Question

The graph of a sinusoidal function has a minimum point at (0,3)(0,3) and then intersects its midline at (5π,5)

Write the formula of the function, where x is entered in radians.

Answer 1

the correct answer is

`f(x) = - 2 \cos( (1)/(10)x ) + 5`

Answer 2

Answer: F(x) = 2*sin(x/10 +(3/2)*pi) + 5

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