MATH SOLVE

2 months ago

Q:
# Use the graph below to answer the question that follows:trig graph with points at 0, 0 and pi over 2, 3 and pi, 0 and 3 pi over 2, negative 3 and 2 pi, 0 and 5 pi over 2, 3 and 3 pi, 0What trigonometric function represents the graph? f(x) = −3 sin(x − pi over 2) f(x) = −3 cos(x − pi over 2) f(x) = 3 cos(x − pi over 2) f(x) = 3 sin(x − pi over 2)

Accepted Solution

A:

Note that the graph contains the point [tex]\displaystyle{ ( \frac{ \pi }{2}, 3)[/tex]

The 3 comes from the multiplication of the y-value in the parent function by.

Not considering the multiplication by 3 we would have [tex]\displaystyle{ (\frac{ \pi}{2},1)[/tex]. Also remember that the available functions are

sin(x-π/2) and cos(x-π/2).

We can check that at π/2, sin(x-π/2) is sin(π/2-π/2)=sin(0)=0,

and, π/2, cos(x-π/2) is cos(π/2-π/2)=cos(0)=1.

Thus, our function is f(x)=3cos(x-π/2).

Check for π rad: f(π)=3cos(π-π/2)=3cos(π/2)=3*0=0.

and so on...

Answer: f(x) = 3 cos(x − pi over 2)

The 3 comes from the multiplication of the y-value in the parent function by.

Not considering the multiplication by 3 we would have [tex]\displaystyle{ (\frac{ \pi}{2},1)[/tex]. Also remember that the available functions are

sin(x-π/2) and cos(x-π/2).

We can check that at π/2, sin(x-π/2) is sin(π/2-π/2)=sin(0)=0,

and, π/2, cos(x-π/2) is cos(π/2-π/2)=cos(0)=1.

Thus, our function is f(x)=3cos(x-π/2).

Check for π rad: f(π)=3cos(π-π/2)=3cos(π/2)=3*0=0.

and so on...

Answer: f(x) = 3 cos(x − pi over 2)