Graphing Exercise
1. For the given function f , find its amplitude and period, and graph the function. On
your graph, for at least one cycle, indicate the x−coordinate of where the the maximum
and minimum value(s) occurs, and where the function intercepts its neutral position.
a. f (x) = −2 sin (x + 1).
Ans: Start Point: x = −1; amplitude = 2; period = 2π; End Point: x = −1 + 2π; neutral
π
3π
position at x = π; min at x = ; max at x =
;
2
2
b. f (x) = sin (−x + 1).
Ans: Start Point: x = 1; amplitude = 1; period = 2π; End Point: x = 1 + 2π; neutral
π
3π
position at x = π; min at x = ; max at x =
;
2
2
c. f (x) = 2 sin (πx + 1).
1
1
Ans: Start Point: x = − ; amplitude = 2; period = 2; End Point: x = − + 2; neutral
π
π
1 3
1 1
1
position at x = − + 1; min at x = − + ; max at x = − + ;
π
π 2
π 2
d. f (x) = −2 sin (2x − π) − 3.
π
π
3π
Ans: Start Point: x = ; amplitude = 2; period = π; End Point: x = + π =
; neutral
2
2
2
π π
π π
3π
5π
position at x = + = π; min at x = + =
; max at x =
;
2 2
2 4
4
4
x
e. f (x) = sin
−1 .
2
Ans: Start Point: x = 2; amplitude = 1; period = 4π; End Point: x = 2 + 4π; neutral
position at x = 2 + 2π; min at x = 2 + 3π; max at x = 2 + π;
!
x
f. f (x) = sin − + 1 − 1.
π
Ans: Start Point: x = π; amplitude = 1; period = 2π 2 ; End Point: x = π + π 2 ; neutral
π2
3π 2
π2
;
position at x = π + ; min at x = π + ; max at x = π +
2
4
4
!
πx
g. f (x) = 2 sin
+2 .
2
!
4
4
Ans: Start Point: x = − ; amplitude = 2; period = 4; End Point: x = − + 4; neutral
π
π
4
4
4
position at x = − + 2; min at x = − + 3; max at x = − + 1;
π
π
π
h. f (x) = cos (−x + 1) + 2.
Ans: Start Point: x = 1; amplitude = 1; period = 2π; End Point: x = 1 + 2π; neutral
3π
π
positions at x = 1 + and x = 1 + ; min at x = 1 + π;
2
2
πx
i. f (x) = −2 cos
−π .
3
Ans: Start Point: x = 3; amplitude = 2; period = 6; End Point: x = 9; neutral positions
9
15
at x = and x = ; max at x = 6
2
2
!
j. f (x) = cos (−x − 2) − 1.
Ans: Start Point: x = −2; amplitude = 1; period = 2π; End Point: x = 2 + 2π; neutral
3π
π
positions at x = −2 + and x = −2 + ; min at x = −2 + π;
2
2
2x
k. f (x) = cos
− 1 + 2.
3
3
3
Ans: Start Point: x = ; amplitude = 1; period = 3π; End Point: x = + 3π; neutral
2
2
3 3π
3 9π
3 + 3π
positions at x = +
and at x = + ; min at x =
;
2
4
2
4
2
!
x
l. f (x) = − cos − + 1 .
2
Ans: Start Point: x = 2; amplitude = 1; period = 4π; End Point: x = 2 + 4π; neutral
positions at x = 2 + π and x = 2 + 3π; max at x = 2 + 2π;
!
m. f (x) = cos (−3x + 1).
1
2π
1 2π
Ans: Start Point: x = ; amplitude = 1; period =
; End Point: x = +
; neutral
3
3
3
3
1 π
1+π
1 π
;
positions at x = + and x = + ; min at x =
3 6
3 2
3
n. f (x) = cos (x − π).
Ans: Start Point: x = π; amplitude = 1; period = 2π; End Point: x = 3π; neutral positions
3π
5π
at x =
and x =
; min at x = 2π;
2
2