what year were the populations about equal?

Answer:

### Comparing Linear, Exponential, and Quadratic Functions 8.6 Exercises

**Vocabulary and Core Concept Check**

Question 1.**WRITING**

Name three types of functions that you can use to model data. Describe the equation and graph of each type of function.

Answer:

Question 2.**WRITING**

How can you decide whether to use a linear, an exponential, or a quadratic function to model a data set?

Answer:

If consecutive y-values have a constant first difference, a linear function must be used to model a data set.

If consecutive y-values have a constant second difference, a quadratic function must be used to model a data set.

If a common ratio exists consecutive y-values, an exponential function must be used to model a data set.

Question 3.**VOCABULARY**

Describe how to find the average rate of change of a function y = f(x) between x = a and x = b.

Answer:

Question 4.**WHICH ONE DOESN’T BELONG?**

Which graph does not belong with the other three? Explain your reasoning.

Answer:

The third graph from the left does not belong to the group. This is because the three other graphs are parabolas which are graphs of quadratic functions. The third graph involves an exponential function.

**Monitoring Progress and Modeling with Mathematics**

**In Exercises 5–8, tell whether the points appear to represent a linear, an exponential, or a quadratic function.**

Question 5.

Answer:

Question 6.

Answer:

The graph is a curve that is decreasing from left to right therefore it must involve an exponential function.

Also, notice that the y-values are 16, 8, 4, … which shows a common ratio of 1/2.

, Question 7.

Answer:

Question 8.

Answer:

The points from a straight line.

Notice also that for every 1 unit increase in the value of x, the value of y increases by 3 units.

Thus the points represent a linear function.

**In Exercises 9–14, plot the points. Tell whether the points appear to represent a linear, an exponential, or a quadratic function.**

Question 9.

(-2, -1), (-1, 0), (1, 2), (2, 3), (0, 1)

Answer:

Question 10.

( 0, \(\frac{1}{4}\)), (1, 1), (2, 4), (3, 16), (-1, \(\frac{1}{16}\))

Answer:

The points appear to represent an exponential function.

Question 11.

(0, -3), (1, 0), (2, 9), (-2, 9), (-1, 0)

Answer:

Question 12.

(-1, -3), (-3, 5), (0, -1), (1, 5), (2, 15)

Answer:

The point appears to represent a quadratic function with an axis of symmetry of x = -1

Question 13.

(-4, -4), (-2, -3.4), (0, -), (2, -2.6), (4, -2)

Answer:

Question 14.

(0, 8), (-4, 0.25), (-3, 0.4), (-2, 1), (-1, 3)

Answer:

**In Exercises 15–18, tell whether the table of values represents a linear, an exponential, or a quadratic function.**

Question 15.

Answer:

Question 16.

Answer:

The consecutive y-values have a common ratio of 5.

Therefore the table of values represents an exponential function.

Question 17.

Answer:

Question 18.

Answer:

The y-values in the table have:

First Difference: 2.5, 3.5, 4.5, 5, 5.5

Second Difference: 1, 1, 1

The y-values have a common 2nd difference therefore the table of values represents a quadratic function.

Question 19.**MODELING WITH MATHEMATICS**

A student takes a subway to a public library. The table shows the distances d (in miles) the student travels in t minutes. Let the time t represent the independent variable. Tell whether the data can be modeled by a linear, an exponential, or a quadratic function. Explain.

Answer:

Question 20.**MODELING WITH MATHEMATICS**

A store sells custom circular rugs. The table shows the costs c (in dollars) of rugs that have diameters of d feet. Let the diameter d represent the independent variable. Tell whether the data can be modeled by a linear, an exponential, or a quadratic function. Explain.

Answer:

Since second differences for consecutive x-values are constant, data can be modelled as quadratic function.

First difference: 49.7, 63.9, 78.1

Second difference: 14.2, 14.2

**In Exercises 21–26, tell whether the data represent a linear, an exponential, or a quadratic function. Then write the function.**

Question 21.

(-2, 8), (-1, 0), (0, -4), (1, -4), (2, 0), (3, 8)

Answer:

Question 22.

(-3, 8), (-2, 4), (-1, 2), (0, 1), (1, 0.5)

Answer:

The data has no constant first nor second difference.

Note that there is a common ratio of 0.5 that you can multiply to get the next y-value. Therefore the data must represent an exponential function with a base of 0.5 or 1/2.

f(x) = a . (1/2)^{x}

Use any point say (0, 1) to find the value of a

1 = a . (1/2)^{0}

1 = a . 1

a = 1

Thus the function that models the given data is f(x) = (1/2)^{x}

Question 23.

Answer:

Question 24.

Answer:

The data has no constant first nor second difference.

Note that there is a common ratio of 2 that you can multiply to get the next y-value. Therefore the data must represent an exponential function with a base of 2.

f(x) = a . (2)^{x}

Use any point say (0, 5) to find the value of a

5 = a . (2)^{0}

5 = a . 1

a = 5

Thus the function that models the given data is f(x) = 5 . 2^{x}

Question 25.

Answer:

Question 26.

Answer:

Notice that the graph show points that form a line therefore the data must represent a linear function.

The graph shows that the functions y-intercept is -2.

f(x) = mx + b

f(x) = mx – 2

Use any point say (1, -4) to find the value of m

-4 = m(1) – 2

-4 = m – 2

m = -4 + 2

m = -2

Now substitute the above value

f(x) = -2x – 2

Question 27.**ERROR ANALYSIS**

Describe and correct the error in determining whether the table represents a linear, an exponential, or a quadratic function.

Answer:

Question 28.**ERROR ANALYSIS**

Describe and correct the error in writing the function represented by the table.

Answer:

The error is in the first line of the solving for a.

The operation should have been addition

f(x) = a(x + 2)(x – 1)

Using the point (-3, 4) to find the value of a

4 = a(-3 + 2)(-3 – 1)

4 = 4a

a = 1

f(x) = 1(x + 2)(x – 1)

f(x) = x² + x – 2

Question 29.**REASONING**

The table shows the numbers of people attending the first five football games at a high school.

a. Plot the points. Let the game g represent the independent variable.

b. Can a linear, an exponential, or a quadratic function represent this situation? Explain.

Answer:

Question 30.**MODELING WITH MATHEMATICS**

The table shows the breathing rates y (in liters of air per minute) of a cyclist traveling at different speeds x (in miles per hour).

a. Plot the points. Let the speed x represent the independent variable. Then determine the type of function that best represents this situation.

Answer:

Although the points approximate a line, there is no common difference using the above table but there is a common ratio of about 1.11 between consecutive points so the solution is represented by an exponential function.

b. Write a function that models the data.

y = ab^{x}

The common ration is 1.11 so b = 1.11

Using any of the points say (20, 51.4)

51.4 = a(1.11)^{20}

a ≈ 6.38

So, the exponential function is

f(x) = 6.38(1.11)^{x}

c. Find the breathing rate of a cyclist traveling 18 miles per hour. Round your answer to the nearest tenth.

Answer:

Substitute x = 18 for 18 mph

f(18) = 6.38(1.11)^{18}

f18) = 41.7 liters of air per minute

Question 31.**ANALYZING RATES OF CHANGE**

The function f(t) = -16t^{2} + 48t + 3 represents the height (in feet) of a volleyball t seconds after it is hit into the air.

a. Copy and complete the table.

b. Plot the ordered pairs and draw a smooth curve through the points.

c. Describe where the function is increasing and decreasing.

d. Find the average rate of change for each 0.5-second interval in the table. What do you notice about the average rates of change when the function is increasing? decreasing?

Answer:

Question 32.**ANALYZING RELATIONSHIPS**

The population of Town A in 1970 was 3000. The population of Town A increased by 20% every decade. Let x represent the number of decades since 1970. The graph shows the population of Town B.

a. Compare the populations of the towns by calculating and interpreting the average rates of change from 1990 to 2010.

Answer:

Town A can be modeled using y = a(1 + r)^{t}

where a = 3000, r = 0.2

y = 3000(1.2)^{t}

At 1990 (t = 2): 3000(1.2)^{2 }= 4320

At 2010 (t = 4): 3000(1.2)^{4 }= 6221

The average rate of change = (6221 – 4320)/4 – 2 = 950.5

Town B population based on the graph are

At 1990 (t = 2): y = 5000

At 2010 (t = 4): y = 6500

Average rate of change = (6500 – 5000)/4 – 2 = 750

The average rates of change are the rate of increase of the population from 1990 to 2010.

b. Predict which town will have a greater population after 2020. Explain.

Answer:

Based on the rates of change from Town A will have a greater population after 2030.

Question 33.**ANALYZING RELATIONSHIPS**

Three organizations are collecting donations for a cause. Organization A begins with one donation, and the number of donations quadruples each hour. The table shows the numbers of donations collected by Organization B. The graph shows the numbers of donations collected by Organization C.

a. What type of function represents the numbers of donations collected by Organization A? B? C?

b. Find the average rates of change of each function for each 1-hour interval from t = 0 to t = 6.

c. For which function does the average rate of change increase most quickly? What does this tell you about the numbers of donations collected by the three organizations?

Answer:

Question 34.**COMPARING FUNCTIONS**

The room expenses for two different resorts are shown.

a. For what length of vacation does each resort cost about the same?

Answer:

Construct a table comparing the two resorts. Blue Water’s terms increase by 112 starting at t = 3 while see breeze’s terms increase by a factor of 1.1:

From the table, a vacation of 9 days will amount to the price.

b. Suppose Blue Water Resort charges $1450 for the first three nights and $105 for each additional night. Would Sea Breeze Resort ever be more expensive than Blue Water Resort? Explain.

Answer:

Yes, starting at 13 days of vacation, the price for sea breeze is greater than blue water.

c. Suppose Sea Breeze Resort charges $1200 for the first three nights. The charge increases 10% for each additional night. Would Blue Water Resort ever be more expensive than Sea Breeze Resort? Explain.

Answer:

No, Sea Breeze’s prices will always be greater than blue water for all number of vacations.

Question 35.**REASONING**

Explain why the average rate of change of a linear function is constant and the average rate of change of a quadratic or exponential function is not constant.

Answer:

Question 36.**HOW DO YOU SEE IT?**

Match each graph with its function. Explain your reasoning.

Answer:

a. The graph is a line so the function is: Function D

b. The graph is an exponential decay so b < 0 which is Function C.

c. The graph is an exponential growth so b > 0 which is Function B.

d. The graph is a parabola so the function is Function A.

Question 37.**CRITICAL THINKING**

In the ordered pairs below, the y-values are given in terms of n. Tell whether the ordered pairs represent a linear, an exponential, or a quadratic function. Explain.

(1, 3n – 1), (2, 10n + 2), (3, 26n),

(4, 51n – 7), (5, 85n – 19)

Answer:

Question 38.**USING STRUCTURE**

Write a function that has constant second differences of 3.

Answer:

If the second difference is constant, then it must be a quadratic function.

Let the first difference be: 1, 4, 7, 10

Therefore, a possible sequence with the first difference shown above is

0, 1, 5, 12, 22

Find a quadratic function that passes through:

(0, 0), (1, 1), (2, 5), (3, 12), (4, 22)

Choose any 3 points:

(0, 0), (1, 1), (2, 5)

Us the model y = ax² + bx + c

At (0, 0): 0 = a(0)² + b(0) + c

c = 0

At (0, 0): 0 = a(1)² + b(1) + c

a + b + c = 1

At (2, 5): 0 = a(2)² + b20) + c

4a + 2b + c = 5

Substitute c = 0

a + b + c = 1

a + b = 1

4a + 2b = 5

Solving a and b, a = 1.5, b = -0.5

y = 1.5x² – 0.5x

Question 39.**CRITICAL THINKING**

Is the graph of a set of points enough to determine whether the points represent a linear, an exponential, or a quadratic function? Justify your answer.

Answer:

Question 40.**THOUGHT PROVOKING**

Find four different patterns in the figure. Determine whether each pattern represents a linear, an exponential, or a quadratic function. Write a model for each pattern.

Answer:

In a given pattern corresponding to n, the number of vertices with xth distance from root is given by,

f(x) = 2^{x-1}

where, x ∈ N ∩ [1, n]

Thus each pattern represents exponential function with common factor 2.

Question 41.**MAKING AN ARGUMENT**

Function p is an exponential function and function q is a quadratic function. Your friend says that after about x = 3, function q will always have a greater y-value than function p. Is your friend correct? Explain.

Answer:

Question 42.**USING TOOLS**

The table shows the amount a (in billions of dollars) United States residents spent on pets or pet-related products and services each year for a 5-year period. Let the year x represent the independent variable. Using technology, find a function that models the data. How did you choose the model? Predict how much residents will spend on pets or pet-related products and services in Year 7.

Answer:

By plotting the point using a graphing utility, a parabola seems to fit the data so we use the quadratic function.

Using the graphing utility, the model is:

y = -0.4143x² + 6.166x + 46.98

**Maintaining Mathematical Proficiency**

**Evaluate the expression.**

Question 43.

\(\sqrt{121}\)

Answer:

Question 44.

\(\sqrt [ 3 ]{ 125 }\)

Answer: 5

Question 45.

\(\sqrt [ 3 ]{ 512 }\)

Answer:

Question 46.

\(\sqrt [ 5 ]{ 243 }\)

Answer: 3

**Find the product.**

Question 47.

(x + 8)(x – 8)

Answer:

Question 48.

(4y + 2)(4y – 2)

Answer:

Given,

(4y + 2)(4y – 2)

4y(4y – 2) + 2(4y – 2)

16y² – 8y + 8y – 4

16y² – 4

Question 49.

(3a – 5b)(3a + 5b)

Answer:

Question 50.

(-2r + 6s)(-2r – 6s)

Answer:

Given,

(-2r + 6s)(-2r – 6s)

-2r(-2r – 6s) + 6s(-2r – 6s)

4r² + 12rs – 12rs – 36s²

4r² – 36s²

### Graphing Quadratic Functions Performance Task: Asteroid Aim

**8.4–8.6 What Did You Learn?**

**Core Vocabulary**

**Core Concepts**

**Mathematical Practices**

Question 1.

How can you use technology to confirm your answer in Exercise 64 on page 448?

Answer:

Question 2.

How did you use the structure of the equation in Exercise 85 on page 457 to solve the problem?

Answer:

Question 3.

Describe why your answer makes sense considering the context of the data in Exercise 20 on page 466.

Answer:

**Performance Task: Asteroid Aim**

Apps take a long time to design and program. One app in development is a game in which players shoot lasers at asteroids. They score points based on the number of hits per shot. The designer wants your feedback. Do you think students will like the game and want to play it? What changes would improve it?

To explore the answers to this question and more, go to

Answer:

### Graphing Quadratic Functions Chapter Review

**8.1 Graphing f(x) = ax ^{2} (pp. 419–424)**

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 1.

p(x) = 7x

^{2}

Answer:

The function y = ax

^{2}is derived from the parent function y = x

^{2}by various transformation: vertical stretch (|a| > 1) or contraction (0 < |a| < 1) and reflection about the x-axis.

In the given function q(x) = 7x

^{2}

we have a vertical stretch by a factor of 7 of the parent function y = x

^{2}

Question 2.

q(x) = \(\frac{1}{2}\)x^{2}

Answer:

The function y = ax^{2} is derived from the parent function y = x^{2} by various transformation: vertical stretch (|a| > 1) or contraction (0 < |a| < 1) and reflection about the x-axis.

In the given function q(x) = \(\frac{1}{2}\)x^{2}

we have a vertical stretch by a factor of \(\frac{1}{2}\) of the parent function y = x^{2}

Question 3.

g(x) = – \(\frac{3}{4}\)x^{2}

Answer:

The function y = ax^{2} is derived from the parent function y = x^{2} by various transformation: vertical stretch (|a| > 1) or contraction (0 < |a| < 1) and reflection about the x-axis.

In the given function g(x) = – \(\frac{3}{4}\)x^{2}

we have a vertical stretch by a factor of \(\frac{3}{4}\) of the parent function y = x^{2}

Question 4.

h(x) = -6x^{2}

Answer:

The function y = ax^{2} is derived from the parent function y = x^{2} by various transformation: vertical stretch (|a| > 1) or contraction (0 < |a| < 1) and reflection about the x-axis.

In the given function h(x) = – 6x^{2}

we have a vertical stretch by a factor of 6 of the parent function y = x^{2}

Question 5.

Identify characteristics of the quadratic function and its graph.

Answer:

From the graph, we can see that the quadratic function opens upward and the vertex is (1, -3)

Also, the axis of symmetry is x=1 and points on the graph are (0, -1) and (2, -1)

This graph is a vertical shift of 3 units downward and a horizontal shift of 1 unit to the right.

**8.2 Graphing f(x) = ax ^{2} + c (pp. 425–430)**

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 6.

g(x) = x

^{2}+ 5

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis

In the given function g(x) = x

^{2}+ 5 we have a vertical shift of 5 units upward of the parent function y = x²

Question 7.

h(x) = -x^{2} – 4

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis.

In the given function h(x) = -x^{2} – 4

a. a vertical shift of 4 units downward

b. reflection about the x-axis of the parent function y = x²

Question 8.

m(x) = -2x^{2} + 6

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis.

In the given function m(x) = -2x^{2} + 6

a. a vertical stretch by a factor of 2

b. vertical shift of 6 units upward

c. reflection about the x-axis of the parent function y = x²

Question 9.

n(x) = \(\frac{1}{3}\)x^{2} – 5

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis.

In the given function n(x) = \(\frac{1}{3}\)x^{2} – 5

a. a vertical contraction by a factor of \(\frac{1}{3}\)

b. vertical shift of 5 units downward

c. reflection about the x-axis of the parent function y = x²

**8.3 Graphing f(x) = ax ^{2} + bx + c (pp. 431–438)**

**Graph the function. Describe the domain and range.**

Question 10.

y = x^{2} – 2x + 7

Answer:

The domain of the function is all real numbers.

If we write a graph in the form y = x^{2} – 2x + 7

x^{2} – 2x + 1 + 6

= (x – 1)^{2 }+ 6

(x – 1)^{2 }is always ≥ , so the range is y ≥ 6

Question 11.

f(x) = -3x^{2} + 3x – 4

Answer:

First we will graph the function and find the axis of symmetry

x = \(\frac{-b}{2a}\)

= -3/2(-3) = 1/2 = 0.5

Question 12.

y = \(\frac{1}{2}\)x^{2} – 6x + 10

Answer:

Step 1: Find axis of symmetry:

x = \(\frac{-b}{2a}\)

x = -(-6)/2 . 1/2 = 6

Step 2: Find the plot the vertex.

The x-coordinate of the vertex is 6 so find the y-coordinate by solving for f(6):

f(6) = \(\frac{1}{2}\)(6)² – 6(6) + 10

= -8

Thus the vertex is at (6, -8)

Step 3: Use the y-intercept to find two more points on the graph.

With c = 10, the y-intercept is 10 therefore the parabola passes through the point (0, 10).

The axis of symmetry is x = 6

Therefore the point (12, 10) must also be on the graph.

Step 4: Draw a smooth curve through the points.

The domain is the set of all real numbers.

The range is [-8, +∞)

Question 13.

The function f(t) = -16t^{2} + 88t + 12 represents the height (in feet) of a pumpkin t seconds after it is launched from a catapult. When does the pumpkin reach its maximum height? What is the maximum height of the pumpkin?

Answer:

The trajectory of the pumpkin will be a parabola, so the time required for the pumpkin to reach maximum speed is actually the x-coordinate of the vertex of the parabola and the maximum height is the y-coordinate of the vertex.

f(t) = -16t^{2} + 88t + 12

f(t) = -16(t^{2} + 2 . t . \(\frac{11}{4}\)+ (\(\frac{11}{4}\))² – (\(\frac{11}{4}\))² – \(\frac{3}{4}\))

f(t) = -16((t – \(\frac{11}{4}\))² – \(\frac{133}{16}\))

f(t) = -16(t – \(\frac{11}{4}\))² + 133

f(t) = -16(t – 2.75)² + 133

Pumpkin reaches a maximum speed 2.75 seconds after it is launched and the maximum height is 133 feet.

**8.4 Graphing f(x) = a(x − h) ^{2} + k (pp. 441–448)**

**Determine whether the function is even, odd, or neither.**

Question 14.

w(x) = 5^{x}

Answer:

Function is even if f(-x) = f(x) and function is odd if f(-x) = -f(x)

So we replace x with w(x) with -x

w(-x) = 5^{-x}= \(\frac{1}{5^{x}}\)

notice that w(-x) is not equal to -w(x) = -5^{x} or w(x)

So, the function w(x) is neither odd nor even.

Question 15.

r(x) = -8x

Answer:

Function is even if f(-x) = f(x) and function is odd if f(-x) = -f(x)

So we replace x with r(x) with -x

r(-x) = -8(-x) = 8x

notice that r(-x) is equal to r(x) so function r(x) is odd function.

Question 16.

h(x) = 3x^{2} – 2x

Answer:

Function is even if f(-x) = f(x) and function is odd if f(-x) = -f(x)

So we replace x with h(x) with -x

h(-x) =3(-x)^{2} – 2(-x) = 3x^{2} + 2x

notice that h(-x) is not equal to -h(x) or h(x)

So, the function w(x) is neither odd nor even.

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 17.

h(x) = 2(x – 4)

^{2}

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis

In the given function h(x) = 2(x – 4)

^{2}we have

– a vertical stretch by a factor of 2

– 4 shift unit to the right of the parent function y = x²

Question 18.

g(x) = \(\frac{1}{2}\)(x – 1)^{2} + 1

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis

In the given function g(x) = \(\frac{1}{2}\)(x – 1)^{2} + 1 we have

– a vertical stretch by a factor of \(\frac{1}{2}\)

– 1 unit shift upward of the parent function y = x²

Question 19.

q(x) = -(x + 4)^{2} + 7

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis

In the given function q(x) = -(x + 4)^{2} + 7 we have

– Reflection about the x-axis

– a 4 unit shift to the left

– 7 unit shift upward of the parent function y = x²

Question 20.

Consider the function g(x) = -3(x + 2)^{2} – 4. Graph h(x) = g(x = 1).

Answer:

The function y = a(x-h)² + k has vertex as (h, k) and derived from the parent function y = x² by various transformations: vertical shift of |k| units, horizontal of |h| units, vertical stretch or contraction (0 < |a| < 1) and reflection about the x-axis

In the given function g(x) = -3(x + 2)^{2} – 4we have

– Reflection about the x-axis

– vertical stretch by a factor of 3

– a 2 unit shift to the left

– 4 unit shift downward

Question 21.

Write a quadratic function whose graph has a vertex of (3, 2) and passes through the point (4, 7).

Answer:

The vertex form of the quadratic function is

y = a(x – h)² + k

where (h, k) is the vertex.

It is given that the vertex is (3, 2)

y = a(x – 3)² + 2

As the graph passes through the point (4, 7). We can insert a point in the equation to get a

7 = a(4 – 3)² + 2

7 – 2 = a . 1

a = 5

Therefore the equation that satisfies the given vertex and point is

y = 5(x – 3)² + 2

**8.5 Using Intercept Form (pp. 449–458)**

**Graph the quadratic function. Label the vertex, axis of symmetry, and x-intercepts. Describe the domain and range of the function.**

Question 22.

y = (x – 4)(x + 2)

Answer:

0 = (x – 4) (x + 2)

x – 4 = 0 or x + 2 = 0

x = 4 or x = -2

The y coordinates are y = 0

The x-intercepts are 4 and -2

The axis of symmetry is given by x = (4 + (-2))/2 = 1

So, the vertex is (1, -9)

Also, domain is all real numbers and range, as we can see from is y ≥ -9

Question 23.

f(x) = -3(x + 3)(x + 1)

Answer:

Given,

f(x) = -3(x + 3)(x + 1)

0 = -3(x + 3)(x + 1)

x + 3 = 0 or x + 1 = 0

x = -3 or x = – 1

The y coordinates are y = 0

The x-intercepts are -3 and -1

The axis of symmetry is given by x = (-3 + (-1))/2 = -2

So, the vertex is (-2, 3)

Also, domain is all real numbers and range, as we can see from is y ≤ 3

Question 24.

y = x^{2} – 8x + 15

Answer:

y = x^{2} – 8x + 15

0 = x^{2} – 8x + 15

(x – 5) (x – 3) = 0

x – 5 = 0 or x – 3 = 0

x = 5 or x = 3

The y coordinates are y = 0

The x-intercepts are 5 and 3

The axis of symmetry is given by x = (5 + 3)/2 = 4

So, the vertex is (4, -1)

Also, domain is all real numbers and range, as we can see from is y ≥ -1

Use zeros to graph the function.

Question 25.

y = -2x^{2} + 6x + 8

Answer:

y = -2x^{2} + 6x + 8

0 = -2x^{2} + 6x + 8

0 = -2(x – 4)(x + 1)

x – 4 = 0 or x + 1 = 0

x = 4 or x = -1

The axis of symmetry is given by x = (-1 + 4)/2 = 1.5

Question 26.

f(x) = x^{2} + x – 2

Answer:

f(x) = x^{2} + x – 2

0 = x^{2} + x – 2

0 = (x – 1)(x + 2)

x = 1 or x = -2

The axis of symmetry is given by x = (1 – 2)/2 = -0.5

Question 27.

f(x) = 2x^{2} – 18x

Answer:

Question 28.

Write a quadratic function in standard form whose graph passes through (4, 0) and (6, 0).

Answer:

We can write quadratic equation in the form of a factor product

f(x) = (x – 4)(x – 6)

where (4, 0) and (6, 0) are zeros

So, the equation is f(x) = x² – 10x + 24

**8.6 Comparing Linear, Exponential, and Quadratic Functions (pp. 459−468)**

Question 29.

Tell whether the table of values represents a linear, an exponential, or a quadratic function. Then write the function.

Answer:

We can see that in the consecutive y-values, each subsequent y-value is divided by 4.

So, the table represents an exponential function.

Write the equation using the form f(x) = a . b^{x}, where b is common ratio.

We also can see that the functions y-intercepts is 123,

y = 128 . b^{x},

8 = 128 . b^{2},

b = 1/4

f(x) = 128 . (1/4)^{x},

Question 30.

The balance y (in dollars) of your savings account after t years is represented by y = 200(1.1)t. The beginning balance of your friend’s account is $250, and the balance increases by $20 each year. (a) Compare the account balances by calculating and interpreting the average rates of change from t = 2 to t = 7. (b) Predict which account will have a greater balance after 10 years. Explain.

Answer:

average rates of change from t = 2 to t = 7

t(7) – t(2)/7 – 2 = 29.55

My friend’s balance is given by

y = 20t + 250

t(7) – t(2)/7 – 2 = (250 + 140 – 250 + 40)/5

= 100/5 = 200

My saving account has more money and is growing faster than my friend’s saving account.

### Graphing Quadratic Functions Chapter Test

**Graph the function. Compare the graph to the graph of f(x) = x ^{2}.**

Question 1.

h(x) = 2x

^{2}– 3

Answer:

The function h(x) = 2x² – 3 involves a vertical stretch by a factor of 2 and a 3 unit downward shift of the function f(x) = x

^{2}

Question 2.

g(x) = –\(\frac{1}{2}\)x^{2}

Answer:

The function g(x) = –\(\frac{1}{2}\)x^{2 }involves a reflection about the x-axis and a vertical contraction by a factor of \(\frac{1}{2}\) of the function f(x) = x^{2}

Question 3.

p(x) = \(\frac{1}{2}\)(x + 1)^{2} – 1

Answer:

The function p(x) = \(\frac{1}{2}\)(x + 1)^{2} – 1 involves:

i. a horizontal shift of 1 unit to the left

ii. a vertical contraction by a factor of \(\frac{1}{2}\) and

iii. a 1 unit downward shift

Question 4.

Consider the graph of the function f.

a. Find the domain, range, and zeros of the function.

Answer:

Domain: All real numbers

Range: y ≤ 8

The zeros also the x-intercepts: 3, 7

b. Write the function f in standard form.

Answer:

Use the vertex form:

y = (x-h)² + k

At vertex (h, k) = (5, 8)

y = (x-5)² + 8

Substitute either intercepts to find a:

0 = a(3-5)² + 8

0 = 4a + 8

4a = -8

a = -2

y = -2(x-5)² + 8

y -2x^{2} + 20x – 42

c. Compare the graph of f to the graph of g(x) = x^{2}.

Answer:

Using the vertex form, the graph of f(x) is the graph of g(x) is:

– Reflected in the x-axis

– Vertically stretched by a factor of 2

– Vertically shifted 8 units upward

– Horizontally shifted 5 units to the right.

d. Graph h(x) = f (x – 6).

Answer:

The graph of h(x) is the graph of f(x) that is horizontally shifted 6 units to the right so the intercepts will be (9, 0) and (13, 0)

The vertex will now be at (11, 8)

**Use zeros to graph the function. Describe the domain and range of the function.**

Question 5.

f(x) = 2x^{2} – 8x + 8

Answer:

f(x) = 2x^{2} – 8x + 8

y = 2x^{2} – 8x + 8

y = 2(x-2)(x-2)

y = 2(x-2)^{2}

Thus the x-intercept of the given function is 2 and the axis of symmetry is x = 2.

y = 2(2)^{2} – 8(2) + 8

y = 8 – 16 + 8

y = 0

Thus the vertex of the parabola is at (2, 0)

Question 6.

y = -(x + 5)(x – 1)

Answer:

x = (-5+1)/2 = -2

This means that the vertex of the parabola has an x-coordiate of -2.

y = -(x + 5)(x – 1)

y = -(-2+5)(-2-1)

y = -(3)(-3)

y = 9

Therefore the vertex of the parabola is at (-2, 9)

Question 7.

h(x) = 16x^{2} – 4

Answer:

Given,

h(x) = 16x^{2} – 4

Write the given function in intercept form to have:

h(x) = 16(x^{2} – 1/4)

h(x) = 16(x – 1/2)(x + 1/2)

The axis of symmetry is x = (-0.5+0.5)/2 = 0

This means that the vertex of the parabola has an x-coordiate of 0.

y = 16(0) – 4

y = 0 – 4

y = -4

Therefore the vertex of the parabola is at (0, -4)

Tell whether the table of values represents a linear, an exponential, or a quadratic function. Explain your reasoning. Then write the function.

Question 8.

Answer:

Among the given y-values, which are consecutive, the next y-value can be found by multiplying 2 to the previous y-value.

Therefore the table of values represents an exponential function.

Exponential function can be written as f(x) = a . b^{x}

where b is the common ratio.

Thus the tentative equation of the function that represents the given table of values

f(x) = a . b^{x}

Use the point (0, 8) to find the value of a:

8 = a . 2^{0}

8 = a(1)

8 = a

Therefore the function that represents the given table of values is f(x) = 8 . 2^{x}

Thus is equivalent to the function f(x) = 2³ . 2^{x+3}

Question 9.

Answer:

Since the second difference is constant, then the function is quadratic.

Notice that the y-values are symmetric with respect to the point (0, 0). This means that the vertex of the parabola is (0, 0)

y = a . x^{2}

Use the point (0, 8) to find the value of a:

-2 = a . 1^{2}

-2 = a(1)

-2 = a

Therefore the equation of the quadratic function is y = -2 x^{2}

**Write a quadratic function in standard form whose graph satisfies the given conditions. Explain the process you used.**

Question 10.

passes through (-8, 0), (-2, 0), and (-6, 4)

Answer:

Given the x-intercepts (-8, -2) use the intercept form:

f(x) = a(x – p)(x – q)

Substitute the intercepts and simplify:

f(x) = a(x – (-8))(x – (-2))

f(x) = a(x + 8)(x+2)

Use the other point, (-6, 4) to find the value of a:

4 = a(-6 + 8)(-6 + 2)

4 = -8a

Thus the function

f(x) = -1/2(x + 8)(x + 2)

f(x) = -1/2x² – 5x – 8

Question 11.

passes through (0, 0), (10, 0), and (9, -27)

Answer:

Given the x-intercepts (0, 10) use the intercept form

f(x) = a(x – p)(x – q)

f(x) = a(x – 0)(x – 10)

f(x) = a(x)(x – 10)

Use the other point, (9, -27) to find the value of a:

-27 = a(9)(9 – 10)

-27= -9a

a = 3

Therefore the function is

f(x) = 3(x)(x – 10)

f(x) = 3x² – 30x

Question 12.

is even and has a range of y ≥ 3

Answer:

The quadratic is even if it is symmetric about the y-axis so the symmetry is the y-axis.

The range implies that the parabola opens upward and the vertex is at (0, 3)

Use the vertex form:

y = a(x – h)² + k

At vertex (h, k) = (0, 3)

y = a(x – 0)² + 3

Let a = 1

y = (1)(x – 0)² + 3

y = x² + 3

Question 13.

passes through (4, 0) and (1, 9)

Answer:

Let the parabola be in the form

y = ax² + bx

Set up 2 equations using the points

At(4, 0): 0 = 16a + 4b

At (1, 9): 9 = a + b

b = 9 – a

0 = 16a + 4(9 – a)

0 = 16a + 36 – 4a

-12a = 36

a = -3

b = 9 – (-3)

y = -3x² + 12x

Question 14.

The table shows the distances d (in miles) that Earth moves in its orbit around the Sun after t seconds. Let the time t be the independent variable. Tell whether the data can be modeled by a linear, an exponential, or a quadratic function. Explain. Then write a function that models the data.

Answer:

The first differences are constant at 19 so the function is linear.

Use any 2 points,

(t1, d1) → (1, 19)

(t2, d2) → (2, 38)

Use the point-slope formula:

y – y1 = m(x – x1)

t – t1 = m(t – t1)

m = (d2-d1)/(t2-t1) = (38-19)/2-1 = 19

d – 19 = 19(t – 1)

d – 19 = 19t – 19

d = 19t

Question 15.

You are playing tennis with a friend. The path of the tennis ball after you return a serve can be modeled by the function y = -0.005x^{2} + 0.17x + 3, where x is the horizontal distance (in feet) from where you hit the ball and y is the height (in feet) of the ball.

a. What is the maximum height of the tennis ball?

Answer:

The maximum height is the y-coordinate of the vertex.

x = -b/2a

where a = -0.005 and b = 0.17

x = –\(\frac{0.07}{2(-0.005)}\) = 17

The y-coordinate is

y = -0.005(17)² + 0.17(17) + 3

y = 4.445

b. You are standing 30 feet from the net, which is 3 feet high. Will the ball clear the net? Explain your reasoning.

Answer:

The ball will clear the net if at x = 30, the height, y, is greater than 3 feet.

y = 0.005(30)² + 0.17(30) + 3

y = 3.6 ft

Since 3.6ft > 3 ft, then the ball will clear the net.

Question 16.

Find values of a, b, and c so that the function f(x) = ax^{2} + bx + c is (a) even, (b) odd, and (c) neither even nor odd.

Answer:

a. The function is even is it satisfies:

f(-x) = f(x)

a(-x)² + b(-x) + c = ax^{2} + bx + c

ax^{2} – bx + c = ax^{2} + bx + c

For this equation to be true, we can let b = 0 so that a and c any value.

A possible answer is a = 1, b = 0, c = 2

b.

The function is odd if it satisfies:

f(-x) = -f(x)

a(-x)² + b(-x) + c = -(ax^{2} + bx + c)

ax^{2} – bx + c = -ax^{2} – bx – c

For this equation to be true, we can let a = c = 0 so that b can be any value.

A possible answer is a = 0, b = 1, c = 0

c.

The function is neither odd nor even when a, b and c are non-zeros

A possible answer is: a = 1, b = 2, c = 3

Question 17.

Consider the function f(x) = x^{2} + 4. Find the average rate of change from x = 0 to x = 1, from x = 1 to x = 2, and from x = 2 to x = 3. What do you notice about the average rates of change when the function is increasing?

Answer:

Construct a table from x = 0 and x = 3

Compare the rate of change using the average rate of change

= f(b) – f(a)/b-a

From t = 0 to t = 1:

(5-4)/1-0 = 1

From t = 1 to t = 2:

(8-5)/2-1 = 3

From t = 2 to t = 3:

(13-8)/3-2 = 5

As the function is increasing, the rate of change also increases.

### Graphing Quadratic Functions Cumulative Assessment

Question 1.

Which function is represented by the graph?

Answer:

In the graph, we can see that vertex is (0, 0).

Thus, parabola is of the form f(x) =ax^{2}

Since (2, -2) lie on the graph,

f(2) = -2

ax^{2 }= -2

a = -1/2

Thus the correct answer is option C.

Question 2.

Find all numbers between 0 and 100 that are in the range of the function defined below.(HSF-IF.A.3)

f(1) = 1, f(2) = 1, f(n) = f(n – 1) + f(n – 2)

Answer:

The recursive rule is the Fibonacci sequence where each term is the sum of the previous terms.

Use the recursive rule until the term is greater than 100 where it is not included:

f(1) = 1

f(2) = 1

f(3) = f(2) + f(1) = 1 + 1 = 2

f(4) = f(3) + f(2) = 2 + 1 = 3

f(5) = f(4) + f(3) = 3 + 2 = 5

f(6) = f(5) + f(4) = 5 + 3 = 8

f(7) = f(6) + f(5) = 8 + 5 = 13

f(8) = f(7) + f(6) = 13 + 8 = 21

f(9) = f(8) + f(7) = 21 + 13 = 34

f(10) = f(9) + f(8) = 34 + 21 = 55

f(11) = f(10) + f(9) = 55 + 34 = 89

f(12) = f(11) + f(10) = 89 + 55 = 144

Therefore, the numbers that are in between 0 and 100 are: 1, 2, 3, 5, 8, 13, 21, 34, 55 and 89.

Question 3.

The function f(t) = -16t^{2} + v_{0}t + s_{0} represents the height (in feet) of a ball t seconds after it is thrown from an initial height s0 (in feet) with an initial vertical velocity v0 (in feet per second). The ball reaches its maximum height after \(\frac{7}{8}\) second when it is thrown with an initial vertical velocity of ______ feet per second.

Answer:

The maximum height corresponds to the vertex where the x-coordinate is x = -b/2a

t = – v_{0}/2(-16)

7/8 = v_{0}/32

v_{0 }= 28

Thus the initial velocity is 28ft/s.

Question 4.

Classify each system of equations by the number of solutions.

Answer:

Convert each equation to slope-intercept form then compare the slopes and intercepts to determine the number of solutions.

System A:

The slopes are negative reciprocal so they are perpendicular.

Therefore there is 1 solution.

The system is consistent.

System B:

y = –\(\frac{7}{4}\)x + 3

y = \(\frac{7}{8}\)x + \(\frac{3}{2}\)

The slopes are different so they intersect. Therefore there is 1 solution.

The system is consistent.

System C:

y = –\(\frac{1}{2}\)x – \(\frac{1}{2}\)

y = –\(\frac{5}{2}\)x – \(\frac{1}{2}\)

The slopes are different so they intersect. Therefore there is 1 solution.

The system is consistent.

System D:

y = -3x + 5

y = -3x + 5

The 2 equations are the same line. Therefore, there are infinitely many solutions.

The system is consistent.

System E:

y = 2x + 3/2

y = 2x + 3/2

The 2 equations are the same line. Therefore, there are infinitely many solutions.

The system is consistent.

System F:

The equations have the same slope. Therefore, there is no solution.

The system is inconsistent.

Question 5.

Your friend claims that quadratic functions can have two, one, or no real zeros. Do you support your friend’s claim? Use graphs to justify your answer.

Answer:

Yes, he is correct.

There are 2 real zeros when the graph crosses the x-axis. There is only 1 real zero when the graph touches the x-axis.

There is no real zero when the graph does not cross/touch the x-axis.

Question 6.

Which polynomial represents the area (in square feet) of the shaded region of the figure?

Answer:

Each side of the shaded area is x – a.

A = (x – a)²

Using square of binomial difference

A = x² – 2ax + a²

Thus the correct answer is option C.

Question 7.

Consider the functions represented by the tables.

a. Classify each function as linear, exponential, or quadratic.

Answer:

For p(x):

The first differences are: -12, -12, -12

Since the first differences are constant, then it is linear.

For r(x):

The first differences are: 15, 25, 35

The second differences are: 10, 10

Since the second differences are constant, then it is quadratic.

For s(x):

There is a common ratio which is 1/2

Thus, the function is exponential.

For t(x):

The first differences are: -8, -16, -24

The second differences are: -8, -8

Since the second differences are constant, then it is quadratic.

b. Order the functions from least to greatest according to the average rates of change between x = 1 and x = 3.

Answer:

Compare the rate of change using average rate of change = f(b) – f(a)/b-a

p(x) = -40-(-16)/3-1 = -12

r(x) = 40-0/3-1 = 40/2 = 20

s(x) = 18-(72)/3-1 = -27

t(x) = -5-3/3-1 = -4

Compare the absolute values of the rate of change, the order from least to greatest is: t(x), p(x), r(x), s(x)

Question 8.

Complete each function using the symbols + or – , so that the graph of the quadratic function satisfies the given conditions.

Answer:

a. Use the vertex form:

y = a(x-h)² + k

where (h, k) is the verrtex

f(x) = 5(x-(-3))² + 4

f(x) = 5(x + 3)² + 4

b. Use the intercept form:

y = a(x-p)(x-q)

g(x) = -(x-2)(x-(-8))

g(x) = -(x-2)(x+8)

c. The range implies that the parabola opens upward and the y-coordinate is -6

h(x) = 3x² – 6

d. The range implies that the parabola opens downward and the y-coordinate is 4.

Therefore, the 2 factors will have opposite signs:

j(x) = -4(x+1)(x-1)

Question 9.

The graph shows the amounts y (in dollars) that a referee earns for refereeing x high school volleyball games.

a. Does the graph represent a linear or nonlinear function? Explain.

Answer: The graph represents a linear function.

b. Describe the domain of the function. Is the domain discrete or continuous?

Answer: The domain is the set of natural numbers in context with the given problem and is discrete as the number of games must be whole numbers.

c. Write a function that models the data.

Answer:

(x1, y1) → (2, 90)

(x2, y2) → (4, 180)

y – y1 = m(x – x1)

m = 45

substitute:

y – 90 = 45(x – 2)

y – 90 = 45x – 90

y = 45x

d. Can the referee earn exactly $500? Explain.

Answer:

No, because 500 not a multiple of 45 as shown above.

Question 10.

Which expressions are equivalent to (b^{-5})^{-4}?

Answer: