Question

How do you find the amplitude, period and graph `y=3cos(1/2theta)`?

Answer 1

The amplitude is `3`, the period is `4pi`, and the graph is shown below.

Explanation:

This is a graph in the form `y=acos(b(theta-h))+k` where `a=3`, `b=1/2`, and `h=k=0`

For a graph in this form, the amplitude is `|a|` and the period is `(2pi)/|b|`. In this case, the amplitude is `|3|=3` and the period is `(2pi)/|1/2|=4pi`.

To graph it, it is helpful to find certain easy-to-find points. Because `h=k=0`, there is no horizontal or vertical shift, so we can start with a regular cosine curve.
graph{cos(x) [-10, 10, -5, 5]}

Looking at the graph, the period is `2pi` and the graph is maximum at the start of the period, minimum halfway through the period, and `0` one quarter and three quarters through. Keeping these rules in mind but expanding the period to the one we want, `4pi`, we get:
graph{cos(.5x) [-10, 10, -5, 5]}

Currently the amplitude of the graph is `1`, but we need it to be 3 so we expand the maximum and minimum from `1` and `"-"1` to `3` and `"-"3`:
graph{3cos(.5x) [-10, 10, -5, 5]}

This process is equivalent to performing the appropriate transformations to the parent cosine graph: a vertical stretch by 3 and a horizontal stretch by 2.