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Chapter Three

QUADRATIC FUNCTIONS Contents

3.1 Introduction to the Family of Quadratic

Functions . . . . . . . . . . . . . . . . . 116

Finding the Zeros of a Quadratic Function . . 116

Concavity and Rates of Change for Quadratic

Functions . . . . . . . . . . . . . 117

Finding a Formula From the Zeros

and Vertical Intercept . . . 119

Formulas for Quadratic Functions . . . . . . 120

3.2 The Vertex of a Parabola . . . . . . . . . . 122

The Vertex Form of a Quadratic Function . . 123

Finding a Formula Given the Vertex

and Another Point . . . . . 125

Modeling with Quadratic Functions . . . . . 126

REVIEW PROBLEMS . . . . . . . . . . . 129

STRENGTHEN YOUR UNDERSTANDING 130

Skills Refresher for CHAPTER 3:

QUADRATIC EQUATIONS . . . . . . . . 131

Skills for Factoring . . . . . . . . . . . . . 131

Expanding an Expression . . . . . . . . . . 131

Factoring . . . . . . . . . . . . . . . . . 131

Removing a Common Factor . . . . 131

Grouping Terms . . . . . . . . . . 132

Factoring Quadratics . . . . . . . . 132

Perfect Squares and the Difference of

Squares . . . . . . . . . . 132

Solving Quadratic Equations . . . . . . . . 133

Solving by Factoring . . . . . . . . 133

Solving with the Quadratic Formula . 134

Completing the Square . . . . . . . . . . . 136

Visualizing Completing the Square . . 136

Deriving the Quadratic Formula . . . . . . . 137

116 Chapter Three QUADRATIC FUNCTIONS

3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS

A baseball is “popped” straight up by a batter. The height of the ball above the ground is given by

the function y = f(t) = −16t2 + 47t + 3, where t is time in seconds after the ball leaves the bat

and y is in feet. See Figure 3.1. Note that the path of the ball is straight up and down, although the

graph of height against time is a curve. The ball goes up fast at first and then more slowly because

of gravity; thus the graph of its height as a function of time is concave down.

3

3

y = f(t) = −16t2 + 47t+ 3

❘

Initialheight

t (seconds)

y (feet)

Figure 3.1: The height of a ball t seconds after being “popped up”. (Note: This

graph does not represent the ball’s path.)

The baseball height function is an example of a quadratic function whose standard form is

y = ax2 + bx+ c.

The graph of a quadratic function is called a parabola. Notice that the function in Figure 3.1 is con-

cave down and has a maximum corresponding to the time at which the ball stops rising and begins

to fall back to the earth. The maximum point on the parabola is called the vertex. The intersection of

the parabola with the horizontal axis gives the times when the height of the ball is zero; these times

are the zeros of the height function. The horizontal intercepts of a graph occur at the zeros of the

function. In this section we examine the zeros and concavity of quadratic functions, and in the next

section we see how to find the vertex.

Finding the Zeros of a Quadratic Function

A natural question to ask is when the ball hits the ground. The graph suggests that y = 0 when t is

approximately 3. In symbols, the question is: For what value(s) of t does f(t) = 0? These are the

zeros of the function.

It is easy to find the zeros of a quadratic function if it is expressed in factored form,

q(x) = a(x− r)(x− s), where a, r, s are constants, a 6= 0.

Then r and s are zeros of the function q.

The baseball function factors as

y = −1(16t+ 1)(t− 3),

so the zeros of f are t = 3 and t = −1/16. We are interested in positive values of t, so the ball hits

the ground 3 seconds after it was hit. (For more on factoring, see the Skills Review on page 131.)

Example 1 Find the zeros of f(x) = x2 − x− 6.

Solution To find the zeros, set f(x) = 0 and solve for x by factoring:

x2 − x− 6 = 0

(x− 3)(x+ 2) = 0.

Thus the zeros are x = 3 and x = −2.

3.1 INTRODUCTION TO THE FAMILY OF QUADRATIC FUNCTIONS 117

We can also find the zeros of a quadratic function by using the quadratic formula. (See the

Skills Review on page 131 to review the quadratic formula.)

Example 2 Find the zeros of f(x) = x2 − x− 6 by using the quadratic formula.

Solution We solve the equation x2 − x− 6 = 0. For this equation, a = 1, b = −1, and c = −6. Thus

x =−b±

√b2 − 4ac

2a=

−(−1)±√

(−1)2 − 4(1)(−6)

2(1)

=1±

√25

2= 3 or − 2.

The zeros are x = 3 and x = −2, the same as we found by factoring.

Since the zeros of a function occur at the horizontal intercepts of its graph, quadratic functions

without horizontal intercepts (such as in the next example) have no zeros.

Example 3 Figure 3.2 shows a graph of h(x) = −1

2x2 − 2. What happens if we try to use algebra to find the

zeros of h?

✛ yintercept(0,−2)

h(x)

x

y

Figure 3.2: Zeros of h(x) = −(x2/2)− 2?

Solution To find the zeros, we solve the equation

−1

2x2 − 2 = 0

−1

2x2 = 2

x2 = −4

x = ±√−4.

Since√−4 is not a real number, there are no real solutions, so h has no real zeros. This corresponds

to the fact that the graph of h in Figure 3.2 does not cross the x-axis.

Concavity and Rates of Change for Quadratic Functions

A quadratic function has a graph that is either concave up or concave down. Recall that for a graph

that is concave up, rates of change increase as we move right. For a graph that is concave down,

rates of change decrease as we move right.

Example 4 Let f(x) = x2. Find the average rate of change of f over the four consecutive intervals of length 2

between x = −4 and x = 4. What do these rates tell you about the concavity of the graph of f?


Solution Between x = 0 and x = 2, we have

Average rate of change

of f=

f(2)− f(0)

2− 0=

22 − 02

2− 0= 2.

Between x = 2 and x = 4, we have

Average rate of change

of f=

f(4)− f(2)

4− 2=

42 − 22

4− 2= 6.

Similarly, between x = −4 and x = −2, we can show the average rate of change is −6; between

x = −2 and x = 0, the rate of change is −2. See Figure 3.3. Since the rates of change are increasing,

the graph is concave up.

−4 −2 2 4

❄

Slope = −6

❄

Slope = −2

❄

Slope = 2

❄

Slope = 6

f(x) = x2

x

Figure 3.3: Rate of change and concavity of f(x) = x2

Example 5 A high-diver jumps off a 10-meter platform. For t in seconds after the diver leaves the platform until

she hits the water, her height h in meters above the water is given by

h = f(t) = −4.9t2 + 8t+ 10.

The graph of this function is shown in Figure 3.4.

(a) Estimate and interpret the domain and range of the function, and the intercepts of the graph.

(b) Identify the concavity.

1 2 3

5

10f(t)

15

t (seconds)

h (meters)

Figure 3.4: Height of diver above water as a function of time

Solution (a) The diver enters the water when her height above the water is 0. This occurs when

h = f(t) = −4.9t2 + 8t+ 10 = 0.

Using the quadratic formula to solve this equation, we find the only positive solution is t =2.462 seconds. The domain is the interval of time the diver is in the air, which is approximately

0 ≤ t ≤ 2.462. To find the range of f , we look for the largest and smallest outputs for h. From

the graph, the diver’s maximum height appears to occur at about t = 1, so we estimate the

largest output value for f to be about

f(1) = −4.9 · 12 + 8 · 1 + 10 = 13.1 meters.

Thus, the range of f is approximately 0 ≤ f(t) ≤ 13.1. Methods to find the exact maximum

height are in Section 3.2.


The vertical intercept of the graph is

f(0) = −4.9 · 02 + 8 · 0 + 10 = 10 meters.

The diver’s initial height is 10 meters (the height of the diving platform). The horizontal inter-

cept is the point estimated earlier where f(t) = 0. This intercept, t = 2.462 seconds, gives the

time after leaving the platform that the diver enters the water.

(b) In Figure 3.4, we see that the graph is bending downward over its entire domain, so it is concave

down. This is reflected in Table 3.1, where the rate of change, ∆h/∆t, is decreasing.

Table 3.1 Slope of f(t) = −4.9t2 + 8t+ 10

t (sec) h (meters)Rate of change

∆h/∆t

0 10

5.55

0.5 12.775

0.65

1.0 13.100

−4.25

1.5 10.975

−9.15

2.0 6.400

The previous two examples illustrate that the concavity of the graph of y = ax2 + bx + c is

determined by the sign of the coefficient a:

• If a > 0, the parabola is concave up.

• If a < 0, the parabola is concave down.

Example 6 Determine the concavity of the graphs of the following quadratic functions:

(a) y = −3x2 − 3x+ 2(b) y = −3(−2x+ 1)(x− 4)

Solution (a) Since the coefficient of x2 is negative, this function has a graph which is concave down.

(b) Expanding we have

y = −3(−2x+ 1)(x− 4) = −3(−2x2 + 8x+ x− 4) = 6x2 − 27x+ 12.

Since the coefficient of x2 is positive, the graph of this function is concave up.

Finding a Formula From the Zeros and Vertical Intercept

If we know the zeros and the vertical intercept of a quadratic function, we can use the factored form

to find a formula for the function.

Example 7 Find the equation of the parabola in Figure 3.5 using the factored form.

1 3

6

x

f(x)

Figure 3.5: Finding a formula for a quadratic from the zeros


Solution Since the parabola has x-intercepts at x = 1 and x = 3, its formula can be written as

f(x) = a(x− 1)(x− 3).

To find a, we substitute x = 0, y = 6, giving

6 = a · 3a = 2.

Thus, the formula is

f(x) = 2(x− 1)(x− 3).

Multiplying out gives f(x) = 2x2 − 8x+ 6.

Formulas for Quadratic Functions

The function f in Example 7 can be written in at least two different ways. The standard form

f(x) = 2x2 − 8x+ 6 shows that the parabola opens upward, since the coefficient of x2 is positive;

the constant 6 gives the vertical intercept. The factored form f(x) = 2(x − 1)(x − 3) shows that

the parabola crosses the x-axis at x = 1 and x = 3. In general, we have the following:

The graph of a quadratic function is a parabola.

The standard form for a quadratic function is

y = ax2 + bx+ c, where a, b, c are constants, a 6= 0.

The parabola is concave up (opens upward) if a > 0 or concave down (opens downward)

if a < 0, and intersects the y-axis at c.

The factored form, when it exists, is

y = a(x− r)(x− s), where a, r, s are constants, a 6= 0.

The parabola intersects the x-axis at x = r and x = s.

In the next section we look at another form for quadratic functions which shows how to find the

vertex of the graph.

Exercises and Problems for Section 3.1

Skill Refresher

Multiply and write the expressions in Exercises S1–S2 with-

out parentheses. Gather like terms.

S1.(t2 + 1

)50t−

(25t2 + 125

)2t

S2.(A2 −B2

)2

For Exercises S3–S8, factor completely if possible.

S3. u2 − 2u S4. x2 + 3x+ 2

S5. 3x2 − x− 4 S6. (s+ 2t)2 − 4p2

S7. 16x2 − 1 S8. y3 − y2 − 12y

Solve the equations in Exercises S9–S10.

S9. x2 + 7x+ 6 = 0 S10. 2w2 + w − 10 = 0


Exercises

Are the functions in Exercises 1–8 quadratic? If so, write the

function in the form f(x) = ax2 + bx+ c.

1. g(t) = 3(t− 2)2 + 7

2. f(x) = (2x− 3)(5− x)

3. w(n) = n(n− 3)(n− 2)− n2(n− 8)

4. h(t) = −16(t− 3)(t+ 1)

5. R(q) =1

q2(q2 + 1)2

6. K(x) = 132 + 13x

7. T (n) =√5 +

√3n4 −

√

n4

4

8. r(v) =v2 +

√2

3+

v − 3

5+ πv2

9. Find the zeros of Q(r) = 2r2 − 6r − 36 by factoring.

10. Find the zeros of Q(x) = 5x−x2+3 using the quadratic

formula.

11. Solve for x using the quadratic formula and demonstrate

your solution graphically:

(a) 6x− 1

3= 3x2 (b) 2x2 + 7.2 = 5.1x

In Exercises 12–19, find the zeros (if any) of the function al-

gebraically.

12. y = (2− x)(3− 2x) 13. y = 9x2 + 6x+ 1

14. y = 6x2 − 17x+ 12 15. N(t) = t2 − 7t+ 10

16. y = 89x2 + 55x+ 34 17. Q(r) = 2r2 − 6r − 36

18. y = x4 + 5x2 + 6 19. y = x−√x− 12

In Exercises 20–23, is the graph of the quadratic function con-

cave up or concave down?

20. y = 2x− 3x2 + 1 21. y = 5(x− 3)2

22. y = 3(1− 2x)(x− 4) 23. y = −2(5x−1)(4−x)

Problems

24. Use the quadratic formula to find the time at which the

baseball in Figure 3.1 on page 116 hits the ground.

In Problems 25–26, find a formula for the quadratic function

whose graph has the given properties.

25. A y-intercept of y = 7 and the only zero at x = −2.

26. A y-intercept of y = 7 and x-intercepts at x = 1, 4.

27. Is there a quadratic function with zeros x = 1, x = 2and x = 3?

28. Graph a quadratic function which has all the follow-

ing properties: concave up, y-intercept is −6, zeros at

x = −2 and x = 3.

29. Can you graph a quadratic function which has all the fol-

lowing properties: concave down, y-intercept is −6, ze-

ros at x = −2 and x = 3?

30. Determine the concavity of the graph of f(x) = 4 − x2

between x = −1 and x = 5 by calculating average rates

of change over intervals of length 2.

31. The graph of a quadratic function, f(x), passes through

the points (1, 2), (3, 4) and (5, 2). Is the graph of f(x)concave up or concave down? Give a reason for your an-

swer.

32. The quadratic function, f(x), has no zeros and its graph

passes through the point (1, 1). Is the graph of f(x) con-

cave up or concave down? Give a reason for your answer.

33. Without a calculator, graph the following function by fac-

toring and plotting zeros:

y = −4cx+ x2 + 4c2 for c > 0.

34. Without a calculator, graph y = 3x2 − 16x− 12 by fac-

toring and plotting zeros.

In Problems 35–36, find a formula for the parabola.

35.

(−1, 0)

(0,−1)

(3, 0)x

y 36.

(−6, 0)

(0, 5)

(2, 0)x

y

37. Using the factored form, find the formula for the parabola

whose zeros are x = −1 and x = 5, and which passes

through the point (−2, 6).

38. Find two different quadratic functions with zeros x = 1,

x = 2.


39. Write y = 3(0.5x− 4)(4− 20x) in the form

y = k(x− r)(x− s) and give the values of k, r, s.

40. A ball is thrown into the air. Its height (in feet) t seconds

later is given by h(t) = 80t− 16t2.

(a) Evaluate and interpret h(2).(b) Solve the equation h(t) = 80. Interpret your solu-

tions and illustrate them on a graph of h(t).

41. A snowboarder slides up from the bottom of a half-pipe

and comes down again, sliding with little resistance on

the snow. Her height above the top edge of the pipe t sec-

onds after starting up the side is −4.9t2+14t−5 meters.

(a) What is her height at t = 0?

(b) After how many seconds does she reach the top

edge? Return to the edge of the pipe?

(c) How long is she in the air?

42. Let V (t) = t2 − 4t + 4 represent the velocity after tseconds of an object in meters per second.

(a) What is the object’s initial velocity?

(b) When is the object not moving?

(c) Identify the concavity of the velocity graph.

43. Table 3.2 gives the approximate number of cell phone

subscribers, S, in millions, in the US.1 If x is in years

since 2005, show that this data can be approximated by

the quadratic function p(x) = −1.9x2 +24.4x+208.8.

What does this model predict for the year 2015? How

good is this model for predicting the future?

Table 3.2

Year 2005 2006 2007 2008 2009

S (millions) 207.9 233.0 250.0 262.7 276.6

44. Let f(x) = x2 and g(x) = x2 + 2x− 8.

(a) Graph f and g in the window −10 ≤ x ≤ 10,

−10 ≤ y ≤ 10. How are the two graphs similar?

How are they different?

(b) Graph f and g in the window −10 ≤ x ≤ 10,

−10 ≤ y ≤ 100. Why do the two graphs appear

more similar on this window than on the window

from part (a)?

(c) Graph f and g in the window −20 ≤ x ≤ 20,

−10 ≤ y ≤ 400, the window −50 ≤ x ≤ 50,

−10 ≤ y ≤ 2500, and the window −500 ≤ x ≤500, −2500 ≤ y ≤ 250,000. Describe the change

in appearance of f and g on these three successive

windows.

45. A relief package is dropped from a moving airplane.

The package has an initial forward horizontal velocity

and follows a quadratic graph path (instead of dropping

straight down). Figure 3.6 shows the height of the pack-

age, h, in km, as a function of the horizontal distance, d,

in meters, as it drops.

(a) From what height was the package released?

(b) How far away from the spot above which it was re-

leased does the package hit the ground?

(c) Write a formula for h(d). [Hint: The package starts

falling at the highest point on the graph.]

4430

5

4

3

2

1

Airplane →

d, horizontaldistance (meters)

h, height (km)

■Packagedroppedhere

Figure 3.6

3.2 THE VERTEX OF A PARABOLA

In Section 3.1, we looked at the example of a baseball popped upward by a batter. The height of the

ball above the ground is given by the quadratic function y = f(t) = −16t2 + 47t + 3, where t is

time in seconds after the ball leaves the bat, and y is in feet. See Figure 3.7.

The point on the graph with the largest y value appears to be approximately (1.5, 37.5). (See

Example 5 on page 126 for the exact values.) This means that the baseball reaches its maximum

height of about 37.5 feet about 1.5 seconds after being hit. The maximum point on the parabola

is called the vertex. For quadratic functions the vertex shows where the function reaches either its

maximum value or, in the case of a concave-up parabola, its minimum value.

The vertex of a parabola can be determined exactly if the quadratic function is written in the

1http://www.infoplease.com/ipa/A0933563.html. Last accessed January 13, 2014.

3.2 THE VERTEX OF A PARABOLA 123

3

3

40 vertex ≈ (1.5, 37.5)y = f(t) = −16t2 + 47t+ 3

❘

Initialheight

t (seconds)

y (feet)

Figure 3.7: Height of baseball at time t

form

y = a(x− h)2 + k.

This form arises when we shift the parabola y = x2 horizontally by h and vertically by k, moving

the vertex from (0, 0) to (h, k).

Example 1 (a) Sketch f(x) = (x+ 3)2 − 4, and indicate the vertex.

(b) Estimate the coordinates of the vertex from the graph.

(c) Explain how the formula for f can be used to obtain the minimum of f .

(d) Explain how vertical and horizontal shifts can be used to obtain the coordinates of the vertex.

Solution (a) Figure 3.8 shows a sketch of f(x); the vertex gives the minimum value of the function.

(b) The vertex of f appears to be about at the point (−3,−4).(c) Note that (x + 3)2 is always positive or zero, so (x + 3)2 takes on its smallest value when

x+ 3 = 0, that is, at x = −3. At this point (x+ 3)2 − 4 takes on its smallest value,

f(−3) = (−3 + 3)2 − 4 = 0− 4 = −4.

Thus, we see that the vertex is exactly at the point (−3,−4) and the minimum is −4.

(d) The function f(x) = (x+3)2− 4 is a transformation of the function f(x) = x2, with an inside

change of −3 and an outside change of −4. This indicates that f(x) = x2 has been shifted 3units to the left and 4 units down. Thus, the vertex is (−3,−4).

−7

−3

1−4

12

x

f(x)

Figure 3.8: A graph of f(x) = (x+ 3)2 − 4

Notice that if we select x-values that are equally spaced to the left and the right of the vertex, the

y-values of the function are equal. For example, f(−2) = f(−4) = −3. The graph is symmetric

about a vertical line that passes through the vertex. This line is called the axis of symmetry. The

function in Example 1 has axis of symmetry x = −3.

The Vertex Form of a Quadratic Function

Writing the function f in Example 1 in the form

f(x) = (x+ 3)2 − 4

enabled us to find the vertex of the graph and the location and value for the minimum of the function.


In general, we have the following:

The vertex form of a quadratic function is

y = a(x− h)2 + k, where a, h, k are constants, a 6= 0.

The graph of this quadratic function has vertex (h, k) and axis of symmetry x = h.

A quadratic function can always be expressed in both standard form and vertex form.

Example 2 (a) Convert the quadratic f(x) = (x+ 3)2 − 4 in Example 1 to standard form.

(b) Explain how the vertical intercept can be found from the standard form.

(c) Find the zeros of f .

(d) Explain how the axis of symmetry of the parabola is related to the zeros.

Solution (a) Expanding (x+ 3)2 and gathering like terms converts f to standard form:

f(x) = x2 + 6x+ 9− 4 = x2 + 6x+ 5.

Since the coefficient of x2 is a = 1, the parabola opens upward.

(b) The vertical intercept is f(0) = 5. In the standard form y = ax2 + bx+ c, the y-intercept is the

constant c.(c) To find the zeros, we factor f and write it as

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

This form shows that the zeros are x = −1 and x = −5.

(d) The axis of symmetry is x = −3, which is the vertical line through the midpoint of the zeros:

(−1− 5)/2 = −3.

As we see in Example 2, if a quadratic can be written in factored form, its axis of symmetry is

at the midpoint of the zeros. To convert from vertex to standard form, we expand the squared term

and gather like terms. In the next example, we convert from from standard form to vertex form by

completing the square.2

Example 3 Put each quadratic function into vertex form by completing the square and then graph it.

(a) s(x) = x2 − 6x+ 8 (b) t(x) = −4x2 − 12x− 8

Solution (a) To complete the square, find the square of half of the coefficient of the x-term, (−6/2)2 = 9.

Add and subtract this number after the x-term:

s(x) = x2 − 6x+ 9︸︷︷︸

Perfect square

−9 + 8,

so

s(x) = (x− 3)2 − 1.

The vertex of s is (3,−1) and the axis of symmetry is the vertical line x = 3. The parabola

opens upward. See Figure 3.9.

2A more detailed explanation of this method is in the Skills Review on page 136.


−3 6

−3

3

x = 3

x

Figure 3.9: s(x) = x2 − 6x+ 8

−3 6

−3

3

x = −3/2

x

Figure 3.10: t(x) = −4x2−12x−8

(b) To complete the square, first factor out −4, the coefficient of x2, giving

t(x) = −4(x2 + 3x+ 2).

Now add and subtract the square of half the coefficient of the x-term, (3/2)2 = 9/4, inside the

parentheses. This gives

t(x) = −4

(

x2 + 3x+9

4︸︷︷︸

Perfect square

−9

4+ 2

)

t(x) = −4

((

x+3

2

)2

− 1

4

)

t(x) = −4

(

x+3

2

)2

+ 1.

The vertex of t is (−3/2, 1), the axis of symmetry is x = −3/2, and the parabola opens down-

ward. See Figure 3.10.

Finding a Formula Given the Vertex and Another Point

If we know the vertex of a quadratic function and one other point, we can use the vertex form to

find its formula.

Example 4 Find the formula for the quadratic function graphed in Figure 3.11.

(−3, 2)

(0, 5)

x = −3

m(x)

x

y

Figure 3.11: Finding a formula for a quadratic from the vertex

Solution Since the vertex is given, we use the form m(x) = a(x− h)2 + k to find a, h, and k. The vertex is

(−3, 2), so h = −3 and k = 2. Thus,

m(x) = a(x− (−3))2 + 2,

so

m(x) = a(x+ 3)2 + 2.


To find a, use the y-intercept (0, 5). Substitute x = 0 and y = m(0) = 5 into the formula for m(x)and solve for a:

5 = a(0 + 3)2 + 2

3 = 9a

a =1

3.

Thus, the formula is

m(x) =1

3(x+ 3)2 + 2.

If we want the formula in standard form, we multiply out:

m(x) =1

3x2 + 2x+ 5.

Modeling with Quadratic Functions

In applications, it is often useful to find the maximum or minimum value of a quadratic function.

The next example returns to the baseball that started this chapter. Problem 25 on page 128 asks you

to derive the vertex form of the baseball height function; here we see what this form can tell us.

Example 5 For t in seconds, the height of a baseball in feet is given both in standard and in vertex form by

y = f(t) = −16t2 + 47t+ 3 = −16

(

t− 47

32

)2

+2401

64.

Find the maximum height reached by the baseball and the time at which that height is reached.

Solution From the vertex form, we see the vertex is at the point(47

32, 2401

64

)= (1.469, 37.516). This means

that the ball reaches it maximum height of 37.516 feet at t = 1.469 seconds. Note that these values

are close to the graphical estimates (1.5, 37.5) made at the beginning of this section.

Example 6 A city decides to make a park by fencing off a section of riverfront property. Funds are allotted for

80 meters of fence. The area enclosed will be a rectangle, but only three sides will be enclosed by a

fence—the other side will be bounded by the river. What is the maximum area that can be enclosed?

Solution Two sides are perpendicular to the bank of the river and have equal length, which we call h. The

other side is parallel to the bank of the river; its length is b. See Figure 3.12. The area, A, of the park

is the product of the lengths of adjacent sides, so A = bh.

Since the fence is 80 meters long, we have

2h+ b = 80

b = 80− 2h.

h h

b

River

Figure 3.12: A park next to a river


Thus,

A = bh = (80− 2h)h

A = −2h2 + 80h.

The function A = −2h2 + 80h is quadratic. Since the coefficient of h2 is negative, the parabola

opens downward and we have a maximum at the vertex. The factored form of the quadratic is

A = −2h(h− 40), so the zeros are h = 0 and h = 40. The vertex of the parabola occurs on its axis

of symmetry, midway between the zeros, at h = 20. Substituting h = 20 gives the maximum area:

A = −2 · 20(20− 40) = −40(−20) = 800 meters2.

Exercises and Problems for Section 3.2

Skill Refresher

For Exercises S1–S4, complete the square for each expression.

S1. y2 − 12y S2. s2 + 6s− 8

S3. c2 + 3c− 7 S4. 4s2 + s+ 2

In Exercises S5–S7, solve by completing the square.

S5. r2 − 6r + 8 = 0 S6. n2 = 3n+ 18

S7. 5q2 − 8 = 2q

In Exercises S8–S10, solve using factoring, completing the

square, or the quadratic formula.

S8. −3t2 + 4t+ 9 = 0 S9. n2 + 4n− 3 = 2

S10. 2q2 + 4q − 5 = 8

Exercises

For the quadratic functions in Exercises 1–2, state the coor-

dinates of the vertex, the axis of symmetry, and whether the

parabola opens upward or downward.

1. f(x) = 3(x− 1)2 + 2

2. g(x) = −(x+ 3)2 − 4

3. Find the vertex and axis of symmetry of the graph of

v(t) = t2 + 11t− 4.

4. Find the vertex and axis of symmetry for the parabola

whose equation is y = 3x2 − 6x+ 5.

5. Sketch the quadratic functions given in standard form.

Identify the values of the parameters a, b, and c. Label

the zeros, axis of symmetry, vertex, and y-intercept.

(a) g(x) = x2 + 3 (b) f(x) = −2x2+4x+16

In Exercises 6–9, find a formula for the parabola.

6.

1(−1, 0) (3, 0)

(0,−3)

x

y 7.

−15

9(−6, 9)

x

y

8.

(0,−4)

(2, 0)x

y 9.

(10, 8)

(6, 5)x

y

For Exercises 10–13, convert the quadratic functions to ver-

tex form by completing the square. Identify the vertex and the

axis of symmetry.

10. f(x) = x2 + 8x+ 3

11. g(x) = −2x2 + 12x+ 4

12. p(t) = 2t2 − 0.12t+ 0.1

13. w(z) = −3z2 + 9z − 2

In Exercises 14–16, write the quadratic function in its stan-

dard, vertex, and factored forms.

14. y = −6 +x2 − x

2

15. f(t) =5t2 − 20

2

16. g(s) = (s− 5)(2s+ 3)


In Exercises 17–24, write a formula and graph the transforma-

tion of m(n) = 1

2n2.

17. y = m(n) + 1 18. y = m(n+ 1)

19. y = m(n)− 3.7 20. y = m(n− 3.7)

21. y = m(n) +√13 22. y = m(n+ 2

√2)

23. y = m(n+ 3) + 7 24. y = m(n− 17)− 159

Problems

25. Find the vertex form of f(t) = −16t2 + 47t+ 3.

26. Show that y = x2 − x+ 41 has no real zeros.

27. Find the value of k so that the graph of y = (x−3)2+kpasses through the point (6, 13).

28. The parabola y = ax2+k has vertex (0,−2) and passes

through the point (3, 4). Find its equation.

29. Using the vertex form, find a formula for the parabola

with vertex (2, 5) that passes through the point (1, 2).

In Problems 30–34, find a formula for the quadratic function

whose graph has the given properties.

30. Vertex at (4, 2) and y-intercept of y = 6.

31. Vertex at (4, 2) and y-intercept of y = −4.

32. Vertex at (4, 2) and zeros at x = −3, 11.

33. Vertex at (7, 3) and passing through the point (3, 7).

34. A single x-intercept at x = 1/2 and a y-intercept at 3.

35. Let f be a quadratic function whose graph is concave up

with a vertex at (1,−1), and a zero at the origin.

(a) Graph y = f(x).(b) Determine a formula for f(x).(c) Determine the range of f .

(d) Find any other zeros.

36. Find the vertex of y = 26+0.4x−0.01x2 exactly. Graph

the function, labeling all intercepts.

37. Find the vertex of y = 0.03x2+1.8x+2 exactly. Graph

the function, labeling all intercepts.

38. If we know a quadratic function f has a zero at x = −1and vertex at (1, 4), do we have enough information to

find a formula for this function? If your answer is yes,

find it; if not, give your reasons.

39. Figure 3.13 shows f(x) = x2. Define g by shifting the

graph of f to the right 2 units and down 1 unit; see Fig-

ure 3.14. Find a formula for g in terms of f . Find a for-

mula for g in terms of x.

−2 4−1

4 f(x) = x2

x

y

Figure 3.13

−2 4−1

4

g

x

y

Figure 3.14

40. The function g(x) is obtained by shifting the graph of

y = x2. If g(3) = 16, give a possible formula for gwhen

(a) g is the result of applying only a horizontal shift to

y = x2.

(b) g is the result of applying only a vertical shift to

y = x2.

(c) g is the result of applying a horizontal shift right 2units and an appropriate vertical shift of y = x2.

41. Graph y = x2 − 10x + 25 and y = x2. Use a shift

transformation to explain the relationship between the

two graphs.

42. Let f(x) = x2 and let g(x) = (x− 3)2 + 2.

(a) Give the formula for g in terms of f , and describe

the relationship between f and g in words.

(b) Is g a quadratic function? If so, find its standard form

and the parameters a, b, and c.

(c) Graph g, labeling all important features.

43. If you have a string of length 50 cm, what are the di-

mensions of the rectangle of maximum area that you can

enclose with your string? Explain your reasoning. What

about a string of length k cm?

44. A ballet dancer jumps in the air. The height, h(t), in feet,

of the dancer at time t, in seconds since the start of the

jump, is given by3

h(t) = −16t2 + 16Tt,

where T is the total time in seconds that the ballet dancer

is in the air.

(a) Why does this model apply only for 0 ≤ t ≤ T ?

(b) When, in terms of T , does the maximum height of

the jump occur?

3K. Laws, The Physics of Dance (Schirmer, 1984).

REVIEW EXERCISES AND PROBLEMS FOR CHAPTER THREE 129

(c) Show that the time, T , that the dancer is in the air is

related to H , the maximum height of the jump, by

the equation

H = 4T 2.45. A football player kicks a ball at an angle of 37◦ above the

ground with an initial speed of 20 meters/second. The

height in meters, h, as a function of the horizontal dis-

tance traveled, d, is given by:

h = 0.75d− 0.0192d2.

(a) Graph the path the ball follows.

(b) When the ball hits the ground, how far is it from the

spot where the football player kicked it?

(c) What is the maximum height the ball reaches during

its flight?

(d) What is the horizontal distance the ball has traveled

when it reaches its maximum height?4

CHAPTER SUMMARY

• General Formulas for Quadratic Functions

Standard form:

y = ax2 + bx+ c, a 6= 0

Factored form:

y = a(x− r)(x− s)

Vertex form:

y = a(x− h)2 + k

• Graphs of Quadratic Functions

Graphs are parabolas

Vertex (h, k)Axis of symmetry, x = h

Effect of parameter aOpens upward (concave up) if a > 0, minimum at

(h, k)Opens downward (concave down) if a < 0, maxi-

mum at (h, k)Factored form displays zeros at x = r and x = s

• Solving Quadratic Equations

Factoring

Quadratic formula

x =−b±

√b2 − 4ac

2a

Completing the square

REVIEW EXERCISES AND PROBLEMS FOR CHAPTER THREE

Exercises

Are the functions in Exercises 1–3 quadratic? If so, write the

function in standard form.

1. f(x) = 2(7− x)2 + 1

2. L(P ) = (P + 1)(1− P )

3. g(m) = m(m2 − 2m) + 3

(

14− m3

3

)

+√3m

In Exercises 4–9, find the zeros (if any) of the function alge-

braically.

4. y = 2x2 + 5x+ 2 5. y = 4x2 − 4x− 8

6. y = 7x2 + 16x+ 4 7. y = 5x2 + 2x− 1

8. y = −17x2 +23x+19 9. y = 3x2 − 2x+ 6

10. Show that the function y = −x2 + 7x − 13 has no real

zeros.

11. Find the vertex and axis of symmetry of the graph of

w(x) = −3x2 − 30x+ 31.

In Exercises 12–16, find a possible formula for the parabola

with the given conditions.

12. The vertex is (1,−2) and the y-intercept is y = −5.

13. The vertex is (4,−2) and the y-intercept is y = −3.

14. The vertex is (−7,−3) and it passes through the point

(−3,−7).

15. The parabola goes through the origin and its vertex is

(1,−1).

16. The x-intercepts are at x = −1 and x = 2 and (−2, 16)is on the function’s graph.

4Adapted from R. Halliday, D. Resnick, and K. Krane, Physics (New York: Wiley, 1992), p. 58.


In Exercises 17–20, find a formula for the parabola.

17.

(0, 4)

(4, 7)

x

y 18.

(5, 5)

(3, 3)x

y

19.

2

−5(3,−5)

y

x

20.

(−4, 0)

(2, 36)

(5, 0)x

y

Problems

For each parabola in Problems 21–24, state the coordinates of

the vertex, the axis of symmetry, the y-intercept, and whether

the curve is concave up or concave down.

21. y = 2(x−3/4)2−2/3 22. y = −1/2(x+6)2 − 4

23. y = (x− 0.6)2 24. y = −0.3x2 − 7

In Problems 25–28, write each function in factored form or

vertex form and then state its vertex and zeros.

25. y = 0.3x2 − 0.6x− 7.2

26. y = 2x2 − 4x− 2

27. y = −3x2 + 24x− 36

28. y = 2x2 + (7/3)x+ 1

29. For x between x = −2 and x = 4, determine the con-

cavity of the graph of f(x) = (x−1)2+2 by calculating

average rates of change over intervals of length 2.

30. A tomato is thrown vertically into the air at time t = 0.

Its height, d(t) (in feet), above the ground at time t (in

seconds) is given by

d(t) = −16t2 + 48t.

(a) Graph d(t).(b) Find t when d(t) = 0. What is happening to the

tomato the first time d(t) = 0? The second time?

(c) When does the tomato reach its maximum height?

(d) What is the maximum height that the tomato

reaches?

31. When slam-dunking, a basketball player seems to hang

in the air at the height of his jump. The height h(t), in

feet above the ground, of a basketball player at time t, in

seconds since the start of a jump, is given by

h(t) = −16t2 + 16Tt,

where T is the total time in seconds that it takes to com-

plete the jump. For a jump that takes 1 second to com-

plete, how much of this time does the basketball player

spend at the top 25% of the trajectory? [Hint: Find the

maximum height reached. Then find the times at which

the height is 75% of this maximum.]

STRENGTHEN YOUR UNDERSTANDING

Are the statements in Problems 1–15 true or false? Give an

explanation for your answer.

1. The quadratic function f(x) = x(x + 2) is in factored

form.

2. If f(x) = (x+ 1)(x+ 2), then the zeros of f are 1 and

2.

3. A quadratic function whose graph is concave up has a

maximum.

4. All quadratic equations have the form f(x) = ax2.

5. If the height above the ground of an object at time t is

given by s(t) = at2 + bt+ c, then s(0) tells us when the

object hits the ground.

6. To find the zeros of f(x) = ax2 + bx + c, solve the

equation ax2 + bx+ c = 0 for x.

7. Every quadratic equation has two real solutions.

8. There is only one quadratic function with zeros at x =−2 and x = 2.

9. A quadratic function has exactly two zeros.

10. The graph of every quadratic function is a parabola.

11. The maximum or minimum point of a parabola is called

its vertex.

12. If a parabola is concave up its vertex is a maximum point.

13. If the equation of a parabola is written as y = a(x −h)2 + k, then the vertex is located at the point (−h, k).

14. If the equation of a parabola is written as y = a(x −h)2 + k, then the axis of symmetry is found at x = h.

15. If the equation of a parabola is y = ax2 + bx + c and

a < 0, then the parabola opens downward.

131

SKILLS REFRESHER FOR CHAPTER 3: QUADRATIC EQUATIONS

SKILLS FOR FACTORING

Expanding an Expression

The distributive property for real numbers a, b, and c tells us that

a(b+ c) = ab+ ac,

and

(b+ c)a = ba+ ca.

We use the distributive property and the rules of exponents to multiply algebraic expressions involv-

ing parentheses. This process is sometimes referred to as expanding the expression.

Example 1 Multiply the following expressions and simplify.

(a) 3x2

(

x+1

6x−3

)

(b)((2t)2 − 5

)√t

Solution (a) 3x2

(

x+1

6x−3

)

=(3x2)(x) +

(3x2)(1

6x−3

)

= 3x3 +1

2x−1.

(b)((2t)2 − 5

)√t = (2t)2(

√t)− 5

√t =

(4t2) (

t1/2)

− 5t1/2 = 4t5/2 − 5t1/2.

If there are two terms in each factor, then there are four terms in the product:

(a+ b)(c+ d) = a(c+ d) + b(c+ d) = ac+ ad+ bc+ bd.

The following special cases of the above product occur frequently. Learning to recognize their forms

aids in factoring.

(a+ b)(a− b) = a2 − b2

(a+ b)2 = a2 + 2ab+ b2

(a− b)2 = a2 − 2ab+ b2

Example 2 Expand the following and simplify by gathering like terms.

(a) (5x2 + 2)(x− 4) (b) (2√r + 2)(4

√r − 3) (c)

(

3− 1

2x

)2

Solution (a)(5x2 + 2

)(x− 4) =

(5x2)(x) +

(5x2)(−4) + (2)(x) + (2)(−4) = 5x3 − 20x2 + 2x− 8.

(b) (2√r+2)(4

√r−3) = (2)(4)(

√r)2+(2)(−3)(

√r)+(2)(4)(

√r)+(2)(−3) = 8r+2

√r−6.

(c)

(

3− 1

2x

)2

= 32 − 2 (3)

(1

2x

)

+

(

−1

2x

)2

= 9− 3x+1

4x2.

Factoring

To write an expanded expression in factored form, we “un-multiply” the expression. Some tech-

niques for factoring are given in this section. We can check factoring by multiplying the factors.

Removing a Common Factor

It is sometimes useful to factor out the same factor from each of the terms in an expression. This is

basically the distributive law in reverse:

ab+ ac = a(b+ c).

132 SKILLS REFRESHER FOR CHAPTER THREE

One special case is removing a factor of −1, which gives

−a− b = −(a+ b)

Another special case is

(a− b) = −(b− a)

Example 3 Factor the following:

(a)2

3x2y +

4

3xy (b) (2p+ 1)p3 − 3p(2p+ 1) (c) −s2t

8w− st2

16w

Solution (a)2

3x2y +

4

3xy =

2

3xy(x+ 2).

(b) (2p+ 1)p3 − 3p(2p+ 1) = (p3 − 3p)(2p+ 1) = p(p2 − 3)(2p+ 1).(Note that the expression (2p+ 1) was one of the factors common to both terms.)

(c) −s2t

8w− st2

16w= − st

8w

(

s+t

2

)

.

Grouping Terms

Even though all the terms may not have a common factor, we can sometimes factor by first grouping

the terms and then removing a common factor.

Example 4 Factor x2 − hx− x+ h.

Solution x2 − hx− x+ h =(x2 − hx

)− (x− h) = x(x− h)− (x− h) = (x− h)(x− 1).

Factoring Quadratics

One way to factor quadratics is to mentally multiply out the possibilities.

Example 5 Factor t2 − 4t− 12.

Solution If the quadratic factors, it is of the form

t2 − 4t− 12 = (t+ ?)(t+ ?).

We are looking for two numbers whose product is −12 and whose sum is −4. By trying combina-

tions, we find

t2 − 4t− 12 = (t− 6)(t+ 2).

Example 6 Factor 4− 2M − 6M2.

Solution By a similar method to the previous example, we find 4− 2M − 6M2 = (2− 3M)(2 + 2M).

Perfect Squares and the Difference of Squares

Recognition of the special products (x + y)2, (x − y)2 and (x + y)(x − y) in expanded form is

useful in factoring. Reversing the results given on page 131, we have

a2 + 2ab+ b2 = (a+ b)2,

a2 − 2ab+ b2 = (a− b)2,

a2 − b2 = (a− b)(a+ b).

SKILLS FOR FACTORING 133

When we see squared terms in an expression to be factored, it is often useful to look for one of these

forms. The difference of squares identity (the third one listed previously) is especially useful.

Example 7 Factor: (a) 16y2 − 24y + 9 (b) 25S2R4 − T 6 (c) x2(x− 2) + 16(2− x)

Solution (a) 16y2 − 24y + 9 = (4y − 3)2.

(b) 25S2R4 − T 6 =(5SR2

)2 −(T 3)2

=(5SR2 − T 3

) (5SR2 + T 3

).

(c) x2(x−2)+16(2−x) = x2(x−2)−16(x−2) = (x−2)(x2 − 16

)= (x−2)(x−4)(x+4).

Solving Quadratic Equations

Example 8 Give exact and approximate solutions to x2 = 3.

Solution The exact solutions are x = ±√3; approximate ones are x ≈ ±1.73, or x ≈ ±1.732, or x ≈

±1.73205 (since√3 = 1.732050808 . . .). Notice that the equation x2 = 3 has only two exact

solutions, but many possible approximate solutions, depending on how much accuracy is required.

Solving by Factoring

Some equations can be put into factored form such that the product of the factors is zero. Then we

solve by using the fact that if a · b = 0, then either a or b (or both) is zero.

Example 9 Solve (x+ 1)(x+ 3) = 15.

Solution Although it is true that if a · b = 0, then a = 0 or b = 0, it is not true that a · b = 15 means that

a = 15 or b = 15, or that a and b are 3 and 5. To solve this equation, we expand the left-hand side

and rearrange so that the right-hand side is equal to zero:

x2 + 4x+ 3 = 15,

x2 + 4x− 12 = 0.

Then, factoring gives

(x− 2)(x+ 6) = 0.

Thus x = 2 and x = −6 are the solutions.

Example 10 Solve 2(x+ 3)2 = 5(x+ 3).

Solution You might be tempted to divide both sides by (x + 3). However, if you do this you will overlook

one of the solutions. Instead, write

2(x+ 3)2 − 5(x+ 3) = 0

(x+ 3) (2(x+ 3)− 5) = 0

(x+ 3)(2x+ 6− 5) = 0

(x+ 3)(2x+ 1) = 0.

Thus, x = −1/2 and x = −3 are solutions.

Note that if we had divided by (x + 3) at the start, we would have lost the solution x = −3,

which was obtained by setting x+ 3 = 0.


Solving with the Quadratic Formula

Instead of factoring, we can solve the equation ax2 + bx+ c = 0 by using the quadratic formula:

x =−b±

√b2 − 4ac

2a.

The quadratic formula is derived by completing the square for y = ax2 + bx+ c. See page 136.

Example 11 Solve 11 + 2x = x2.

Solution The equation is

−x2 + 2x+ 11 = 0.

The expression on the left does not factor using integers, so we use

x =−2 +

√

4− 4(−1)(11)

2(−1)=

−2 +√48

−2=

−2 +√16 · 3

−2=

−2 + 4√3

−2= 1− 2

√3,

x =−2−

√

4− 4(−1)(11)

2(−1)=

−2−√48

−2=

−2−√16 · 3

−2=

−2− 4√3

−2= 1 + 2

√3.

The exact solutions are x = 1− 2√3 and x = 1 + 2

√3.

The decimal approximations to these numbers x = 1 − 2√3 = −2.464 and x = 1 + 2

√3 =

4.464 are approximate solutions to this equation. The approximate solutions could also be found

directly from a graph or calculator.

Exercises for Skills for Factoring

For Exercises 1–15, expand and simplify.

1. 2(3x− 7) 2. −4(y + 6)

3. 12(x+ y) 4. −7(5x− 8y)

5. x(2x+ 5) 6. 3z(2x− 9z)

7. −10r(5r + 6rs)

8. x(3x− 8) + 2(3x− 8)

9. 5z(x− 2)− 3(x− 2)

10. (x+ 1)(x+ 3)

11. (x− 2)(x+ 6)

12. (5x− 1)(2x− 3)

13. (y + 1)(z + 3)

14. (12y − 5)(8w + 7)

15. (5z − 3)(x− 2)

Multiply and write the expressions in Problems 16–22 without

parentheses. Gather like terms.

16. −(x− 3)− 2(5− x)

17. (x− 5)6− 5(1− (2− x))

18.(3x− 2x2

)4 + (5 + 4x)(3x− 4)

19. P (p− 3q)2 20. 4(x− 3)2 + 7

21. −(√

2x+ 1)2

22. u(u−1 + 2u

)2u

For Exercises 23–67, factor completely if possible.

23. 2x+ 6 24. 3y + 15

25. 5z − 30 26. 4t− 6

27. 10w − 25 28. 3u4 − 4u3

29. 3u7 + 12u2 30. 12x3y2 − 18x

31. 14r4s2 − 21rst 32. x2 + 3x− 2

33. x2 − 3x+ 2 34. x2 − 3x− 2

35. x2 + 2x+ 3 36. x2 − 2x− 3

37. x2 − 2x+ 3 38. x2 + 2x− 3

39. 2x2 + 5x+ 2 40. 2x2 − 10x+ 12

41. x2 + 3x− 28 42. x3 − 2x2 − 3x

43. x3 + 2x2 − 3x

SKILLS FOR FACTORING 135

44. ac+ ad+ bc+ bd

45. x2 + 2xy + 3xz + 6yz

46. x2 − 1.4x− 3.92

47. a2x2 − b2

48. πr2 + 2πrh

49. B2 − 10B + 24

50. c2 + x2 − 2cx

51. x2 + y2

52. a4 − a2 − 12

53. (t+ 3)2 − 16

54. x2 + 4x+ 4− y2

55. a3 − 2a2 + 3a− 6

56. b3 − 3b2 − 9b+ 27

57. c2d2 − 25c2 − 9d2 + 225

58. hx2 + 12− 4hx− 3x

59. r(r − s)− 2(s− r)

60. y2 − 3xy + 2x2

61. x2e−3x + 2xe−3x

62. t2e5t + 3te5t + 2e5t

63. P (1 + r)2 + P (1 + r)2r

64. x2 − 6x+ 9− 4z2

65. dk + 2dm− 3ek − 6em

66. πr2 − 2πr + 3r − 6

67. 8gs− 12hs+ 10gm− 15hm

Solve the equations in Exercises 68–93.

68. y2 − 5y − 6 = 0

69. 4s2 + 3s− 15 = 0

70.2

x+

3

2x= 8

71.3

x− 1+ 1 = 5

72.√y − 1 = 13

73. −16t2 + 96t+ 12 = 60

74. g3 − 4g = 3g2 − 12

75. 8 + 2x− 3x2 = 0

76. 2p3 + p2 − 18p− 9 = 0

77. N2 − 2N − 3 = 2N(N − 3)

78.1

64t3 = t

79. x2 − 1 = 2x

80. 4x2 − 13x− 12 = 0

81. 60 = −16t2 + 96t+ 12

82. n5 + 80 = 5n4 + 16n

83. 5a3 + 50a2 = 4a+ 40

84. y2 + 4y − 2 = 0

85.2

z − 3+

7

z2 − 3z= 0

86.x2 + 1− 2x2

(x2 + 1)2= 0

87. 4− 1

L2= 0

88. 2 +1

q + 1− 1

q − 1= 0

89.√r2 + 24 = 7

90.13√x

= −2

91. 3√x =

1

2x

92. 10 =

√v

7π

93.(3x+ 4)(x− 2)

(x− 5)(x− 1)= 0

In Exercises 94–97, solve for the indicated variable.

94. T = 2π

√

l

g, for l.

95. Ab5 = C, for b.

96. |2x+ 1| = 7, for x.

97.x2 − 5mx+ 4m2

x−m= 0, for x

Solve the systems of equations in Exercises 98–102.

98.

{

y = 2x− x2

y = −399.

{y = 1/xy = 4x

100.

{

x2 + y2 = 36y = x− 3

101.

{

y = 4− x2

y − 2x = 1

102.

{

y = x3 − 1y = ex

103. Let ℓ be the line of slope 3 passing through the ori-

gin. Find the points of intersection of the line ℓ and the

parabola whose equation is y = x2. Sketch the line and

the parabola, and label the points of intersection.

Determine the points of intersection for Problems 104–105.

104.

x2 + y2 = 25 y = x− 1

x

y 105.

y = x2

y = 15− 2x

x

y


COMPLETING THE SQUARE

An example of changing the form of an expression is the conversion of ax2 + bx+ c into the form

a(x−h)2+k. We make this conversion by completing the square, a method for producing a perfect

square within a quadratic expression. A perfect square is an expression of the form:

(x+ n)2 = x2 + 2nx+ n2.

In order to complete the square in an expression, we must find that number n, which is half the

coefficient of x. Before giving a general procedure, let’s work through an example.

Example 1 Complete the square to rewrite x2 − 10x+ 4 in the form a(x− h)2 + k.

Solution Step 1: We divide the coefficient of x by 2, giving 1

2(−10) = −5.

Step 2: We square the result of step 1, giving (−5)2 = 25.

Step 3: Now add and subtract the 25 after the x-term:

x2 − 10x+ 4 = x2 − 10x+ 25− 25 + 4

= (x2 − 10x+ 25)︸︷︷︸

Perfect square

−25 + 4.

Step 4: Notice that we have created a perfect square, x2 − 10x + 25. The next step is to factor the

perfect square and combine the constant terms, −25 + 4, giving the final result:

x2 − 10x+ 4 = (x− 5)2 − 21.

Thus, a = +1, h = +5, and k = −21.

Visualizing the Process of Completing the Square

We can visualize how to find the constant that needs to be added to x2 + bx in order to obtain

a perfect square by thinking of x2 + bx as the area of a rectangle. For example, the rectangle in

Figure 3.15 has area x(x + b) = x2 + bx. Now imagine cutting the rectangle into pieces as in

Figure 3.16 and trying to rearrange them to make a square, as in Figure 3.17. The corner piece,

whose area is (b/2)2, is missing. By adding this piece to our expression, we “complete” the square:

x2 + bx+ (b/2)2 = (x+ b/2)2.

x

x+ b

Figure 3.15: Rectangle with

sides x and x+ b

x

x b/2 b/2

Figure 3.16: Cut off 2 strips

of width b/2

✻

❄

x+ b/2

✲✛ x+ b/2

x x

x b/2

b/2 b/2

Figure 3.17: Rearrange to see a square

with missing corner of area (b/2)2

The procedure we followed can be summarized as follows:

COMPLETING THE SQUARE 137

To complete the square in the expression x2+bx+c, divide the coefficient of x by 2, giving

b/2. Then add and subtract (b/2)2 = b2/4 and factor the perfect square:

x2 + bx+ c =

(

x+b

2

)2

− b2

4+ c.

To complete the square in the expression ax2 + bx+ c, factor out a first.

The next example has a coefficient a with a 6= 1. After factoring out the coefficient, we follow

the same steps as in Example 1.

Example 2 Complete the square in the formula h(x) = 5x2 + 30x− 10.

Solution We first factor out 5:

h(x) = 5(x2 + 6x− 2).

Now we complete the square in the expression x2 + 6x− 2.

Step 1: Divide the coefficient of x by 2, giving 3.

Step 2: Square the result: 32 = 9.

Step 3: Add the result after the x term, then subtract it:

h(x) = 5(x2 + 6x+ 9︸︷︷︸

Perfect square

−9− 2).

Step 4: Factor the perfect square and simplify the rest:

h(x) = 5((x+ 3)2 − 11

).

Now that we have completed the square, we can multiply by the 5:

h(x) = 5(x+ 3)2 − 55.

Deriving the Quadratic Formula

We derive a general formula to find the zeros of q(x) = ax2 + bx + c, with a 6= 0, by completing

the square. To find the zeros, set q(x) = 0:

ax2 + bx+ c = 0.

Before we complete the square, we factor out the coefficient of x2:

a

(

x2 +b

ax+

c

a

)

= 0.

Since a 6= 0, we can divide both sides by a:

x2 +b

ax+

c

a= 0.


To complete the square, we add and then subtract ((b/a)/2)2= b2/(4a2):

x2 +b

ax+

b2

4a2︸︷︷︸

Perfect square

− b2

4a2+

c

a= 0.

We factor the perfect square and simplify the constant term, giving:

(

x+b

2a

)2

−(b2 − 4ac

4a2

)

= 0 since−b

2

4a2+ c

a=

−b2

4a2+ 4ac

4a2= −

(b2−4ac

4a2

)

(

x+b

2a

)2

=b2 − 4ac

4a2adding

b2−4ac

4a2to both sides

x+b

2a= ±

√

b2 − 4ac

4a2=

±√b2 − 4ac

2ataking the square root

x =−b

2a±

√b2 − 4ac

2asubtracting b/2a

x =−b±

√b2 − 4ac

2a.

Exercises for Skills for Completing the Square

For Exercises 1–8, complete the square for each expression.

1. x2 + 8x 2. w2 + 7w

3. 2r2 + 20r 4. 3t2 + 24t− 13

5. a2 − 2a− 4 6. n2 + 4n− 5

7. 3r2 + 9r − 4 8. 12g2 + 8g + 5

In Exercises 9–12, rewrite in the form a(x− h)2 + k.

9. x2 − 2x− 3 10. 10− 6x+ x2

11. −x2 + 6x− 2 12. 3x2 − 12x+ 13

In Exercises 13–22, complete the square to find the vertex of

the parabola.

13. y = x2 + 6x+ 3 14. y = x2 − x+ 4

15. y = −x2 − 8x+ 2 16. y = x2 − 3x− 3

17. y = −x2 + x− 6 18. y = 3x2 + 12x

19. y = −4x2 + 8x− 6 20. y = 5x2 − 5x+ 7

21. y = 2x2 − 7x+ 3 22. y = −3x2 − x− 2

In Exercises 23–29, solve by completing the square.

23. g2 = 2g + 24 24. p2 − 2p = 6

25. d2 − d = 2 26. 2r2 + 4r − 5 = 0

27. 2s2 = 1− 10s 28. 7r2 − 3r − 6 = 0

29. 5p2 + 9p = 1

In Exercises 30–35, solve by using the quadratic formula.

30. n2 − 4n− 12 = 0 31. 2y2 + 5y = −2

32. 6k2 + 11k = −3 33. w2 + w = 4

34. z2 + 4z = 6 35. 2q2 + 6q − 3 = 0

In Exercises 36–46, solve using factoring, completing the

square, or the quadratic formula.

36. r2 − 2r = 8

37. s2 + 3s = 1

38. z3 + 2z2 = 3z + 6

39. 25u2 + 4 = 30u

40. v2 − 4v − 9 = 0

41. 3y2 = 6y + 18

42. 2p2 + 23 = 14p

43. 2w3 + 24 = 6w2 + 8w

44. 4x2 + 16x− 5 = 0

45. 49m2 + 70m+ 22 = 0

46. 8x2 − 1 = 2x

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