Question

What are the amplitude, period, phase shift, and midline of f(x) = 2 sin(x + π) − 4?

Answer 1

$\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ % templates f(x)=Asin(Bx+C)+D \\\\ f(x)=Acos(Bx+C)+D\\\\ f(x)=Atan(Bx+C)+D \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ \bullet \textit{ stretches or shrinks}\\ ~~~~~~\textit{horizontally by amplitude } A\cdot B\\\\ \bullet \textit{ flips it upside-down if }A\textit{ is negative}\\ ~~~~~~\textit{reflection over the x-axis} \\\\ \bullet \textit{ flips it sideways if }B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis}$

$\bf \bullet \textit{ horizontal shift by }\frac{C}{B}\\ ~~~~~~if\ \frac{C}{B}\textit{ is negative, to the right}\\\\ ~~~~~~if\ \frac{C}{B}\textit{ is positive, to the left}\\\\ \bullet \textit{vertical shift by }D\\ ~~~~~~if\ D\textit{ is negative, downwards}\\\\ ~~~~~~if\ D\textit{ is positive, upwards}\\\\ \bullet \textit{function period or frequency}\\ ~~~~~~\frac{2\pi }{B}\ for\ cos(\theta),\ sin(\theta),\ sec(\theta),\ csc(\theta)\\\\ ~~~~~~\frac{\pi }{B}\ for\ tan(\theta),\ cot(\theta)$

with that template in mind, let's take a peek

$\bf f(x)=2sin(x+\pi )-4\implies f(x)=\stackrel{A}{2}sin(\stackrel{B}{1}x\stackrel{C}{+\pi })\stackrel{D}{-4} \\\\[-0.35em] ~\dotfill\\\\ \textit{Amplitude}\implies 2 \\\\\\ \stackrel{phase}{\textit{Horizontal Shift}}\implies \cfrac{C}{B}\implies \cfrac{+\pi }{1}\implies +\pi \impliedby \pi \textit{ units to the left} \\\\\\ \textit{Period}\implies \cfrac{2\pi }{B}\implies \cfrac{2\pi }{1}\implies 2\pi \\\\\\ \textit{Vertical Shift}\implies D\implies -4\impliedby \textit{4 units downwards}$

now, the midline for the parent function of sin(x) is simply the x-axis, namely y = 0.

this shifted/transformed version of it, has a vertical shift of 4 units down, so the midline moved from y = 0, to y = -4.