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Lesson 8.1 Exercises, pages 621–629
A
4. Complete each table of values.
a)
x
-3
-2
-1
0
1
2
3
y ⴝ ⴚ3x ⴙ 2
11
8
5
2
-1
-4
-7
y ⴝ | ⴚ3x ⴙ 2|
11
8
5
2
1
4
7
To complete the table for y ⴝ 円 ⴚ3x ⴙ 2円 , take the absolute value of
each value of y ⴝ ⴚ3x ⴙ 2.
b)
x
-4
-3
-2
y ⴝ 2x ⴚ x ⴚ 4
32
17
6
y ⴝ | 2x 2 ⴚ x ⴚ 4 |
32
17
6
2
0
1
-1 -4
-1
1
3
4
-3
2 11
24
3
2 11
24
4
2
To complete the table for y ⴝ 円 2x 2 ⴚ x ⴚ 4円 , take the absolute value
of each value of y ⴝ 2x2 ⴚ x ⴚ 4.
5. The graph of y = f (x) is given. Sketch the graph of y = ƒ f (x) ƒ .
a)
b)
y
4
4
2
0
y f (x)
y |f (x)|
2
y f (x)
x
2
y |f (x)|
2
y
x
0
2
2
2
5
The value of y is never negative, so reflect, in the x-axis, the part of
each graph that is below the x-axis.
6. Sketch a graph of each absolute value function.
a) y = ƒ -3 ƒ
b) y = ƒ 6 ƒ
4
y
y
y 6 and y |6|
y |3|
4
2
x
4
2
0
2
2
y 3
4
Graph the line y ⴝ ⴚ3.
The graph lies below the x-axis,
so to graph y ⴝ 円 ⴚ3円 , reflect
the entire graph in the x-axis.
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2
4
x
4
2
0
2
4
2
Graph the line y ⴝ 6.
The graph lies above the x-axis,
so the graph of y ⴝ 円 6円 is the same
as the graph of y ⴝ 6.
8.1 Absolute Value Functions—Solutions
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B
7. Complete each table of values, then graph the absolute value
function. Identify its intercepts, domain, and range.
a)
x
-1
-0.5
y ⴝ 4x ⴚ 2
-6
-4
6
4
y ⴝ 円 4x ⴚ 2 円
6
0
0.5
1
1.5
2
-2
0
2
4
6
2
0
2
4
6
Plot the points, then join them.
From the graph, the x-intercept is 0.5 and
the y-intercept is 2.
y |4x 2| The domain of y ⴝ 円4x ⴚ 2 円 is x ç ⺢, and
the range is y » 0.
y
4
2
x
1
1
0
2
2
b)
x
-4
-2
0
2
4
6
8
y ⴝ x ⴚ 4x ⴚ 12
20
0
-12
-16
-12
0
20
y ⴝ 円 x 2 ⴚ 4x ⴚ 12 円
20
0
12
16
12
0
20
2
24
y
16
6
2
x
2
6
10
2
8 y |x 4x 12|
0
Plot the points, then join them
with a smooth curve.
From the graph, the x-intercepts are
ⴚ2 and 6, and the y-intercept is 12.
The domain of y ⴝ 円x2 ⴚ 4x ⴚ 12円 is
x ç ⺢, and the range is y » 0.
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8. Write each absolute value function in piecewise notation.
a)
y 3x 9
y
b)
y (x 1)2 + 1
6
4
x
0
2
6
x
6
4
x-intercept: 3
y ⴝ ⴚ 3x ⴙ 9 when
ⴚ 3x ⴙ 9 » 0, or x ◊ 3
y ⴝ ⴚ(ⴚ3x ⴙ 9), or
y ⴝ 3x ⴚ 9 when
ⴚ3x ⴙ 9<0, or x>3
So, using piecewise notation:
ⴚ3x ⴙ 9, if x ◊ 3
yⴝ e
3x ⴚ 9, if x>3
2
0
2
x-intercepts: ⴚ2 and 0
The graph of y ⴝ ⴚ (x ⴙ 1)2 ⴙ 1
opens down.
ⴚ(x ⴙ 1)2 ⴙ 1 » 0 for ⴚ2 ◊ x ◊ 0.
So, for these domain values,
円 ⴚ(x ⴙ 1)2 ⴙ 1円 ⴝ ⴚ(x ⴙ 1)2 ⴙ 1.
When x<ⴚ2 or x>0, the value of
ⴚ(x ⴙ 1)2 ⴙ 1 is negative. So,
for these domain values,
円ⴚ(x ⴙ 1)2 ⴙ 1円 ⴝ ⴚ ( ⴚ(x ⴙ 1)2 ⴙ 1),
or (x ⴙ 1)2 ⴚ 1.
So, using piecewise notation:
yⴝ e
ⴚ(x ⴙ 1)2 ⴙ 1, if ⴚ2 ◊ x ◊ 0
(x ⴙ 1)2 ⴚ 1, if x<ⴚ2 or x>0
9. Write each absolute value function in piecewise notation.
a) y = ƒ -x + 2 ƒ
y ⴝ ⴚx ⴙ 2 when
ⴚx ⴙ 2 » 0
ⴚx » ⴚ2
x ◊ 2
y ⴝ ⴚ(ⴚx ⴙ 2),
or y ⴝ x ⴚ 2 when
ⴚ x ⴙ 2<0
ⴚx<ⴚ2
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b) y = ƒ 3x - 5 ƒ
y ⴝ 3x ⴚ 5 when
3xⴚ5 » 0
3x » 5
x
»
5
3
y ⴝ ⴚ (3xⴚ5),
or y ⴝ ⴚ3x ⴙ 5 when
3x ⴚ 5<0
5
3
x<
x>2
So, using piecewise notation:
So, using piecewise notation:
ⴚ x ⴙ 2, if x ◊ 2
yⴝ e
x ⴚ 2, if x>2
yⴝ c
DO NOT COPY.
3x ⴚ 5, if x
»
5
3
5
3
ⴚ3x ⴙ 5, if x<
8.1 Absolute Value Functions—Solutions
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c) y = ƒ x(x - 4) ƒ
Determine the x-intercepts of
the graph of y ⴝ x(x ⴚ 4).
x(x ⴚ 4) ⴝ 0
The x-intercepts are:
x ⴝ 0 and x ⴝ 4
The graph opens up, so
between the x-intercepts,
the graph is below
the x-axis. For the graph
of y ⴝ 円x(x ⴚ 4)円 :
For x ◊ 0 or x » 4, the
value of x(x ⴚ 4) » 0
For 0<x<4, the value of
x(x ⴚ 4)<0;
that is, y ⴝ ⴚx(x ⴚ 4).
So, using piecewise notation:
yⴝ e
x(x ⴚ 4), if x ◊ 0 or x » 4
ⴚx(x ⴚ 4), if 0<x<4
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d) y = ƒ -(x - 1)2 + 4 ƒ
Determine the x-intercepts of
the graph of y ⴝ ⴚ(x ⴚ 1)2 ⴙ 4.
ⴚ(x ⴚ 1)2 ⴙ 4 ⴝ 0
ⴚ(x ⴚ 1)2 ⴝ ⴚ 4
(x ⴚ 1)2 ⴝ 4
The x-intercepts are:
x ⴝ ⴚ1 and x ⴝ 3
The graph opens down, so between
the x-intercepts, the graph is above
the x-axis. For the graph of
y ⴝ 円 ⴚ(x ⴚ 1)2 ⴙ 4円 :
For ⴚ1 ◊ x ◊ 3, the value of
ⴚ(x ⴚ 1)2 ⴙ 4 » 0
For x<ⴚ1 or x>3, the value of
ⴚ(x ⴚ 1)2 ⴙ 4<0;
that is, y ⴝ ⴚ ( ⴚ(x ⴚ 1)2 ⴙ 4), or
y ⴝ (x ⴚ 1)2 ⴚ 4,
So, using piecewise notation:
yⴝe
ⴚ (x ⴚ 1)2 ⴙ 4, if ⴚ1 ◊ x ◊ 3
(x ⴚ 1)2 ⴚ 4, if x<ⴚ1 or x>3
10. Sketch a graph of each absolute value function.
Identify the intercepts, domain, and range.
a) y = ƒ 2x + 1 ƒ
4
b) y = ƒ - 3x + 3 ƒ
y
y
y |3x 3|
2
4
4
2
y |2x 1|
x
2
4
0
2
x
4
2
0
2
2
4
4
2
4
Draw the graph of y ⴝ 2x ⴙ 1.
It has x-intercept ⴚ0.5
Reflect, in the x-axis, the part of
the graph that is below the x-axis.
Draw the graph of y ⴝ ⴚ3x ⴙ 3.
It has x-intercept 1.
Reflect, in the x-axis, the part of
the graph that is below the x-axis.
From the graph, the x-intercept
is ⴚ0.5, the y-intercept is 1, the
domain of y ⴝ 円 2x ⴙ 1円 is x ç ⺢,
and the range is y » 0.
From the graph, the x-intercept is 1,
the y-intercept is 3, the domain of
y ⴝ 円 ⴚ 3x ⴙ 3円 is x ç ⺢, and the
range is y » 0.
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c) y = ƒ (x + 2)2 + 1 ƒ
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d) y = ƒ -2x2 + 2 ƒ
y
6
4
4
2
y |(x 6
2)2
2
y |2x 2 2|
1|
4
y
x
2
x
4
2
0
2
4
2
2
The graph of y ⴝ (x ⴙ 2)2 ⴙ 1
opens up with vertex (ⴚ2, 1), so
the graph has no
x-intercepts. It has
y-intercept 5. The graph lies
above the x-axis, so the graph of
y ⴝ 円(x ⴙ 2)2 ⴙ 1 円 is the same
as the graph of y ⴝ (x ⴙ 2)2 ⴙ 1.
From the graph, there are no
x-intercepts, the y-intercept
is 5, the domain of
y ⴝ 円 (x ⴙ 2)2 ⴙ 1 円 is x ç ⺢
and the range is y
2
0
Draw the graph of y ⴝ ⴚ2x2 ⴙ 2.
The graph opens down and has
x-intercepts 1 and ⴚ1. The vertex
is at (0, 2). Reflect, in the x-axis,
the part of the graph that is below
the x-axis.
From the graph, the x-intercepts
are ⴚ1 and 1, the y-intercept is 2.
The domain of y ⴝ 円 ⴚ 2x2 ⴙ 2円 is
x ç ⺢ and the range is y
» 0.
» 1.
11. For each description of a quadratic function y = f (x), sketch a
graph of y = ƒ f (x) ƒ . Describe your thinking.
a) the graph of y = f (x) opens up and its vertex is:
i) on the x-axis
6
y |f (x )|
ii) above the x-axis
y
6
4
4
2
2
y
y |f (x)|
x
4
2
0
2
4
x
4
2
The graph of y ⴝ f(x) is
on or above the x-axis so
the graphs of y ⴝ f(x) and
y ⴝ 円 f(x)円 are the same.
2
2
0
4
2
The graph of y ⴝ f(x) is
above the x-axis so
the graphs of y ⴝ f(x) and
y ⴝ 円f(x)円 are the same.
iii) below the x-axis
6
Part of the graph of y ⴝ f(x) is
below the x-axis so this part
of the graph is reflected in the
x-axis to get the graph of y ⴝ 円f(x)円 .
y
4
2
4
2
0
y |f (x )|
x
2
4
2
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b) the graph of y = f (x) opens down and its vertex is:
i) on the x-axis
6
y |f (x )|
ii) above the x-axis
y
6
4
4
2
2
y
y |f (x )|
x
x
4
2
0
2
4
4
2
0
2
4
2
2
The graph of y ⴝ f(x) is
on or below the x-axis so
the graph of y ⴝ f(x) is reflected
in the x-axis to get the graph
y ⴝ 円f(x)円 .
Part of the graph of y ⴝ f(x) is
below the x-axis so this part
of the graph is reflected in the
x-axis to get the graph of
of y ⴝ 円 f(x)円 .
iii) below the x-axis
The graph of y ⴝ f(x) is below the x-axis so the graph is reflected
in the x-axis to get the graph of y ⴝ 円f(x)円 .
6
y
4
2
y |f (x )|
0
2
x
4
2
4
2
12. The function y = f (x) is linear. Two points on the graph of the
absolute value function y = ƒ f (x) ƒ are (3, 2) and (-1, 0). Determine
the value of y = ƒ f (x) ƒ when x = - 3. Explain your strategy.
6
4
y
y |f (x )|
2
x
4
2
0
2
4
2
(ⴚ1, 0) is the critical point. Plot the points (3, 2) and (ⴚ1, 0) and draw a
line through them. The graph of y ⴝ 円 f(x)円 is symmetrical about the line
x ⴝ ⴚ1. So, reflect the point (3, 2) in the line x ⴝ ⴚ1, then draw a line
through (ⴚ1, 0) and the reflected point, (ⴚ5, 2), to complete the graph of
y ⴝ 円f(x)円 . From the graph, when x ⴝ ⴚ3, the value of y is 1.
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13. This graph represents the volume of water in a 12-L tank over a
period of time.
Volume of Water in a Tank
v
Volume (L)
10
8
6
4
2
t
0
2
4
6
Time (h)
8
a) Write the function represented by this graph in piecewise
notation. Let v represent the volume in litres and t represent the
time in hours.
Equation of line segment for 0 ◊ t ◊ 4:
v-intercept: 12
t-intercept: 4
Equation has form: v ⴝ mt ⴙ 12
Use the points (0,12) and (4, 0) to determine the slope.
mⴝ
0 ⴚ 12
4ⴚ0
ⴝ ⴚ3
So, equation is: v ⴝ ⴚ 3t ⴙ 12
Equation of line segment for 4<t
Passes through: (4, 0) and (8, 12)
Equation has form: v ⴝ mt ⴙ b
Slope: m ⴝ
◊ 8:
12 ⴚ 0
8ⴚ4
ⴝ3
Use: v ⴝ 3t ⴙ b
Substitute: t ⴝ 4, v ⴝ 0
0 ⴝ 3(4) ⴙ b
b ⴝ ⴚ12
So, equation is: v ⴝ 3t ⴚ 12
So, using piecewise notation:
vⴝ e
ⴚ 3t ⴙ 12, if 0 ◊ t
3t ⴚ 12,
if 4<t
◊4
◊8
b) Which absolute value function is represented by this graph?
How do you know?
The equation of the absolute value function is the absolute value of the
left segment or the absolute value of the right segment:
v ⴝ 円 ⴚ3t ⴙ 12 円 or v ⴝ 円 3t ⴚ 12 円
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c) Determine the volume of water in the tank after 5.5 h.
What strategy did you use? What other strategy could you use?
Substitute t ⴝ 5.5 in v ⴝ 円3t ⴚ 12円 :
v ⴝ 円3(5.5)ⴚ12 円
v ⴝ 円4.5円
v ⴝ 4.5
The volume of water in the tank is 4.5 L.
I could also use the graph to estimate the volume of water after 5.5 h,
but this would be an approximate value.
C
14. Two linear functions y = f (x) and y = h(x) have opposite slopes
and the same x-intercept.
How will the graphs of y = ƒ f (x) ƒ and y = ƒ h(x) ƒ compare?
Include sketches in your explanation.
Both functions are linear so use the slope-point
y
y |2x 4| 6
form of a line.
y |2x 4|
y ⴝ f(x): y ⴚ 0 ⴝ m(x ⴚ a) : y ⴝ m(x ⴚ a)
4
y ⴝ g(x): y ⴚ 0 ⴝ ⴚm(x ⴚ a) : y ⴝ ⴚ m(x ⴚ a)
2
Since 円 ⴚm(x ⴚ a)円 ⴝ 円m(x ⴚ a) 円 , the graphs
of y ⴝ 円f(x)円 and y ⴝ 円h(x)円 are the same.
4 2 0
For example, the graphs of y ⴝ 2x ⴚ 4
y 2x 4
and y ⴝ ⴚ2x ⴙ 4 have the same x-intercepts,
but opposite slopes.
To graph the absolute value of each function,
I reflect the part of the graph below the x-axis
in the x-axis. So, the graphs of y ⴝ 円2x ⴚ 4円 and
y ⴝ 円ⴚ2x ⴙ 4円 will be the same.
x
4
y 2x 4
15. Write an equation for each absolute value function. Explain your
strategy.
a)
4
y
x
6
4
2
0
Determine the equation of the line on the right:
y-intercept: 3
x-intercept: ⴚ3
Equation has form: y ⴝ mx ⴙ 3
Use the points (0, 3) and (ⴚ3, 0) to determine the slope.
mⴝ
0ⴚ 3
ⴚ3 ⴚ 0
ⴝ1
So, an equation is: y ⴝ x ⴙ 3 and an equation for the absolute value
function is: y ⴝ 円x ⴙ 3 円
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b)
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y
8
4
2
x
2
0
2
4
Either the middle piece of the graph or the two end pieces were
reflected in the x-axis. I will assume the middle piece was reflected. So,
the graph of the quadratic function opens up and has vertex (1, ⴚ9).
So, the equation has the form y ⴝ a(x ⴚ 1)2 ⴚ 9.
Use the point (4, 0) to determine a. Substitute x ⴝ 4 and y ⴝ 0.
0 ⴝ a(4ⴚ1)2 ⴚ9
9 ⴝ 9a
aⴝ1
So, an equation of the quadratic function is y ⴝ (x ⴚ 1)2 ⴚ 9 and an
equation of the absolute value function is y ⴝ 円(x ⴚ 1)2 ⴚ 9円 .
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8.1 Absolute Value Functions—Solutions
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