Question

I'm not sure how to solve these trigonometric functions, can someone help? Thank you!

Answer 1

Amplitude: 2

Period:
`\pi`

Vertical shift: -2

Horizontal shift:
`(3 \pi)/(8)`

Cycles between 0 and
`2 \pi`: 2

Equation:
`f(x) = 2 \sin (2(x - (3 \pi)/(8))) - 2`

Explanation:

Amplitude: The amplitude is half of the height of the curve. The height of the curve is
`|-4 - 0| = 4`

Period: The period is the distance between any two adjacent peaks of the curve. We see peaks at
`(5 \pi)/(8)` and
`(13 \pi)/(8)`, so the distance between them is
`(13 - 5 \pi)/(8) = \pi`

Vertical shift: The vertical shift is the difference between the vertical location of any peak and the height of the amplitude. Here that is
`0 - 2 = -2`

Horizontal shift: The horizontal shift is the x location of the first (or actually any) vertical peak of the derivative of the curve. Here we can eyeball where the tangent line has the greatest slope, and it seems to be at
`x = (3 \pi)/(8)`

Frequency: We can determine the frequency (number of cycles between 0 and
`2 \pi`) by counting how many periods are in between
`x = 0` and
`x = 2 \pi`. There are
`(2 \pi)/(\pi)` of them (length of the interval divided by the period), which simplifies to 2.

Equation of the graph: We already have all of the things we need to determine the formula of the graph. In general, the graph of an arbitrary sine wave is
`f(x) = A * \sin ( c * (x - h)) + k`

where
`A` is the amplitude,
`c` is the frequency,
`h` is the horizontal shift, and
`k` is the vertical shift. Plugging in what we've already determined, the function becomes

`f(x) = 2 \sin (2(x - (3 \pi)/(8))) - 2`