Answer: F(x) = 2*sin(x/10 +(3/2)*pi) + 5
Step-by-step explanation:
The information that we have is that:
We have a minimum at (0, 3)
the midline is at (5*pi, 5)
This is a sinusoidal function, so we can write one generic one as:
F(x) = A*sin(c*x + p) + B.
where A and B are constants, c is the frequency and p is a phase
First, the minimum of the sine function is when sin(x) = -1, and this happens at (3/2)*pi
We know that this minimum is at x = 0.
sin(c*0 + p) = -1
Then p = 3/2*pi.
So our function is:
F(x) = A*sin(c*x +(3/2)*pi) + B.
Now, we know that F(0) = 3, so:
3 = A*sin(c*0 +(3/2)*pi) + B = -A + B.
now we can use the other hint, the midpoint of the sine function is when sin(x) = 0, and this happens at x = 0 and x = pi, particularlly as we here have a phase of 3/2*pi, we should find x = 2*pi.
then:
c*5*pi + (3/2)*pi = 2*pi
c*5 + 3/2 = 2
c*5 = 2 - 3/2 = 1/2
C = 1/2*5 = 1/10
So our function is
F(x) = A*sin(x/10 +(3/2)*pi) + B
and we know that when x = 5*pi, F(5*pi) = 5, so:
5 = F(x) = A*sin(5*pi/10 +(3/2)*pi) + B
5 = B
and we aready knew that:
- A + B = 3
-A + 5 = 3
A = 5 - 3 = 2
So our equation is:
F(x) = 2*sin(x/10 +(3/2)*pi) + 5
