Answer:
y=4sin[2(x−π2)]−6
Step-by-step explanation:
The standard form of a sine function is
y=asin[b(x−h)]+k
where
•a : is the amplitude,
•2π/b : is the period,
•h : is the phase shift, and
•k : is the vertical displacement.
We start with classic
y=sinx :
graph{(y-sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}
(The circle at (0,0) is for a point of reference.)
The amplitude of this function is
a=1 .To make the amplitude 4, we need
a to be 4 times as large, so we set
a=4
.
Our function is now
y=4sinx ,and looks like:
graph{(y-4sin(x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}
The period of this function—the distance between repetitions—right now is
2π , with b=1
.To make the period π , we need to make the repetitions twice as frequent, so we need
b=normal period/desired period
=2π/π=2
.
Our function is now
y=4sin(2x), and looks like: graph
{(y-4sin(2x))(x^2+y^2-0.075)=0 [-15, 15, -11, 5]}
This function currently has no phase shift, since
h=0
. To induce a phace shift, we need to offset
xby the desired amount, which in this case is
π2 to the right. A phase shift right means a positive
h, so we set
h = π2
.
Our function is now
y=4sin[2(x−π2)] , and looks like:graph
{(y-4sin(2(x-pi/2)))((x-pi/2)^2+y^2-0.075)=0 [-15, 15, -11, 5]}
Finally, the function currently has no vertical displacement, since
k=0
.To displace the graph 6 units down, we set
k=−6
.
Our function is now
y=4sin[2(x−π2)]−6, and looks like:graph {(y-4sin(2(x-pi/2))+6)((x-pi/2)^2+(y+6)^2-0.075)=0 [-15, 15, -11, 5]}
