1. ## Trig graph shifts

So the problem is:
${2}{\cos{{\left({3}{x}+{\left(\frac{\pi}{{2}}\right)}\right)}}}$
Find the amplitude, period and phase shift and then graph the function.
${\left|{A}\right|}={2}$
${P}{e}{r}{i}{o}{d}={2}{\cos{{\left({3}{\left({x}+{\left(\frac{\pi}{{{3}\cdot{2}}}\right)}\right)}{\quad\text{or}\quad}{3}\to\frac{{{2}\pi}}{\omega}=\frac{{{2}\pi}}{{3}}\right.}}}$
${P}{h}{a}{s}{e}=\frac{\phi}{\omega}=\frac{{\frac{\pi}{{2}}}}{{3}}\to{\left(\frac{\pi}{{6}}\right)}$
So the answer in the book has these answers as ${A}={2}{P}{e}{r}{i}{o}{d}=\frac{{{2}\pi}}{{3}}{\quad\text{and}\quad}{p}{h}{a}{s}{e}=-\frac{\pi}{{6}}$
My question is, where did the negative come from for the phase shift?

2. Originally Posted by xsavethesporksx
So the problem is:
${2}{\cos{{\left({3}{x}+{\left(\frac{\pi}{{2}}\right)}\right)}}}$
Find the amplitude, period and phase shift and then graph the function.
${\left|{A}\right|}={2}$
${P}{e}{r}{i}{o}{d}={2}{\cos{{\left({3}{\left({x}+{\left(\frac{\pi}{{{3}\cdot{2}}}\right)}\right)}{\quad\text{or}\quad}{3}\to\frac{{{2}\pi}}{\omega}=\frac{{{2}\pi}}{{3}}\right.}}}$
${P}{h}{a}{s}{e}=\frac{\phi}{\omega}=\frac{{\frac{\pi}{{2}}}}{{3}}\to{\left(\frac{\pi}{{6}}\right)}$
So the answer in the book has these answers as ${A}={2}{P}{e}{r}{i}{o}{d}=\frac{{{2}\pi}}{{3}}{\quad\text{and}\quad}{p}{h}{a}{s}{e}=-\frac{\pi}{{6}}$
My question is, where did the negative come from for the phase shift?
A simple way to understand it is to consider this:

A parabola with equation $y = (x-3)^2$ is shifted 3 units right from the origin.

A parabola with equation $y = (x+3)^2$ is shifted 3 units left fom the origin.