Answers and Explanations 177
Answers
101–200
145.
amplitude: 1
2
; period: 2π; phase shift: − π
2 ; midline y = 0
For the function f x
A
B x C
B
D
( )
sin
=
−
( )
+ , the amplitude is A , the
period is 2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
sin
s in
=
+
( ) =
− −
( )
+
1
2
2
1
2
1
2
0
π
π
gives you the amplitude as 1
2
1
2
= ,
the period as 2
1
2
π
π
= , the phase shift as − π
2
, and the midline as y = 0.
146.
amplitude: 1
4
; period: 2; phase shift: 4
π
; midline y = 0
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period is
2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
cos
c os
= −
−
(
)= −
−
( )
+
1
4
4
1
4
4
0
π
π
π
gives you the amplitude as
− =
1
4
1
4
, the period as 2
2
π
π
= , the phase shift as 4
π
, and the midline as y = 0.
147.
amplitude: 3; period: 2; phase shift: 6
π
; midline y = 2
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period
is 2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
cos(
)
cos
= −
− = −
−
( )
+
2 3
6
3
6
2
π
π
π
gives you the amplitude as − =
3 3,
the period as 2
2
π
π
= , the phase shift as 6
π
, and the midline as y = 2.
148.
amplitude: 1; period: 4π; phase shift: –π; midline y = 1
2
For the function f x
A
B x C
B
D
( )
sin
=
−
( )
+ , the amplitude is A , the
period is 2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
sin
s in (
( ))
= −
+
(
) = −
− −
(
) +
1
2
1
2
2
1
1
2
1
2
π
π
gives you the amplitude as
− =
1 1, the period as 2
1 2
4
π
π
= , the phase shift as –π, and the midline as y = 1
2
.
Answers
101–200
145.
amplitude: 1
2
; period: 2π; phase shift: − π
2 ; midline y = 0
For the function f x
A
B x C
B
D
( )
sin
=
−
( )
+ , the amplitude is A , the
period is 2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
sin
s in
=
+
( ) =
− −
( )
+
1
2
2
1
2
1
2
0
π
π
gives you the amplitude as 1
2
1
2
= ,
the period as 2
1
2
π
π
= , the phase shift as − π
2
, and the midline as y = 0.
146.
amplitude: 1
4
; period: 2; phase shift: 4
π
; midline y = 0
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period is
2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
cos
c os
= −
−
(
)= −
−
( )
+
1
4
4
1
4
4
0
π
π
π
gives you the amplitude as
− =
1
4
1
4
, the period as 2
2
π
π
= , the phase shift as 4
π
, and the midline as y = 0.
147.
amplitude: 3; period: 2; phase shift: 6
π
; midline y = 2
For the function f x
A
B x C
B
D
( )
cos
=
−
( )
+ , the amplitude is A , the period
is 2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
cos(
)
cos
= −
− = −
−
( )
+
2 3
6
3
6
2
π
π
π
gives you the amplitude as − =
3 3,
the period as 2
2
π
π
= , the phase shift as 6
π
, and the midline as y = 2.
148.
amplitude: 1; period: 4π; phase shift: –π; midline y = 1
2
For the function f x
A
B x C
B
D
( )
sin
=
−
( )
+ , the amplitude is A , the
period is 2π
B
, the phase shift is C
B
, and the midline is y = D. Writing
f x
x
x
( )
sin
s in (
( ))
= −
+
(
) = −
− −
(
) +
1
2
1
2
2
1
1
2
1
2
π
π
gives you the amplitude as
− =
1 1, the period as 2
1 2
4
π
π
= , the phase shift as –π, and the midline as y = 1
2
.