Download 0-3 Quadratic Functions and Equations

0-3 Quadratic Functions and Equations
Graph each equation by making a table of values.
1. f (x) = x2 + 5x + 6
SOLUTION: Sample table:
x
f (x) = x 2 + 5x + 6
2
–6
f (–6) = (–6) + 5(–6) + 6
f (x)
(x, f (x))
12
(–6, 12)
2
–4
f (–4) = (–4) + 5(–4) + 6
2
(–4, 2)
–2
f (–2) = (–2) + 5(–2) + 6
2
0
(–2, 0)
2
f (0) = 0 + 5(0) + 6
6
(0, 6)
2
20
(2, 20)
0
2
f (2) = 2 + 5(2) + 6
Graph these ordered pairs and connect the points with a smooth curve.
2. f (x) = x2 – x − 2
SOLUTION: Sample table:
2
x
f (x) = x – x − 2
2
–3
f (–3) = (–3) – ( –3) − 2
f (x)
(x, f (x))
10
( –3, 10)
2
4
( –2, 4)
2
–2
f ( –2) = (–2) – ( –2) − 2
–1
f ( –1) = (–1) – ( –1) − 2
0
( –1, 0)
2
0
f (0) = 0 – 0 − 2
–2
(0, –2)
1
f (1) = 1 – 1 − 2
2
–2
(1, –2)
2
f (2) = 2 – 2 − 2
2
0
(2, 0)
3
2
f (3) = 3 – 3 − 2
4
(3, 4)
4
f (4) = 4 – 4 − 2
2
10
(4, 10)
Graph these ordered pairs and connect the points with a smooth curve.
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3. f (x) = 2x2 + x − 3
SOLUTION: Page 1
0-3 Quadratic Functions and Equations
2. f (x) = x2 – x − 2
SOLUTION: Sample table:
2
x
f (x) = x – x − 2
2
–3
f (–3) = (–3) – ( –3) − 2
–2
f (x)
(x, f (x))
10
( –3, 10)
2
4
( –2, 4)
2
0
( –1, 0)
f ( –2) = (–2) – ( –2) − 2
–1
f ( –1) = (–1) – ( –1) − 2
0
f (0) = 0 – 0 − 2
2
–2
(0, –2)
1
2
f (1) = 1 – 1 − 2
–2
(1, –2)
2
2
f (2) = 2 – 2 − 2
0
(2, 0)
3
2
f (3) = 3 – 3 − 2
4
(3, 4)
4
2
10
(4, 10)
f (4) = 4 – 4 − 2
Graph these ordered pairs and connect the points with a smooth curve.
3. f (x) = 2x2 + x − 3
SOLUTION: Sample table:
2
x
f (x) = 2x + x − 3
2
–3
f (–3) = 2(–3) + (–3) − 3
f (x)
(x, f (x))
12
(–3, 12)
2
3
(–2, 3)
2
–2
(–1, –2)
2
–3
(0, –3)
2
0
(1, 0)
2
7
(2, 7)
–2
f (–2) = 2(–2) + (–2) − 3
–1
f (–1) = 2(–1) + (–1) − 3
0
f (0) = 2(0) + 0 − 3
1
f (1) = 2(1) + 1 − 3
2
f (2) = 2(2) + 2 − 3
Graph these ordered pairs and connect the points with a smooth curve.
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4. f (x) = 3x + 4x − 5 SOLUTION: Page 2
0-3 Quadratic Functions and Equations
3. f (x) = 2x2 + x − 3
SOLUTION: Sample table:
2
x
f (x) = 2x + x − 3
2
–3
f (–3) = 2(–3) + (–3) − 3
–2
f (x)
(x, f (x))
12
(–3, 12)
2
3
(–2, 3)
2
–2
(–1, –2)
2
–3
(0, –3)
2
0
(1, 0)
f (–2) = 2(–2) + (–2) − 3
–1
f (–1) = 2(–1) + (–1) − 3
0
f (0) = 2(0) + 0 − 3
1
f (1) = 2(1) + 1 − 3
2
2
7
(2, 7)
f (2) = 2(2) + 2 − 3
Graph these ordered pairs and connect the points with a smooth curve.
4. f (x) = 3x2 + 4x − 5 SOLUTION: Sample table:
2
x
f (x) = 3x + 4x − 5 2
–3
f (–3) = 3(–3) + 4(–3) − 5
f (x)
(x, f (x))
10
(–3, 10)
2
–1
(–2, –1)
2
–2
f (–2) = 3(–2) + 4(–2) − 5
–1
f (–1) = 3(–1) + 4(–1) − 5
–6
(–1, –6)
2
0
f (0) = 3(0) + 4(0) − 5
–5
(0, –5)
1
f (1) = 3(1) + 4(1) − 5
2
2
(1, 2)
2
f (2) = 3(2) + 4(2) − 5
2
15
(2, 15)
Graph these ordered pairs and connect the points with a smooth curve.
5. f (x) = x2 – x − 6
SOLUTION: eSolutions
Manual - Powered by Cognero
Sample table:
2
x
f (x) = x – x − 6
Page 3
f (x)
(x, f (x))
0-3 Quadratic Functions and Equations
5. f (x) = x2 – x − 6
SOLUTION: Sample table:
2
x
f (x) = x – x − 6
2
–3
f (–3) = (–3) – (–3) − 6
–2
f (x)
(x, f (x))
6
(–3, 6)
2
0
(–2, 0)
2
–4
(–1, –4)
f (–2) = (–2) – (–2) − 6
–1
f (–1) = (–1) – (–1) − 6
0
f (0) = 0 – 0 − 6
2
–6
(0, –6)
1
2
f (1) = 1 – 1 − 6
–6
(1, –6)
2
2
f (2) = 2 – 2 − 6
–4
(2, –4)
3
2
f (3) = 3 – 3 − 6
0
(3, 0)
4
2
6
(4, 6)
f (4) = 4 – 4 − 6
Graph these ordered pairs and connect the points with a smooth curve.
6. f (x) = −x2 – 3x − 1 SOLUTION: Sample table:
2
x
f (x) = −x – 3x − 1 f (x)
(x, f (x))
–11
(–5, –11)
–5
(–4, –5)
–1
(–3, –1)
1
(–2, 1)
1
(–1, 1)
–1
(0, –1)
2
–5
f (–5) = −(–5) – 3(–5) −
1 –4
f (–4) = −(–4) – 3(–4) −
1 –3
f (–3) = −(–3) – 3(–3) −
1 –2
f (–2) = −(–2) – 3(–2) −
1 –1
f (–1) = −(–1) – 3(–1) −
1 0
f (0) = −0 – 3(0) − 1 1
f (1) = −(1) – 3(1) − 1 2
–5
(1, –5)
2
2
–11
(2, –11)
2
2
2
2
2
f (2) = −(2) – 3(2) − 1 Graph these ordered pairs and connect the points with a smooth curve.
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Page 4
0-3 Quadratic Functions and Equations
6. f (x) = −x2 – 3x − 1 SOLUTION: Sample table:
2
x
f (x) = −x – 3x − 1 f (x)
(x, f (x))
–11
(–5, –11)
–5
(–4, –5)
–1
(–3, –1)
1
(–2, 1)
1
(–1, 1)
–1
(0, –1)
2
–5
f (–5) = −(–5) – 3(–5) −
1 –4
f (–4) = −(–4) – 3(–4) −
1 –3
f (–3) = −(–3) – 3(–3) −
1 –2
f (–2) = −(–2) – 3(–2) −
1 –1
f (–1) = −(–1) – 3(–1) −
1 0
f (0) = −0 – 3(0) − 1 1
f (1) = −(1) – 3(1) − 1 2
–5
(1, –5)
2
2
–11
(2, –11)
2
2
2
2
2
f (2) = −(2) – 3(2) − 1 Graph these ordered pairs and connect the points with a smooth curve.
7. BASEBALL A batter hits a baseball with an initial velocity of 80 feet per second. If the initial height of the ball is
2
3.5 feet above the ground, the function d(t) = 80t – 16t + 3.5 models the ball’s height above the ground in feet as a
function of time in seconds. Graph the function using a table of values.
SOLUTION: Sample table:
2
y
d(t) = 80t – 16t + 3.5 2
0
d(0) = 80(0) – 16(0) + 3.5
1
d(t)
(t, d(t))
3.5
(0, 3.5)
2
67.5
(1, 67.5)
2
99.5
(2, 99.5)
2
99.5
(3, 99.5)
2
67.5
(4, 67.5)
2
3.5
(5, 3.5)
d(1) = 80(1) – 16(1) + 3.5
2
d(2) = 80(2) – 16(2) + 3.5
3
d(3) = 80(3) – 16(3) + 3.5
4
5
d(4) = 80(4) – 16(4) + 3.5
d(5) = 80(5) – 16(5) + 3.5
Graph these ordered pairs and connect the points with a smooth curve.
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0-3 Quadratic Functions and Equations
7. BASEBALL A batter hits a baseball with an initial velocity of 80 feet per second. If the initial height of the ball is
2
3.5 feet above the ground, the function d(t) = 80t – 16t + 3.5 models the ball’s height above the ground in feet as a
function of time in seconds. Graph the function using a table of values.
SOLUTION: Sample table:
2
y
d(t) = 80t – 16t + 3.5 2
0
d(0) = 80(0) – 16(0) + 3.5
1
2
d(t)
(t, d(t))
3.5
(0, 3.5)
2
67.5
(1, 67.5)
2
99.5
(2, 99.5)
2
99.5
(3, 99.5)
2
67.5
(4, 67.5)
2
3.5
(5, 3.5)
d(1) = 80(1) – 16(1) + 3.5
d(2) = 80(2) – 16(2) + 3.5
3
d(3) = 80(3) – 16(3) + 3.5
4
d(4) = 80(4) – 16(4) + 3.5
5
d(5) = 80(5) – 16(5) + 3.5
Graph these ordered pairs and connect the points with a smooth curve.
Use the axis of symmetry, y–intercept, and vertex to graph each function.
8. f (x) = x2 +3x + 2
SOLUTION: Determine a, b, and c
a = 1, b = 3, and c = 2.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line
vertex is
. Since the vertex lies on the axis of symmetry, the x–coordinate of the
. Use the original equation to find the y–coordinate of the vertex.
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0-3 Quadratic Functions and Equations
Use the axis of symmetry, y–intercept, and vertex to graph each function.
8. f (x) = x2 +3x + 2
SOLUTION: Determine a, b, and c
a = 1, b = 3, and c = 2.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line
vertex is
. Since the vertex lies on the axis of symmetry, the x–coordinate of the
. Use the original equation to find the y–coordinate of the vertex.
The vertex is at
or
. The y–intercept is c or 2. The coordinates of the y–intercept are (0, 2).
The reflection of the y–intercept in the line
is .
Graph the axis of symmetry, vertex, y–intercept and its reflection. Then connect the points with a smooth curve.
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.
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0-3 Quadratic Functions and Equations
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 or 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.
10. f (x) = x2 – 2x + 1
SOLUTION: Determine a, b, and c
a = Manual
1, b = -–2,
and cby=Cognero
1.
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Use a and b to find the equation of the axis of symmetry.
Page 8
0-3 Quadratic Functions and Equations
10. f (x) = x2 – 2x + 1
SOLUTION: Determine a, b, and c
a = 1, b = –2, and c = 1.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line x = 1. Since the vertex lies on the axis of symmetry, the x–coordinate of the vertex
is 1. Use the original equation to find the y–coordinate of the vertex.
The vertex is at (1, 0). The y–intercept is c or 1. The coordinates of the y–intercept are (0, 1). The reflection of the
y–intercept in the line x = 1 is
which is (2, 1).
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
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.
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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 or –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.
12. f (x) = 2x2 – 8x − 5
SOLUTION: Determine a, b, and c
a = 2, b = –8, and c = –5.
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.
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0-3 Quadratic Functions and Equations
12. f (x) = 2x2 – 8x − 5
SOLUTION: Determine a, b, and c
a = 2, b = –8, and c = –5.
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, –13). The y–intercept is c or –5. The coordinates of the y–intercept are (0, –5). The reflection
of the y–intercept in the line x = 2 is (2 + 2, –5) which is (4, –5).
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.
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.
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The vertex is at (–2, –16). The y–intercept is c or –4. The coordinates of the y–intercept are (0, –4). The reflection
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 or –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.
14. HEALTH The normal systolic pressure P in millimeters of mercury (mm Hg) for a woman can be modeled by P
2
(x) = 0.01x + 0.05x + 107, where x is age in years.
a. Find the axis of symmetry, y–intercept, and vertex for the graph of P.
b. Graph P using the values you found in part a.
SOLUTION: a. Determine a, b, and c
a = 0.01, b = 0.05, and c = 107.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line x = –2.5. Since the vertex lies on the axis of symmetry, the x–coordinate of the
equation to find the y–coordinate of the vertex.
Page 12
vertex
is –2.5.
Use by
theCognero
original
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0-3 Quadratic Functions and Equations
14. HEALTH The normal systolic pressure P in millimeters of mercury (mm Hg) for a woman can be modeled by P
2
(x) = 0.01x + 0.05x + 107, where x is age in years.
a. Find the axis of symmetry, y–intercept, and vertex for the graph of P.
b. Graph P using the values you found in part a.
SOLUTION: a. Determine a, b, and c
a = 0.01, b = 0.05, and c = 107.
Use a and b to find the equation of the axis of symmetry.
The axis of symmetry is the line x = –2.5. Since the vertex lies on the axis of symmetry, the x–coordinate of the
vertex is –2.5. Use the original equation to find the y–coordinate of the vertex.
The vertex is at (–2.5, 106.9375). The y–intercept is c or 107.
b. The coordinates of the y–intercept are (0, 107). The reflection of the y–intercept in the line x = –2.5 is (–2.5 + (–
2.5), 107) which is (–5, 107). Find another point on the graph and it’s reflection in the axis of symmetry.
The coordinates of another point on the graph are therefore (150, 339.5). This point is 150 – (–2.5) or 152.5 units
away from the axis of symmetry, x = –2.5, so a reflection of this point in the line x = –2.5 has coordinates (–2.5 –
152.5, 339.5) or (–155, 339.5).
Graph the axis of symmetry, vertex, y–intercept and its reflection, and the additional points found. 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.
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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. 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 R. The range of the function is all real numbers less
than or equal to the maximum value
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, so the range is y ≤
, for y
R.
Page 14
Therefore, f (x) has a maximum value at
.
The domain of the function is all real numbers, so the domain is R. 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
R.
16. SOLUTION: The graph opens up, so it has a minimum value.
2
For the function f (x) = 5x + 3x – 2, a = 5, b = 3, c = –2.
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 = –0.3, the x–coordinate of the vertex is –0.3. Find the y–
coordinate of the vertex.
Therefore, f (x) has a minimum value at (–0.3, –2.45). The domain of the function is all real numbers, so the domain
is R. The range of the function is all real numbers greater than or equal to the minimum value –2.45, so the range is
y ≥ –2.45, for y R.
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.
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0-3
Therefore, f (x) has a minimum value at (–0.3, –2.45). The domain of the function is all real numbers, so the domain
is
R. The rangeFunctions
of the functionand
is allEquations
real numbers greater than or equal to the minimum value –2.45, so the range is
Quadratic
y ≥ –2.45, for y R.
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 R. The range of the function is all real numbers less
than or equal to the maximum value
, so the range is y ≤ –2
, for y
R.
18. f (x) = x2 – 5x + 6
SOLUTION: 2
For the function f (x) = x – 5x + 6, a = 1. 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 =
, the x–coordinate of the vertex is
. Find the y–coordinate
of the vertex.
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Page 16
Therefore, f (x) has a maximum value at
.
The domain of the function is all real numbers, so the domain is R. 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 ≤ –2
, for y
R.
18. f (x) = x2 – 5x + 6
SOLUTION: 2
For the function f (x) = x – 5x + 6, a = 1. 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 =
, the x–coordinate of the vertex is
. Find the y–coordinate
of the vertex.
Therefore, f (x) has a minimum value at
or .
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers
greater than or equal to the minimum value
, so the range is y ≥
, for y
R.
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.
eSolutions
Manual -fPowered
Cognero
Therefore,
(x) has by
a minimum
Page 17
value at (–1, 5).
The domain of the function is all real numbers, so the domain is R. 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 R.
Therefore, f (x) has a minimum value at
or .
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers
0-3 greater
Quadratic
Functions
and Equations
than or equal
to the minimum
value
, so the range is y ≥
, for y
R.
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 R. 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 R.
20. f (x) = 6x2 + 3x − 1
SOLUTION: 2
For the function f (x) = 6x + 3x − 1, a = 6. 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 =
, the x–coordinate of the vertex is
. Find the y–
coordinate of the vertex.
Therefore, f (x) has a minimum value at
TheManual
domain
of the function
eSolutions
- Powered
by Cognerois
.
all real numbers, so the domain is R. The range of the function is all real numbersPage 18
greater than or equal to the minimum value
, so the range is y ≥ –1 , for y
R.
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 R. 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 R.
20. f (x) = 6x2 + 3x − 1
SOLUTION: 2
For the function f (x) = 6x + 3x − 1, a = 6. 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 =
, the x–coordinate of the vertex is
. Find the y–
coordinate of the vertex.
Therefore, f (x) has a minimum value at
.
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers
greater than or equal to the minimum value
, so the range is y ≥ –1 , for y
R.
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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Page 19
Therefore, f (x) has a minimum value at
.
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers
0-3 greater
Quadratic
Functions
and Equations
than or equal
to the minimum
value
, so the range is y ≥ –1
, for y
R.
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 R. The range of the function is all real numbers less
than or equal to the maximum value
, so the range is y ≤
, for y
R.
22. f (x) = −5x2 + 10x − 6
SOLUTION: 2
For the function f (x) = −5x + 10x − 6, a = –5. 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 = 1, the x–coordinate of the vertex is 1. Find the y–coordinate of
the vertex.
Therefore, f (x) has a maximum value at (1, –1).
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers less
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Page 20
than or equal to the maximum value–1, so the range is y ≤ –1, for y R.
Solve each equation by factoring.
Therefore, f (x) has a maximum value at
.
The domain of the function is all real numbers, so the domain is R. 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
R.
22. f (x) = −5x2 + 10x − 6
SOLUTION: 2
For the function f (x) = −5x + 10x − 6, a = –5. 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 = 1, the x–coordinate of the vertex is 1. Find the y–coordinate of
the vertex.
Therefore, f (x) has a maximum value at (1, –1).
The domain of the function is all real numbers, so the domain is R. The range of the function is all real numbers less
than or equal to the maximum value–1, so the range is y ≤ –1, for y R.
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.
24. p 2 − 6p + 5 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
The solutions are 1 and 5.
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 Quadratic Functions and Equations
The solutions are 1 and 5.
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.
26. 4w2 + 19w – 5 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
The solutions are
and –5.
27. 4r2 – r = 5
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
The solutions are
and –1.
28. g 2 + 6g – 16 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
The solutions are –8 and 2.
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Solve each equation by completing the square.
29. x2 + 8x – 20 = 0
Page 22
0-3 The
Quadratic
Functions
solutions are
and –1. and Equations
28. g 2 + 6g – 16 = 0
SOLUTION: Factor the quadratic expression on the left side of the equation and then apply the Zero Product Property.
The solutions are –8 and 2.
Solve each equation by completing the square.
29. x2 + 8x – 20 = 0
SOLUTION: The solutions are –10 and 2.
30. 2a 2 + 11a – 21 = 0
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0-3 The
Quadratic
Functions
solutions are
–10 and 2. and Equations
30. 2a 2 + 11a – 21 = 0
SOLUTION: The solutions are –7 and
.
31. z 2 – 2z – 24 = 0
SOLUTION: The solutions are –4 and 6.
32. p 2 – 3p – 88 = 0
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0-3 Quadratic Functions and Equations
The solutions are –4 and 6.
32. p 2 – 3p – 88 = 0
SOLUTION: The solutions are –8 and 11.
33. t2 – 3t – 7 = 0
SOLUTION: The solutions are
and , or
.
34. 3g 2 – 12g = −4
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0-3 The
Quadratic
Functions
and Equations
solutions are
, or
and .
34. 3g 2 – 12g = −4
SOLUTION: and The solutions are
, or
.
Solve each equation by using the Quadratic Formula.
35. m2 + 12m + 36 = 0
SOLUTION: 2
In the equation m + 12m + 36 = 0 , a = 1, b = 12, and c = 36. Apply the Quadratic Formula.
36. t2 – 6t + 13 = 0
SOLUTION: 2
In the equation t – 6t + 13 = 0, a = 1, b = –6, and c = 13. Apply the Quadratic Formula.
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0-3 Quadratic Functions and Equations
36. t2 – 6t + 13 = 0
SOLUTION: 2
In the equation t – 6t + 13 = 0, a = 1, b = –6, and c = 13. Apply the Quadratic Formula.
37. 6m2 + 7m – 3 = 0
SOLUTION: 2
In the equation 6m + 7m – 3 = 0, a = 6, b = 7, and c = –3. Apply the Quadratic Formula.
38. c2 – 5c + 9 = 0
SOLUTION: 2
In the equation c – 5c + 9 = 0, a = 1, b = –5, and c = 9. Apply the Quadratic Formula.
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0-3 Quadratic Functions and Equations
38. c2 – 5c + 9 = 0
SOLUTION: 2
In the equation c – 5c + 9 = 0, a = 1, b = –5, and c = 9. Apply 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.
40. 3p 2 + 4p = 8
SOLUTION: 2
2
The equation 3p + 4p = 8 can be rewritten as 3p + 4p – 8 = 0. In this form, a = 3, b = 4, and c = –8. Apply the
Quadratic Formula.
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0-3 Quadratic Functions and Equations
40. 3p 2 + 4p = 8
SOLUTION: 2
2
The equation 3p + 4p = 8 can be rewritten as 3p + 4p – 8 = 0. In this form, a = 3, b = 4, and c = –8. Apply the
Quadratic Formula.
41. PHOTOGRAPHY Jocelyn wants to frame a 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 l times the width w of the photo, both in inches, give the area of the photo, which is 20 square inches.
l ⋅ w = 20
Since the photo is positioned so that it is the same distance x from each side of the 8–inch by 10 inch matting, l = 8 −
x − x or 8 − 2x and w = 10 − x − x or 10 − 2x. Substitute these expressions for l and w into the equation for the area
of the picture.
(8 − 2x) (10 − 2x) = 20
b. The related function is f (x)= (8 − 2x)(10 − 2x) − 20.
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(8 − 2x) (10 − 2x) = 20
b. The related function is f (x)= (8 − 2x)(10 − 2x) − 20.
0-3 Quadratic Functions and Equations
c. To find the width of the exposed part of the matting, solve the equation
Formula. In this equation, a = 4, b = –36, and c = 60.
= 0 using the Quadratic
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.
42. x2 + 5x – 6 = 0
SOLUTION: 43. a 2 – 13a + 40 = 0
SOLUTION: 44. x2 – 11x + 24 = 0
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0-3 Quadratic Functions and Equations
44. x2 – 11x + 24 = 0
SOLUTION: 45. q 2 – 12q + 36 = 0
SOLUTION: 46. –x2 + 4x – 6 = 0
SOLUTION: 47. 7x2 + 3 = 0
SOLUTION: 48. x2 – 4x + 7 = 0
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SOLUTION: 2
In the equation x – 4x + 7, a = 1, b = –4, and c = 7. Apply the Quadratic Formula.
Page 31
0-3 Quadratic Functions and Equations
48. x2 – 4x + 7 = 0
SOLUTION: 2
In the equation x – 4x + 7, a = 1, b = –4, and c = 7. Apply the Quadratic Formula.
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.
50. PETS A rectangular turtle pen is 6 feet long by 4 feet wide. The pen is enlarged by increasing the length and width
by an equal amount in order to double its area. What are the dimensions of the new pen?
SOLUTION: A rectangular turtle pen is 6 feet long by 4 feet wide. The pen is enlarged by increasing the length and width by an
equal amount in order to double its area. What are the dimensions of the new pen?
Let x be the equal amount by which both the length and width of the pen are increased. Then the length of the new
2
pen is 6 + x and the width of the new pen is 4 + x. The area of the original pen is 6(4) or 24 ft . The new pen is to
2
have an area double this measure, 2(24) or 48 ft . Therefore, an equation relating the length and width of the new
pen to its new are is (6 + x)(4 + x) = 48. Solve this equation for x.
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0-3 Quadratic Functions and Equations
50. PETS A rectangular turtle pen is 6 feet long by 4 feet wide. The pen is enlarged by increasing the length and width
by an equal amount in order to double its area. What are the dimensions of the new pen?
SOLUTION: A rectangular turtle pen is 6 feet long by 4 feet wide. The pen is enlarged by increasing the length and width by an
equal amount in order to double its area. What are the dimensions of the new pen?
Let x be the equal amount by which both the length and width of the pen are increased. Then the length of the new
2
pen is 6 + x and the width of the new pen is 4 + x. The area of the original pen is 6(4) or 24 ft . The new pen is to
2
have an area double this measure, 2(24) or 48 ft . Therefore, an equation relating the length and width of the new
pen to its new are is (6 + x)(4 + x) = 48. Solve this equation for x.
Since an increase in width of –12 feet does not make sense, x = 2. Therefore, the dimensions of the new pen are 6 +
2 or 8 ft long by 4 + 2 or 6 ft, or 8 ft by 6 ft.
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.
52. Their sum is 7 and their product is 14.
SOLUTION: Let x be the first number and 7 – x be the other number.
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Page 33
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
0-3 second
Quadratic
and
numberFunctions
is –17 – (–9) or
Therefore the two numbers are –8 and –9.
–8.Equations
52. Their sum is 7 and their product is 14.
SOLUTION: Let x be the first number and 7 – x be the other number.
Because the graph of the related function does not intersect the x–axis, this equation has no real solutions.
Therefore, no such numbers exist.
53. Their sum is −9 and their product is 24.
SOLUTION: Let x be the first number and −9 – x be the other number.
Because the graph of the related function does not intersect the x–axis, this equation has no real solutions.
Therefore, no such numbers exist.
54. Their sum is 12 and their product is −28.
SOLUTION: Let x be the first number and 12 – x be the other number.
If –2 is the first number, then the second number is 12 – (–2) or 14. If we let the first number be 14, then the second
number is 12 – 14 or –2. Therefore the two numbers are –2 and 14.
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Page 34