Section 11.3 Quadratic Functions and Their Graphs
The x–intercepts are
1
and 3.
Set x = 0 and solve for y to obtain the y–intercept.
2
01 4 3y
2.
2
21fx x
Since
1a
is positive, the parabola opens upward.
The vertex of the parabola is
,2,1hk
.
Replace
fx
with 0 to find x–intercepts.
2
2
021
21
x
x
3.
2
281fx x x
The x–coordinate of the vertex of the parabola is
88
2
2224
b
a
, and the y–coordinate of
the vertex of the parabola is
4.
2
41fx x x
Since
1a
is negative, the parabola opens
downward. The x–coordinate of the vertex of the
parabola is
44
2
221 2
b
a
and the
y-coordinate of the vertex of the parabola is
2
041xx
2
2
4
2
444(1)(1)
2( 1)
25
0.2 or 4.2
bb ac
xa
x
x
xx
The x–intercepts are 0.2 and 4.2.
a. Because
0,a
the function has a minimum value.
b. The minimum value occurs at
16 16 2
224 8
b
a
Chapter 11 Quadratic Equations and Functions
6.
2
0.04 2.1 6.1gx x x
Because
0,a
the function has a maximum value
that occurs at
7. Let the two numbers be represented by x and y, and
let the product be represented by P.
We must minimize .Pxy
Because the difference of the two numbers is 8, then
8.xy
Solve for y in terms of x.
8
8
xy
yx
8. Let the two dimensions be represented by x and y,
and let the area be represented by A.
We must maximize .
A
xy
Because the perimeter, P, is 120, then
22
120 2 2
Pxy
xy
Solve for y in terms of x.
Because
0,a
the function has a maximum value
that occurs at
60 30.
221
b
xa
Substitute to find the other dimension.
11.3 Concept and Vocabulary Check
1. parabola; upward; downward
2. lowest/minimum
3. highest/maximum
4.
(,)hk
3. The vertex of the graph is the point
1, 1
. This
means that the equation is
2
11jx x
.
4. Since the vertex of the graph is the point
(–1, –1), the equation is
2
11.fx x
Section 11.3 Quadratic Functions and Their Graphs
7. The vertex of the graph is the point (1, 0). This
means that the equation is
2
22
10
121
fx x
xxx
12. The vertex is (–4, –8).
13.
2
283fx x x
The x–coordinate of the vertex of the parabola is
88
2
2224
b
a
, and the y–coordinate of
14.
12 12 2
223 6
b
a
2
232 1221
b
ff
15.
2
28fx x x
The x–coordinate of the vertex of the parabola is
22
1
221 2
b
a
,
and the y–coordinate of the vertex of the parabola is
24 16 1
8161 7
The vertex is (2, 7).
17.
2
41fx x
Apply the square root property.
411
41 5 or 3
x
x
The x–intercepts are 5 and 3.
Chapter 11 Quadratic Equations and Functions
18. The vertex of the parabola is
,1,2hk
. The x–
intercepts are –0.41 and 2.41. The y–intercept is –1.
The axis of symmetry is the line x = 1.
19.
2
12fx x
Since
1a
is positive, the parabola opens upward.
The vertex of the parabola is
,1,2hk
.
The y–intercept is 3.
20. The vertex of the parabola is
,3,2hk
. There
are no x–intercepts. The y–intercept is 11. The axis
of symmetry is the line x = 3.
21.
2
2
13
31
yx
yx
Because the solutions to the equation are imaginary,
there are no x–intercepts. Set x = 0 and solve for y
to obtain the y–intercept.
22
03 1 3 19110y
Axis of symmetry: x = 3.
Range:
1,
.
22. The vertex of the parabola is
,1,3hk
. There
Range:
3,
Section 11.3 Quadratic Functions and Their Graphs
23.
2
221fx x
Since
2a
is positive, the parabola opens upward.
The vertex of the parabola is
,2,1hk
.
The x–intercepts are –1.3 and –2.7.
Set x = 0 and solve for y to obtain the y–intercept.
2
2
20 2 1
22 1 24 1 8 1 7
y
The y–intercept is 7.
24. The vertex of the parabola is
15
,,
24
hk
. The
x–intercepts are –0.62 and 1.62. The y–intercept is 1.
1
2
14
fx x
Since
1a
is negative, the parabola opens
downward. The vertex of the parabola is
,1,4hk
. Replace
fx
with 0 to find x–
intercepts.
2
2
2
014
41
41
x
x
x
Chapter 11 Quadratic Equations and Functions
26. The vertex of the parabola is
,3,1hk
. The x–
intercepts are 2 and 4. The y–intercept is –8. The
axis of symmetry is the line x = 3.
27.
2
23fx x x
Since
1a
is positive, the parabola opens upward.
The x–coordinate of the vertex of the parabola is
22
1
2212
b
a
and the
y–coordinate of the vertex of the parabola is
The x–intercepts are
3
and –1. Set
0x
and solve
for y to obtain the y–intercept.
2
02033y
28.
22
1
2212
b
a
2
112115
2
1215 16
b
ff
a
Range:
16,
29.
2
310fx x x
24
Replace
fx
with 0 to find the x–intercepts.
2
0310
052
xx
xx
Apply the zero product principle.
Section 11.3 Quadratic Functions and Their Graphs
Set x = 0 and solve for y to obtain the y–intercept.
2
03010 10y
30.
77
2224
b
a
2
77 7 81
274
24448
b
ff
a
Range:
81 ,
8
y-coordinate of the vertex of the parabola is
2
11213
2
123 4.
b
ff
a
x = 0 and solve for y to obtain the
y–intercept.
2
02033y
32.
2
2
54
45
fx x x
fx x x
44
2
221 2
b
a
2
2542 2 9
2
b
ff
a
The vertex is
2, 9 .
The x–intercepts are –5 and
1. The y–intercept is 5. The axis of symmetry is
the line x = –2.
Chapter 11 Quadratic Equations and Functions
33.
2
63fx x x
63
221
b
a
Range:
6, .
34.
2
41fx x x
42
221
b
a
The x–intercepts are 4.2 and 0.2. The y–
intercept is
1.
To find x–intercepts let 0.y
2
2
2
243
4
2
444(2)(3)
2(2)
10
12
2.6 or 0.6
fx x x
bb ac
xa
x
x
xx
The x–intercepts are 2.6 and 0.6. The y–
intercept is
3.
Section 11.3 Quadratic Functions and Their Graphs
36.
2
324fx x x
21
2233
b
a
113
33
0.9 or 1.5
x
xx
The x–intercepts are 0.9 and 1.5. The y–
intercept is
4.
37.
2
2
22
22
fx x x
fx x x
Since
1a
is negative, the parabola opens
The vertex is
1, 1 .
Replace
fx
with 0 to find
x–intercepts.
2
2
022
22
xx
xx
Axis of symmetry: x = 1.
Range:
,1
.
The vertex is
2,2 .
There are no x–intercepts.
The y–intercept is 6. The axis of symmetry is the
line x = 2.
Chapter 11 Quadratic Equations and Functions
39.
2
() 3 12 1fx x x
a. Since
0,a
the parabola opens upward and has a minimum.
40.
2
() 2 8 3fx x x
a. Since
0,a
the parabola opens upward and has a minimum.
41.
2
483fx x x
a. Since
0,a
the parabola opens downward and has a maximum.
42.
2
2123fx x x
a. Since
0,a
the parabola opens downward and has a maximum.
Section 11.3 Quadratic Functions and Their Graphs
c. Domain:
,
.
Range:
,21
.
43.
2
55fx x x
a. Since
0,a
the parabola opens upward and has a minimum.
b. The x–coordinate of the minimum is
551
225102
b
a
and the y–coordinate of the minimum is
44.
2
66fx x x
a. Since
0,a
the parabola opens upward and has a minimum.
2
45. Since the parabola opens up, the vertex
1, 2
is a minimum point.
The domain is
,
. The range is
2,
.
50.
,7,4hk
22
2274fx x h k x
Chapter 11 Quadratic Equations and Functions
51.
,10,5hk
52.
,8,6hk
53. Since the vertex is a maximum, the parabola opens
down and
3a
.
,2,4hk
54. Since the vertex is a maximum, the parabola opens
down and
3a
.
,5,7hk
2
2
2
3
35 7
357
fx x h k
x
x
56. Since the vertex is a minimum, the parabola opens
up and
3a
.
57.
2
16 64 160st t t
2
2 16 2 64 2 160
s
b.
2
2
016 64160
0410
1 4 10
tt
tt
ab c
2
45647.48
22
Evaluate the expression to obtain two solutions.
47.48 47.48
or
22
11.48 3.48
22
xx
xx
At t = 0, the ball has not yet been thrown and is
at a height of 160 feet. This is the height of the
58. a.
64 64 2
2216 32
b
ta
b.
2
2
0 16 64 200
0 4 12.5
1 4 12.5
tt
tt
ab c
c.
2
0 16 0 64 0 200
16 0 0 200 200
s
At t = 0, the ball has not yet been thrown and is
at a height of 200 feet. This is the height of the
building.
59. Let x = one of the numbers.
Let
16 x
= the other number.
The vertex is (8, 64). The maximum product is 64.
This occurs when the two number are 8 and
16 8 8
.
2
88168
64 128 64
f
The vertex is
8, 64
. The minimum product is
64
. This occurs when the two number are
8
and
816 8
.
62. Let x = the larger number. Then
24x
is the
smaller number. The product of these two numbers
Chapter 11 Quadratic Equations and Functions
63. Maximize the area of a rectangle constructed along
a river with 600 feet of fencing.
Let x = the width of the rectangle.
When the width is
150x
feet, the length is
600 2 150 600 300 300
feet.
64. From the diagram, we have that x is the width of the
rectangular plot and
200 2x
is the length. Thus,
A
65. Maximize the area of a rectangle constructed with
50 yards of fencing.
Let x = the length of the rectangle. Let y = the width
of the rectangle.
A as a function of x.
2
25 25
A
xx x x x
Since
1a
is negative, the function opens
downward and has a maximum at
66. Let x = the length of the rectangle.
Let y = the width of the rectangle.
22 80
xy
2
40 40
A
xx x x x
downward and has a maximum at
20 20 5.
222 4
b
xa
When the height x is 5, the width is
Section 11.3 Quadratic Functions and Their Graphs
68.
2
2
12 2 12 2
212
A
xx x xx
xx
69. a.
525 0.55Cx x
b.
Px Rx Cx
When the number of units x is 1225, the profit is
2
1225
0.001 1225 2.45 1225 525
P
70. a.
3000 20Cx x
Px Rx Cx
c.
980 490
221
b
xa
coordinate of the vertex of the parabola is
2
20.5
2
2 20.5 82 20.5 720
b
ff
a
Chapter 11 Quadratic Equations and Functions
78. The x–coordinate of the vertex of the parabola is
40 40 80
2 2 0.25 0.5
b
a
79.
2
420160yx x
The x–coordinate of the vertex of the parabola is
20 20 2.5
224 8
b
a
The vertex is (2.5, 185).
80. The x–coordinate of the vertex of the parabola is
40 40 4
22510
b
a
81.
2
0.01 0.6 100yxx
The x–coordinate of the vertex of the parabola is
0.6 0.6 30
b
The vertex is
30,91 .
Section 11.3 Quadratic Functions and Their Graphs
85. does not make sense; Explanations will vary.
Sample explanation: If thrown straight up, the path
will be a straight line.
88. false; Changes to make the statement true will vary.
A sample change is: The graph has no x–intercepts.
To find x–intercepts, set y = 0 and solve for x.
89. false; Changes to make the statement true will vary.
A sample change is: The x–coordinate of the
maximum is
111
221 22
b
a
and the
90.
2
325fx x
Since the vertex is
2, 5 ,
the axis of symmetry is
91.
2
32fx x
92. Start with the form
2
fx ax h k
.
Since the vertex is
,3,4hk
, we have
2
2
4134
84
a
a
93. The vertex is
,3,4hk
, so the equation is of
the form
2
2
31
fx ax h k
ax
2
a
Thus, the equation of the parabola is
2
231fx x
.
Chapter 11 Quadratic Equations and Functions
94.
34
1000 3 4
1000 3 4
1000 3
4
Pxy
xy
xy
xy
1000 3
4
x
Ax x
95. Let x = the number of trees over 50 that will be planted.
The function describing the annual yield per lemon tree when
50x
trees are planted per acre is
50 320 4
fx x x
96.
2
21 16
55 25
21 16
55 55
xx x
xx xx
Mid-Chapter Check Point
97.
22
22
22
22
11 2
44
4
11
xxx
xx
xx
98.
23
24 13 83 11
14
D
63
64 143
14 4
24 42 66
x
D
99.
2
890
(1)(9)0
uu
uu
100.
2
2100
(2)(25)0
uu
uu
20 or 2 50
uu
101.
21
33
51120
xx
Mid-Chapter Check Point – Chapter 11
1.
2
35 36
x
33
2.
2
2
527
5270
(5 7)( 1) 0
xx
xx
xx
Apply the zero-product principle.
570or 10
xx
3.
2
3620
xx
Apply the quadratic formula.
3 6 2
ab c
2
660
6
64156215315
Chapter 11 Quadratic Equations and Functions
4.
2
2
62
620
xx
xx
Apply the quadratic formula.
5.
2
2
2
5137
536
36
5
x
x
x
6.
2
580
xx
Apply the quadratic formula.
1 5 8
ab c
2
(5) (5) 4(1)(8)
2(1)
52532
x
7.
2
2
2
2260
226
13
x
x
x
1
2
x
The solution set is
1
4, 2
.
326
x
The solution set is
326
.
10.
2
14
10
x
x
Multiply both sides of the equation by the least
common denominator
2
x
.
22
14
10
xx