Study Guide and Review - Chapter 4
State the amplitude, period, frequency, phase
shift, and vertical shift of each function. Then
graph two periods of the function.
44. SOLUTION: 43. In this function, a = 3, b = 1, c = SOLUTION: In this function, a =
, b = 1, c = , and d = 0.
, and d = 0.
Because d = 0, there is no vertical shift.
Because d = 0, there is no vertical shift.
Graph y = 3 sin x shifted
units to the left.
Graph y =
cos x shifted
units to the left.
Locate the vertical asymptotes, and sketch the
graph of each function.
45. y = 3 tan x
SOLUTION: The graph of y = 3 tan x is the graph of y = tan x
44. expanded vertically. The period is
SOLUTION: In this function, a = 3, b = 1, c = or . Find
the location of two consecutive vertical asymptotes.
, and d = 0.
Because d = 0, there is no vertical shift.
and Create a table listing the coordinates of key points
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for y = 3 tan x for one period on
Function
.
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Study Guide and Review - Chapter 4
Locate the vertical asymptotes, and sketch the
graph of each function.
45. y = 3 tan x
51. y = sec (x –
)
SOLUTION: SOLUTION: The graph of y = sec (x −
The graph of y = 3 tan x is the graph of y = tan x
x translated π units to the right. The period is
expanded vertically. The period is
or . Find
) is the graph of y = sec
or 2 . Find the location of two vertical asymptotes.
the location of two consecutive vertical asymptotes.
and and Create a table listing the coordinates of key points
Create a table listing the coordinates of key points
for y = 3 tan x for one period on
.
for y = sec (x − π) for one period on
Function
Function
y = tan x
Vertical
Asymptote
Intermediate
Point
x-int
(0, 0)
Vertical
Asymptote
Intermediate
Point
x-int
y = sec x
y = sec (x −
π)
(0, 1)
(0, 0)
Intermediate
Point
Vertical
Asymptote
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
51. y = sec (x –
y = 3 tan x
Intermediate
Point
Vertical
Asymptote
.
Sketch the curve through the indicated key points for
the function. Then repeat the pattern to sketch at
least one more cycle to the left and right of the first
curve.
)
Find the exact value of each expression, if itPage 2
exists.
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SOLUTION: The graph of y = sec (x −
) is the graph of y = sec
53. sin−1 (−1)
= Study Guide and Review - Chapter 4
Find the exact value of each expression, if it
exists.
.
57. arctan −1
SOLUTION: 53. sin−1 (−1)
Find a point on the unit circle on the interval
SOLUTION: such that Find a point on the unit circle on the interval
=
with a y-coordinate of –1.
When t =
When t =
, sin t = –1. Therefore, sin
–1
, tan t =
. Therefore, arctan
–1 =
= .
.
59. 55. SOLUTION: SOLUTION: The inverse property applies, because
lies on Find a point on the unit circle on the interval
such that =
the interval [–1, 1]. Therefore,
=
.
When t =
, tan t =
= . Therefore, tan
–1
.
57. arctan −1
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SOLUTION: Find a point on the unit circle on the interval
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