2.2 quadratic functions - honors precalculus and...

Quadratic Functions

2.2

286 Chapter 2 Polynomial and Rational Functions

Quadratic Functions

Many sports involve objects that are thrown,kicked, or hit, and then proceed with no addi-

tional force of their own. Such objects are calledprojectiles. Paths of projectiles, as well as theirheights over time, can be modeled by quadraticfunctions. We have seen that a quadratic function isany function of the form

where and are real numbers, with Aquadratic function is a polynomial function whosegreatest exponent is 2. In this section, you will learnto use graphs of quadratic functions to gain a visualunderstanding of the algebra that describes football,baseball, basketball, the shot put, and otherprojectile sports.

Graphs of Quadratic FunctionsThe graph of any quadratic function is called a parabola. Parabolas are shaped likebowls or inverted bowls, as shown in Figure 2.2. If the coefficient of (the value of

in ) is positive, the parabola opens upward. If the coefficient of isnegative, the parabola opens downward. The vertex (or turning point) of theparabola is the lowest point on the graph when it opens upward and the highestpoint on the graph when it opens downward.

x2ax2 + bx + cax2

a Z 0.ca, b,

f1x2 = ax2 + bx + c,

Objectives

! Recognize characteristics ofparabolas.

" Graph parabolas.

# Determine a quadraticfunction’s minimum ormaximum value.

$ Solve problems involving aquadratic function’s minimumor maximum value.

Sec t i on


x

y

a � 0: Parabola opens upward. a � 0: Parabola opens downward.

Vertex (minimum point)

f(x ) � ax 2 � bx � c (a � 0)

Axis of symmetry

Vertex (maximum point)

f(x ) � ax 2 � bx � c (a � 0)

Axis of symmetryx

y

Figure 2.2 Characteristics of graphs of quadratic functions

Look at the unusual image of the word mirror shown below.The artist, Scott Kim,has created the image so that the twohalves of the whole are mirror imagesof each other. A parabola shares thiskind of symmetry, in which a verticalline through the vertex divides thefigure in half. Parabolas are sym-metric with respect to this line, calledthe axis of symmetry. If a parabola isfolded along its axis of symmetry, thetwo halves match exactly.

P-BLTZMC02_277-386-hr 19-11-2008 11:37 Page 286


In the graphs that follow, we will show each axis of symmetry as a dashedvertical line. Because this vertical line passes through the vertex, its equationis The line is dashed because it is not part of the parabola.

Graphing a Quadratic Function in Standard Form

Graph the quadratic function

Solution We can graph this function by following the steps in the preceding box.We begin by identifying values for and

Step 1 Determine how the parabola opens. Note that the coefficient of is Thus, this negative value tells us that the parabola opens downward.Step 2 Find the vertex. The vertex of the parabola is at Because and

the parabola has its vertex at (3, 8).Step 3 Find the by solving Replace with 0 in

Find setting equal to 0.

Solve for Add to bothsides of the equation.

Divide both sides by 2.

Apply the square root property.

Add 3 to both sides in each equation.

The are 1 and 5. The parabola passes through (1, 0) and (5, 0).Step 4 Find the by computing Replace with 0 in

The is The parabola passes through Step 5 Graph the parabola. With a vertex at (3, 8), at 5 and 1, anda at the graph of is shown in Figure 2.4. The axis of symmetryis the vertical line whose equation is x = 3.

f-10,y-interceptx-intercepts

10, -102.-10.y-intercept

f102 = -210 - 322 + 8 = -21-322 + 8 = -2192 + 8 = -10

f1x2 = -21x - 322 + 8.xf(0).y-intercept

x-intercepts

x = 5 x = 124 = 2 x - 3 = 2 x - 3 = -2

x - 3 = 24� or� x - 3 = - 24 1x - 322 = 4

21x - 322x. 21x - 322 = 8f1x2x-intercepts, 0 = -21x - 322 + 8

f1x2 = -21x - 322 + 8.f1x2f(x) ! 0.x-intercepts

k = 8,h = 31h, k2.a 6 0;-2.

x2,a,

f(x)=a(x-h)2+k

f(x)=–2(x-3)2+8

a � 2 h � 3 k � 8

Standard form

Given function

k.a, h,

f1x2 = -21x - 322 + 8.

EXAMPLE 1

x = h.1h, k2,

The sign of in determines whether the parabola opens upward or downward. Furthermore, if is small, the parabola opens moreflatly than if is large. Here is a general procedure for graphing parabolas whoseequations are in standard form:

ƒ a ƒƒ a ƒ

f1x2 = a1x - h22 + ka

Graphing Quadratic Functions with Equations in Standard FormTo graph

1. Determine whether the parabola opens upward or downward. If itopens upward. If it opens downward.

2. Determine the vertex of the parabola. The vertex is 3. Find any by solving The function’s real zeros are the

4. Find the by computing 5. Plot the intercepts, the vertex, and additional points as necessary. Connect these

points with a smooth curve that is shaped like a bowl or an inverted bowl.

f102.y-interceptx-intercepts.

f1x2 = 0.x-intercepts1h, k2.a 6 0,

a 7 0,

f1x2 = a1x - h22 + k,

2

68

10

4 6 8

10

1 2 3 4 8765 1 2

y

x

Axis of symmetry: x � 3y-intercept: 10

Vertex: (3, 8)

x-intercept: 5x-intercept: 1

Figure 2.4 The graph of f1x2 = -21x - 322 + 8

P-BLTZMC02_277-386-hr 19-11-2008 11:37 Page 288288 Chapter 2 Polynomial and Rational Functions

In the graphs that follow, we will show each axis of symmetry as a dashedvertical line. Because this vertical line passes through the vertex, its equationis The line is dashed because it is not part of the parabola.



Solution We can graph this function by following the steps in the preceding box.We begin by identifying values for and

Step 1 Determine how the parabola opens. Note that the coefficient of is Thus, this negative value tells us that the parabola opens downward.Step 2 Find the vertex. The vertex of the parabola is at Because and

the parabola has its vertex at (3, 8).Step 3 Find the by solving Replace with 0 in

Find setting equal to 0.

Solve for Add to bothsides of the equation.

Divide both sides by 2.


Add 3 to both sides in each equation.

The are 1 and 5. The parabola passes through (1, 0) and (5, 0).Step 4 Find the by computing Replace with 0 in

The is The parabola passes through Step 5 Graph the parabola. With a vertex at (3, 8), at 5 and 1, anda at the graph of is shown in Figure 2.4. The axis of symmetryis the vertical line whose equation is x = 3.

f-10,y-interceptx-intercepts

10, -102.-10.y-intercept

f102 = -210 - 322 + 8 = -21-322 + 8 = -2192 + 8 = -10

f1x2 = -21x - 322 + 8.xf(0).y-intercept

x-intercepts

x = 5 x = 124 = 2 x - 3 = 2 x - 3 = -2

x - 3 = 24� or� x - 3 = - 24 1x - 322 = 4

21x - 322x. 21x - 322 = 8f1x2x-intercepts, 0 = -21x - 322 + 8

f1x2 = -21x - 322 + 8.f1x2f(x) ! 0.x-intercepts

k = 8,h = 31h, k2.a 6 0;-2.

x2,a,

f(x)=a(x-h)2+k

f(x)=–2(x-3)2+8

a � 2 h � 3 k � 8

Standard form

Given function

k.a, h,

f1x2 = -21x - 322 + 8.

EXAMPLE 1

x = h.1h, k2,

The sign of in determines whether the parabola opens upward or downward. Furthermore, if is small, the parabola opens moreflatly than if is large. Here is a general procedure for graphing parabolas whoseequations are in standard form:

ƒ a ƒƒ a ƒ

f1x2 = a1x - h22 + ka

Graphing Quadratic Functions with Equations in Standard FormTo graph

1. Determine whether the parabola opens upward or downward. If itopens upward. If it opens downward.

2. Determine the vertex of the parabola. The vertex is 3. Find any by solving The function’s real zeros are the

4. Find the by computing 5. Plot the intercepts, the vertex, and additional points as necessary. Connect these

points with a smooth curve that is shaped like a bowl or an inverted bowl.

f102.y-interceptx-intercepts.

f1x2 = 0.x-intercepts1h, k2.a 6 0,

a 7 0,

f1x2 = a1x - h22 + k,

2

68

10

4 6 8

10

1 2 3 4 8765 1 2

y

x

Axis of symmetry: x � 3y-intercept: 10

Vertex: (3, 8)

x-intercept: 5x-intercept: 1

Figure 2.4 The graph of f1x2 = -21x - 322 + 8

P-BLTZMC02_277-386-hr 19-11-2008 11:37 Page 288

Section 2.2 Quadratic Functions 289

Check Point 1 Graph the quadratic function



Solution We begin by finding values for and

Step 1 Determine how the parabola opens. Note that the coefficient of is 1.Thus, this positive value tells us that the parabola opens upward.Step 2 Find the vertex. The vertex of the parabola is at Because and the parabola has its vertex at Step 3 Find the by solving Replace with 0 in

Because the vertex is at which lies above the and the parabola opens upward, it appears that this parabola has no We can verify this observation algebraically.

Find possible settingequal to 0.

Solve for Subtract 1 from bothsides.


The solutions are

Because this equation has no real solutions, the parabola has no Step 4 Find the by computing Replace with 0 in

The is 10. The parabola passes through (0, 10).Step 5 Graph the parabola. With a vertex at no and a

at 10, the graph of is shown in Figure 2.5. The axis of symmetry is thevertical line whose equation is


Graphing Quadratic Functions in the Form

Quadratic functions are frequently expressed in the form How can we identify the vertex of a parabola whose equation is in this form?Completing the square provides the answer to this question.

f1x2 = ax2 + bx + c.

f(x ) ! ax 2 " bx " c

f1x2 = 1x - 222 + 1.

x = -3.fy-intercept

x-intercepts,1-3, 12,y-intercept

f102 = 10 + 322 + 1 = 32 + 1 = 9 + 1 = 10

f1x2 = 1x + 322 + 1.xf(0).y-intercept

x-intercepts.

-3 ; i. x = -3 + i x = -3 - i

2- 1 = i x + 3 = i x + 3 = - i

x + 3 = 2-1� or� x + 3 = - 2-1

x. -1 = 1x + 322 f1x2 x-intercepts, 0 = 1x + 322 + 1

x-intercepts.x-axis,1-3, 12,f1x2 = 1x + 322 + 1.

f1x2f(x) ! 0.x-intercepts1-3, 12.k = 1,

h = -31h, k2.a 7 0;x2,a,

f(x)=1(x-(–3))2+1

f(x)=(x+3)2+1

f(x)=a(x-h)2+k

a � 1 h � 3 k � 1

Standard form of quadratic function

Given function

k.a, h,

f1x2 = 1x + 322 + 1.

EXAMPLE 2

f1x2 = -1x - 122 + 4.

1

1234

13121110

98765

1 2 3 1 3 4 6 7 5

y

x

Vertex: ( 3, 1)

Axis of symmetry: x � 3

y-intercept: 10

Figure 2.5 The graph of f1x2 = 1x + 322 + 1

P-BLTZMC02_277-386-hr 19-11-2008 11:37 Page 289

2.2


Graphing Quadratic Functions in Standard FormIn our earlier work with transformations, we applied a series of transformations tothe graph of The graph of this function is a parabola. The vertex for this parabola is (0, 0). In Figure 2.3(a), the graph of for is shown inblack; it opens upward. In Figure 2.3(b), the graph of for is shownin black; it opens downward.

a 6 0f1x2 = ax2a 7 0f1x2 = ax2

f1x2 = x2.

! Graph parabolas.

g (x) � a(x h)2 � k

Vertex: (0, 0)

Vertex: (0, 0)

f(x) � ax2

f(x) � ax2

Axis of symmetry: x � h Axis of symmetry: x � h

Vertex: (h, k)x

x

y y

g (x) � a(x h)2 � kVertex: (h, k)

Figure 2.3(a) Parabola opens upward.a 7 0: Figure 2.3(b) Parabola opens downward.a 6 0:

Figures 2.3(a) and 2.3(b) also show the graph of in blue.Compare these graphs to those of Observe that determines ahorizontal shift and determines a vertical shift of the graph of

Consequently, the vertex (0, 0) on the black graph of moves to the pointon the blue graph of The axis of symmetry is the

vertical line whose equation is The form of the expression for is convenient because it immediately identifies

the vertex of the parabola as This is the standard form of a quadratic function.1h, k2. gx = h.

g1x2 = a1x - h22 + k.1h, k2 f1x2 = ax2

g(x)=a(x-h)2 +k.

If h � 0, the graph off(x) � ax2 is shifted h

units to the right.

If k � 0, the graph ofy � a(x h)2 is shifted

k units up.

f1x2 = ax2:khf1x2 = ax2.

g1x2 = a1x - h22 + k

Transformations of f1x2 = ax2

The Standard Form of a Quadratic FunctionThe quadratic function

is in standard form. The graph of is a parabola whose vertex is the point The parabola is symmetric with respect to the line If the parabolaopens upward; if the parabola opens downward.a 6 0,

a 7 0,x = h.1h, k2.f

f1x2 = a1x - h22 + k, a Z 0

P-BLTZMC02_277-386-hr 19-11-2008 11:37 Page 287

2.2


Quadratic Functions

Many sports involve objects that are thrown,kicked, or hit, and then proceed with no addi-

tional force of their own. Such objects are calledprojectiles. Paths of projectiles, as well as theirheights over time, can be modeled by quadraticfunctions. We have seen that a quadratic function isany function of the form

where and are real numbers, with Aquadratic function is a polynomial function whosegreatest exponent is 2. In this section, you will learnto use graphs of quadratic functions to gain a visualunderstanding of the algebra that describes football,baseball, basketball, the shot put, and otherprojectile sports.

Graphs of Quadratic FunctionsThe graph of any quadratic function is called a parabola. Parabolas are shaped likebowls or inverted bowls, as shown in Figure 2.2. If the coefficient of (the value of

in ) is positive, the parabola opens upward. If the coefficient of isnegative, the parabola opens downward. The vertex (or turning point) of theparabola is the lowest point on the graph when it opens upward and the highestpoint on the graph when it opens downward.

x2ax2 + bx + cax2

a Z 0.ca, b,

f1x2 = ax2 + bx + c,

Objectives


" Graph parabolas.

# Determine a quadraticfunction’s minimum ormaximum value.

$ Solve problems involving aquadratic function’s minimumor maximum value.

Sec t i on


x

y

a � 0: Parabola opens upward. a � 0: Parabola opens downward.

Vertex (minimum point)

f(x ) � ax 2 � bx � c (a � 0)

Axis of symmetry

Vertex (maximum point)

f(x ) � ax 2 � bx � c (a � 0)

Axis of symmetryx

y

Figure 2.2 Characteristics of graphs of quadratic functions

Look at the unusual image of the word mirror shown below.The artist, Scott Kim,has created the image so that the twohalves of the whole are mirror imagesof each other. A parabola shares thiskind of symmetry, in which a verticalline through the vertex divides thefigure in half. Parabolas are sym-metric with respect to this line, calledthe axis of symmetry. If a parabola isfolded along its axis of symmetry, thetwo halves match exactly.

P-BLTZMC02_277-386-hr 19-11-2008 11:37 Page 286





Solution We begin by finding values for and

Step 1 Determine how the parabola opens. Note that the coefficient of is 1.Thus, this positive value tells us that the parabola opens upward.Step 2 Find the vertex. The vertex of the parabola is at Because and the parabola has its vertex at Step 3 Find the by solving Replace with 0 in

Because the vertex is at which lies above the and the parabola opens upward, it appears that this parabola has no We can verify this observation algebraically.

Find possible settingequal to 0.

Solve for Subtract 1 from bothsides.


The solutions are

Because this equation has no real solutions, the parabola has no Step 4 Find the by computing Replace with 0 in

The is 10. The parabola passes through (0, 10).Step 5 Graph the parabola. With a vertex at no and a

at 10, the graph of is shown in Figure 2.5. The axis of symmetry is thevertical line whose equation is


Graphing Quadratic Functions in the Form

Quadratic functions are frequently expressed in the form How can we identify the vertex of a parabola whose equation is in this form?Completing the square provides the answer to this question.

f1x2 = ax2 + bx + c.

f(x ) ! ax 2 " bx " c

f1x2 = 1x - 222 + 1.

x = -3.fy-intercept

x-intercepts,1-3, 12,y-intercept

f102 = 10 + 322 + 1 = 32 + 1 = 9 + 1 = 10

f1x2 = 1x + 322 + 1.xf(0).y-intercept

x-intercepts.

-3 ; i. x = -3 + i x = -3 - i

2- 1 = i x + 3 = i x + 3 = - i

x + 3 = 2-1� or� x + 3 = - 2-1

x. -1 = 1x + 322 f1x2 x-intercepts, 0 = 1x + 322 + 1

x-intercepts.x-axis,1-3, 12,f1x2 = 1x + 322 + 1.

f1x2f(x) ! 0.x-intercepts1-3, 12.k = 1,

h = -31h, k2.a 7 0;x2,a,

f(x)=1(x-(–3))2+1

f(x)=(x+3)2+1

f(x)=a(x-h)2+k

a � 1 h � 3 k � 1

Standard form of quadratic function

Given function

k.a, h,

f1x2 = 1x + 322 + 1.

EXAMPLE 2

f1x2 = -1x - 122 + 4.

1

1234

13121110

98765

1 2 3 1 3 4 6 7 5

y

x

Vertex: ( 3, 1)

Axis of symmetry: x � 3

y-intercept: 10

Figure 2.5 The graph of f1x2 = 1x + 322 + 1

P-BLTZMC02_277-386-hr 19-11-2008 11:37 89


Graphing a Quadratic Function in the Form

Graph the quadratic function Use the graph to identify thefunction’s domain and its range.

SolutionStep 1 Determine how the parabola opens. Note that the coefficient of is Thus, this negative value tells us that the parabola opens downward.

Step 2 Find the vertex. We know that the of the vertex is

We identify and in

Substitute the values of and into the equation for the

The of the vertex is We substitute for in the equation of thefunction, to find the

The vertex is at

Step 3 Find the by solving Replace with 0 in We obtain This equation cannot be

solved by factoring. We will use the quadratic formula to solve it.0 = -x2 - 2x + 1.f1x2 = -x2 - 2x + 1.

f1x2f(x) ! 0.x-intercepts

1-1, 22.f1-12 = -1-122 - 21-12 + 1 = -1 + 2 + 1 = 2.

y-coordinate:f1x2 = -x2 - 2x + 1,x-1-1.x-coordinate

x = - b2a

= - -2

21-12 = - a -2-2b = -1.

x-coordinate:ba

f(x)=–x2-2x+1

a � 1 c � 1b � 2

f1x2 = ax2 + bx + c.ca, b,

x = - b2a

.x-coordinate

a 6 0;-1.x2,a,

f1x2 = -x2 - 2x + 1.

f(x ) ! ax 2 " bx " cEXAMPLE 3

To locate the x-intercepts, weneed decimal approximations.

Thus, there is no need to simplifythe radical form of the solutions.

–b_!b2-4acx= =

2a–(–2)_!(–2)2-4(–1)(1)

=2(–1)

2_!4-(–4)–2

x= or≠–2.42+!8

–2x= ≠0.4

2-!8–2

–x2-2x+1=0

a � 1 c � 1b � 2

The are approximately and 0.4. The parabola passes throughand (0.4, 0).

Step 4 Find the by computing Replace with 0 in

The is 1, which is the constant term in the function’s equation. Theparabola passes through (0, 1).

y-intercept

f102 = -02 - 2 # 0 + 1 = 1

f1x2 = -x2 - 2x + 1.xf(0).y-intercept

1-2.4, 02 -2.4x-intercepts

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Obtaining Information about a Quadratic Function from Its Equation

Consider the quadratic function a. Determine, without graphing, whether the function has a minimum value or a

maximum value.b. Find the minimum or maximum value and determine where it occurs.c. Identify the function’s domain and its range.

Solution We begin by identifying and in the function’s equation:

a. Because the function has a maximum value.b. The maximum value occurs at

The maximum value occurs at and the maximum value ofis

We see that the maximum is at c. Like all quadratic functions, the domain is Because the function’s

maximum value is the range includes all real numbers at or below The range is

We can use the graph of to visualize the results ofExample 4. Figure 2.7 shows the graph in a by viewingrectangle. The maximum function feature verifies that the function’s maximum isat Notice that gives the location of the maximum and gives the maximumvalue. Notice, too, that the maximum value is and not the ordered pair

Check Point 4 Repeat parts (a) through (c) of Example 4 using the quadraticfunction

Applications of Quadratic FunctionsMany applied problems involve finding the maximum or minimum value of aquadratic function, as well as where this value occurs.

The Parabolic Path of a Punted Football

Figure 2.8 shows that when a football is kicked, the nearestdefensive player is 6 feet from the point of impact with thekicker’s foot. The height of the punted football, in feet,can be modeled by

where is the ball’s horizontal distance, in feet, from the pointof impact with the kicker’s foot.

a. What is the maximum height of the punt and how far fromthe point of impact does this occur?

b. How far must the nearest defensive player, who is 6 feetfrom the kicker’s point of impact, reach to block the punt?

x

f1x2 = -0.01x2 + 1.18x + 2,

f1x2,EXAMPLE 5

f1x2 = 4x2 - 16x + 1000.

11, -102.-10yxx = 1.

-103-50, 20, 1043-6, 6, 14f1x2 = -3x2 + 6x - 13

1- q , -104. -10.-10,1- q , q2.x = 1.-10

f112 = -3 # 12 + 6 # 1 - 13 = -3 + 6 - 13 = -10.

f1x2 = -3x2 + 6x - 13x = 1

x = - b2a

= - 6

21-32 = - 6

-6= -1-12 = 1.

a 6 0,

f(x)=–3x2 +6x-13.

b � 6a � 3 c � 13

ca, b,

f1x2 = -3x2 + 6x - 13.

EXAMPLE 4

[ 6, 6, 1] by [ 50, 20, 10]

Range is ( d, 10].

Figure 2.7

! Solve problems involving aquadratic function’s minimum ormaximum value.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

4

6

8

10

Hei

ght o

f the

Pun

ted

Foot

ball

(fee

t)

y

x

Distance from the Point of Impact (feet)

Figure 2.8

P-BLTZMC02_277-386-hr 19-11-2008 11:38 Page 293


Obtaining Information about a Quadratic Function from Its Equation

Consider the quadratic function a. Determine, without graphing, whether the function has a minimum value or a

maximum value.b. Find the minimum or maximum value and determine where it occurs.c. Identify the function’s domain and its range.

Solution We begin by identifying and in the function’s equation:

a. Because the function has a maximum value.b. The maximum value occurs at

The maximum value occurs at and the maximum value ofis

We see that the maximum is at c. Like all quadratic functions, the domain is Because the function’s

maximum value is the range includes all real numbers at or below The range is

We can use the graph of to visualize the results ofExample 4. Figure 2.7 shows the graph in a by viewingrectangle. The maximum function feature verifies that the function’s maximum isat Notice that gives the location of the maximum and gives the maximumvalue. Notice, too, that the maximum value is and not the ordered pair

Check Point 4 Repeat parts (a) through (c) of Example 4 using the quadraticfunction

Applications of Quadratic FunctionsMany applied problems involve finding the maximum or minimum value of aquadratic function, as well as where this value occurs.

The Parabolic Path of a Punted Football

Figure 2.8 shows that when a football is kicked, the nearestdefensive player is 6 feet from the point of impact with thekicker’s foot. The height of the punted football, in feet,can be modeled by

where is the ball’s horizontal distance, in feet, from the pointof impact with the kicker’s foot.

a. What is the maximum height of the punt and how far fromthe point of impact does this occur?

b. How far must the nearest defensive player, who is 6 feetfrom the kicker’s point of impact, reach to block the punt?

x

f1x2 = -0.01x2 + 1.18x + 2,

f1x2,EXAMPLE 5

f1x2 = 4x2 - 16x + 1000.

11, -102.-10yxx = 1.

-103-50, 20, 1043-6, 6, 14f1x2 = -3x2 + 6x - 13

1- q , -104. -10.-10,1- q , q2.x = 1.-10

f112 = -3 # 12 + 6 # 1 - 13 = -3 + 6 - 13 = -10.

f1x2 = -3x2 + 6x - 13x = 1

x = - b2a

= - 6

21-32 = - 6

-6= -1-12 = 1.

a 6 0,

f(x)=–3x2 +6x-13.

b � 6a � 3 c � 13

ca, b,

f1x2 = -3x2 + 6x - 13.

EXAMPLE 4

[ 6, 6, 1] by [ 50, 20, 10]

Range is ( d, 10].

Figure 2.7

! Solve problems involving aquadratic function’s minimum ormaximum value.

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

4

6

8

10

Hei

ght o

f the

Pun

ted

Foot

ball

(fee

t)

y

x


Figure 2.8

P-BLTZMC02_277-386-hr 19-11-2008 11:38 Page 293


c. If the ball is not blocked by the defensive player, how fardown the field will it go before hitting the ground?

d. Graph the function that models the football’s parabolicpath.

Solutiona. We begin by identifying the numbers and in the

function’s equation.

Because the function has a maximum that occurs at

This means that the maximum height of the punt occurs 59 feet from thekicker’s point of impact. The maximum height of the punt is

or 36.81 feet.b. Figure 2.8 shows that the defensive player is 6 feet from the kicker’s point of

impact. To block the punt, he must touch the football along its parabolic path.This means that we must find the height of the ball 6 feet from the kicker.Replace with 6 in the given function,

The defensive player must reach 8.72 feet above the ground to block the punt.c. Assuming that the ball is not blocked by the defensive player, we are interested

in how far down the field it will go before hitting the ground. We are lookingfor the ball’s horizontal distance, when its height above the ground, is 0feet. To find this replace with 0 in

We obtain or The equation cannot be solved by factoring. We will use the quadratic formulato solve it.

Use a calculator and round to the nearest tenth.

If the football is not blocked by the defensive player, it will go approximately119.7 feet down the field before hitting the ground.

d. In terms of graphing the model for the football’s parabolic path,we have already determined the vertex and the

approximate

vertex: (59, 36.81)The ball’s maximum height,36.81 feet, occurs at a

horizontal distance of 59 feet.

x-intercept: 119.7The ball’s maximum horizontal

distance is approximately119.7 feet.

x-intercept.f1x2 = -0.01x2 + 1.18x + 2,

Reject this value. We are interested inthe football‘s height corresponding to horizontal

distances from its point of impact onward, or x r 0.

x≠–1.7 x≠119.7

Use a calculator to evaluate the radicand.

x= =–b — !b2-4ac

2a–1.18 — !(1.18)2-4(–0.01)(2)

2(–0.01)=

–1.18 — !1.4724–0.02

x= or x=–1.18+!1.4724

–0.02–1.18-!1.4724

–0.02

-0.01x2 + 1.18x + 2 = 0.0 = -0.01x2 + 1.18x + 2,1.18x + 2.f1x2 = -0.01x2� +f1x2x-intercept,

f1x2,x,

f162 = -0.011622 + 1.18162 + 2 = -0.36 + 7.08 + 2 = 8.72f1x2 = -0.01x2 + 1.18x + 2.x

f1592 = -0.0115922 + 1.181592 + 2 = 36.81,

x = - b2a

= - 1.18

21-0.012 = -1-592 = 59

x = - b2a

.a 6 0,

f(x)=–0.01x2+1.18x+2

a � 0.01 b � 1.18 c � 2

ca, b,

–0.01x2+1.18x+2=0

a � 0.01 b � 1.18 c � 2

The equation for determining the ball’smaximum horizontal distance

0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

4

6

8

10

Hei

ght o

f the

Pun

ted

Foot

ball

(fee

t)

y

x


Figure 2.8 (repeated)

P-BLTZMC02_277-386-hr 19-11-2008 11:38 Page 294

2.2 quadratic functions - honors precalculus and...

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