# Can you graph?

Mar 31, 2018

$\frac{3}{4} y = \frac{2}{3} \cos \left(\frac{3}{5} \theta\right)$

We have to know what the cosine graph looks

$\cos \left(\theta\right)$

Min~-1
Max~1
Period= $2 \pi$
Amplitude=1

graph{cos(x) [-10, 10, -5, 5]}

Translation form is $f \left(x\right) = A \cos \left[B \left(x - C\right)\right] + D$

A~ Horizontal stretch, amplitude streches by A
B~ Vertical stretch, Period stretches by $\frac{1}{B}$
C~Vertical translation, x values move over by C
D~Horizontal translation, y values move up by D

But this can't help us until we have y by itself so multiply both sides by $\frac{4}{3}$ to get rid of it from the LHS (left hand side)

$y = \frac{4}{3} \cdot \frac{2}{3} \cos \left(\frac{2}{3} \theta\right)$

$y = \frac{8}{9} \cos \left(\frac{2}{3} \theta\right)$

So the 2/3 is the vertical stretch and it stretches the period by 3/2 so the new period is $3 \pi$

The $\frac{8}{9}$ is the horizontal stretch so the amplitude is $\frac{8}{9}$ so the max is $\frac{8}{9}$ and the min is $- \frac{8}{9}$

graph{8/9cos(2/3x) [-10, 10, -5, 5]}