0-3 Quadratic Functions and Equations
Use the axis of symmetry, y–intercept, and vertex to graph each function.
9. f (x) = x2 – 9x + 8
SOLUTION: Determine a, b, and c
a = 1, b = –9, and c = 8.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line
is
. Since the vertex lies on the axis of symmetry, the x–coordinate of the vertex
. Use the original equation to find the y–coordinate of the vertex.
The vertex is at
or
The reflection of the y–intercept in the line
. The y–intercept is c = 8. The coordinates of the y–intercept are (0, 8).
is which is (9, 8).
Graph the axis of symmetry, vertex, y–intercept and its reflection. Then connect the points with a smooth curve.
11. f (x) = x2 – 6x − 16
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SOLUTION: Determine a, b, and c
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0-3 Quadratic Functions and Equations
11. f (x) = x2 – 6x − 16
SOLUTION: Determine a, b, and c
a = 1, b = –6, and c = –16.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line x = 3. Since the vertex lies on the axis of symmetry, the x–coordinate of the vertex
is 3. Use the original equation to find the y–coordinate of the vertex.
The vertex is at (3, –25). The y–intercept is c = –16. The coordinates of the y–intercept are (0, –16). The reflection
of the y–intercept in the line x = 3 is (3 + 3, –16) which is (6, –16).
Graph the axis of symmetry, vertex, y–intercept and its reflection. Then connect the points with a smooth curve.
13. f (x) = 3x2 + 12x − 4
SOLUTION: Determine a, b, and c
a = 3, b = 12, and c = –4.
Use a and b to find the equation of the axis of symmetry.
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0-3 Quadratic Functions and Equations
13. f (x) = 3x2 + 12x − 4
SOLUTION: Determine a, b, and c
a = 3, b = 12, and c = –4.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line x = –2. Since the vertex lies on the axis of symmetry, the x–coordinate of the
vertex is –2. Use the original equation to find the y–coordinate of the vertex.
The vertex is at (–2, –16). The y–intercept is c = –4. The coordinates of the y–intercept are (0, –4). The reflection
of the y–intercept in the line x = –2 is (–2 + (–2), –4) which is (–4, –4).
Graph the axis of symmetry, vertex, y–intercept and its reflection. Then connect the points with a smooth curve.
Determine whether each function has a maximum or minimum value. Then find the value of the
maximum or minimum, and state the domain and range of the function.
15. SOLUTION: The graph opens down, so it has a maximum value.
2
For Manual
the function
f (x)by=Cognero
–2x + 3x – 5, a = –2, b = 3, c = –5.
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The maximum value of the function is the y–coordinate of the vertex. Find the equation of the axis of symmetry.
0-3 Quadratic Functions and Equations
Determine whether each function has a maximum or minimum value. Then find the value of the
maximum or minimum, and state the domain and range of the function.
15. SOLUTION: The graph opens down, so it has a maximum value.
2
For the function f (x) = –2x + 3x – 5, a = –2, b = 3, c = –5.
The maximum value of the function is the y–coordinate of the vertex. Find the equation of the axis of symmetry.
Because the equation of the axis of symmetry is x =
, the x–coordinate of the vertex is
. Find the y–coordinate
of the vertex.
Therefore, f (x) has a maximum value at
.
The domain of the function is all real numbers, so the domain is . The range of the function is all real numbers less
than or equal to the maximum value
, so the range is y ≤
, for y
.
17. f (x) = −x2 + 3x − 5
SOLUTION: eSolutions Manual - Powered by Cognero
2
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For the function f (x) = −x + 3x − 5, a = –1. Because a is negative, the graph opens down and the function has a
maximum value.
The domain of the function is all real numbers, so the domain is . The range of the function is all real numbers less
or equal toFunctions
the maximumand
valueEquations
, so the range is y ≤
0-3 than
Quadratic
, for y
.
17. f (x) = −x2 + 3x − 5
SOLUTION: 2
For the function f (x) = −x + 3x − 5, a = –1. Because a is negative, the graph opens down and the function has a
maximum value.
The maximum value of the function is the y–coordinate of the vertex. To find the vertex, first find equation of the
axis of symmetry.
Because the equation of the axis of symmetry is x =
, the x–coordinate of the vertex is
. Find the y–coordinate
of the vertex.
Therefore, f (x) has a maximum value at
.
The domain of the function is all real numbers, so the domain is . The range of the function is all real numbers less
than or equal to the maximum value
, so the range is y ≤
, for y
.
19. f (x) = 2x2 + 4x + 7
SOLUTION: 2
For the function f (x) = 2x + 4x + 7, a = 2. Because a is positive, the graph opens up and the function has a
minimum value.
The minimum value of the function is the y–coordinate of the vertex. To find the vertex, first find equation of the
axis of symmetry.
Because
equation
of the axis
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of the vertex.
of symmetry is x = –1, the x–coordinate of the vertex is –1. Find the y–coordinate
Page 5
The domain of the function is all real numbers, so the domain is . The range of the function is all real numbers less
0-3 than
Quadratic
or equal toFunctions
the maximumand
valueEquations
, so the range is y ≤
, for y
.
19. f (x) = 2x2 + 4x + 7
SOLUTION: 2
For the function f (x) = 2x + 4x + 7, a = 2. Because a is positive, the graph opens up and the function has a
minimum value.
The minimum value of the function is the y–coordinate of the vertex. To find the vertex, first find equation of the
axis of symmetry.
Because the equation of the axis of symmetry is x = –1, the x–coordinate of the vertex is –1. Find the y–coordinate
of the vertex.
Therefore, f (x) has a minimum value at (–1, 5).
The domain of the function is all real numbers, so the domain is . The range of the function is all real numbers
greater than or equal to the minimum value 5, so the range is y ≥ 5, for y .
21. f (x) = −3x2 – 2x − 1
SOLUTION: 2
For the function f (x) = −3x – 2x − 1, a = –3. Because a is negative, the graph opens down and the function has a
maximum value.
The maximum value of the function is the y–coordinate of the vertex. To find the vertex, first find equation of the
axis of symmetry.
Because the equation of the axis of symmetry is x =
, the x–coordinate of the vertex is
. Find the y–
coordinate of the vertex.
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0-3
Therefore, f (x) has a minimum value at (–1, 5).
The
domain of Functions
the function is and
all realEquations
numbers, so the domain is . The range of the function is all real numbers
Quadratic
greater than or equal to the minimum value 5, so the range is y ≥ 5, for y .
21. f (x) = −3x2 – 2x − 1
SOLUTION: 2
For the function f (x) = −3x – 2x − 1, a = –3. Because a is negative, the graph opens down and the function has a
maximum value.
The maximum value of the function is the y–coordinate of the vertex. To find the vertex, first find equation of the
axis of symmetry.
Because the equation of the axis of symmetry is x =
, the x–coordinate of the vertex is
. Find the y–
coordinate of the vertex.
Therefore, f (x) has a maximum value at
.
The domain of the function is all real numbers, so the domain is . The range of the function is all real numbers less
than or equal to the maximum value
, so the range is y ≤
, for y
.
Solve each equation by factoring.
23. x2 − 10x + 21 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
The solutions are 3 and 7.
25. x2 – 3x – 28 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
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0-3 The
Quadratic
Functions
solutions are
3 and 7. and Equations
25. x2 – 3x – 28 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
The solutions are –4 and 7.
27. 4r2 – r = 5
SOLUTION: Subtract the 5 from both sides. Then factor the quadratic expression on the left side of the equation and then apply
the Zero Product Property.
The solutions are
and –1.
Solve each equation by completing the square.
29. x2 + 8x – 20 = 0
SOLUTION: The solutions are –10 and 2.
31. z 2 – 2z – 24 = 0
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0-3 The
Quadratic
Functions
solutions are
–10 and 2. and Equations
31. z 2 – 2z – 24 = 0
SOLUTION: The solutions are –4 and 6.
33. t2 – 3t – 7 = 0
SOLUTION: and The solutions are
, or
.
Solve each equation by using the Quadratic Formula.
39. 4x2 – 2x + 9 = 0
SOLUTION: 2
In the equation 4x – 2x + 9 = 0, a = 4, b = –2, and c = 9. Apply the Quadratic Formula.
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0-3 The
Quadratic
Functions
and Equations
solutions are
, or
and .
Solve each equation by using the Quadratic Formula.
39. 4x2 – 2x + 9 = 0
SOLUTION: 2
In the equation 4x – 2x + 9 = 0, a = 4, b = –2, and c = 9. Apply the Quadratic Formula.
41. PHOTOGRAPHY Jocelyn wants to frame a cropped photograph that has an area of 20 square inches with a
uniform width of matting between the photograph and the edge of the frame as shown.
a. Write an equation to model the situation if the length and width of the matting must be 8 inches by 10 inches,
respectively, to fit in the frame.
b. Graph the related function.
c. What is the width of the exposed part of the matting x?
SOLUTION: a. The length
⋅ w = 20
times the width w of the photo, both in inches, give the area of the photo, which is 20 square inches.
Since the photo is positioned so that it is the same distance x from each side of the 8-inch by 10-inch matting,
− 2x and w = 10 − 2x.
=8
Substitute these expressions for
(8 − 2x) (10 − 2x) = 20
and w into the equation for the area of the picture.
b. The related function is f (x)= (8 − 2x)(10 − 2x) − 20.
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b. The related function is f (x)= (8 − 2x)(10 − 2x) − 20.
0-3 Quadratic Functions and Equations
2
c. To find the width of the exposed part of the matting, solve the equation 4x – 36x + 60 = 0 using the Quadratic
Formula. In this equation, a = 4, b = –36, and c = 60.
If x ≈ 6.79, then the width of the photo would be 10 – 2(6.79) or –3.58 inches, which is not possible. If x ≈ 2.21, then
the width of the photo would be 10 – 2(2.21) or 5.58 inches, which is reasonable. Therefore, the width of the
exposed part of the matting is about 2.21 inches.
Solve each equation.
43. a 2 – 13a + 40 = 0
SOLUTION: 45. q 2 – 12q + 36 = 0
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47. 7x2 + 3 = 0
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0-3 Quadratic Functions and Equations
45. q 2 – 12q + 36 = 0
SOLUTION: 47. 7x2 + 3 = 0
SOLUTION: 49. 2x2 + 6x – 3 = 0
SOLUTION: 2
In the equation 2x + 6x – 3 = 0, a = 2, b = 6, and c = –3. Apply the Quadratic Formula.
NUMBER THEORY Use a quadratic equation to find two real numbers that satisfy each situation or
show that no such numbers exist.
51. Their sum is −17 and their product is 72.
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SOLUTION: Let x be the first number and –17 – x be the other number.
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0-3 Quadratic Functions and Equations
NUMBER THEORY Use a quadratic equation to find two real numbers that satisfy each situation or
show that no such numbers exist.
51. Their sum is −17 and their product is 72.
SOLUTION: Let x be the first number and –17 – x be the other number.
If –8 is the first number, then the second number is –17 – (–8) or –9. If we let the first number be –9, then the
second number is –17 – (–9) or –8. Therefore the two numbers are –8 and –9.
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