Respuesta :
[tex]\bf ~~~~~~~~~~~~\textit{function transformations}
\\\\\\
% function transformations for trigonometric functions
% templates
f(x)=Asin(Bx+C)+D
\\\\
f(x)=Acos(Bx+C)+D\\\\
f(x)=Atan(Bx+C)+D
\\\\
-------------------[/tex]
[tex]\bf \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}[/tex]
[tex]\bf ~~~~~~\textit{reflection over the y-axis} \\\\ \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}[/tex]
[tex]\bf \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)[/tex]
with that template in mind, let's see this one.
[tex]\bf f(x)=\stackrel{A}{-4}tan(\stackrel{B}{1}x\stackrel{C}{-\pi })+\stackrel{D}{0} \\\\\\ \stackrel{period}{\cfrac{2\pi }{B}\implies \cfrac{2\pi }{1}\implies 2\pi } \\\\\\ \stackrel{phase/horizontal~shift}{\cfrac{C}{B}\implies \cfrac{-\pi }{1}}\implies -\pi \qquad \textit{shifted to the right by }\pi \textit{ units}[/tex]
[tex]\bf \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}[/tex]
[tex]\bf ~~~~~~\textit{reflection over the y-axis} \\\\ \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}[/tex]
[tex]\bf \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)[/tex]
with that template in mind, let's see this one.
[tex]\bf f(x)=\stackrel{A}{-4}tan(\stackrel{B}{1}x\stackrel{C}{-\pi })+\stackrel{D}{0} \\\\\\ \stackrel{period}{\cfrac{2\pi }{B}\implies \cfrac{2\pi }{1}\implies 2\pi } \\\\\\ \stackrel{phase/horizontal~shift}{\cfrac{C}{B}\implies \cfrac{-\pi }{1}}\implies -\pi \qquad \textit{shifted to the right by }\pi \textit{ units}[/tex]
The general form for the tangent equation is y = (a)tan(bx-c)+d where a is the amplitude, b is the period and c is the phase shift. The period of the tangent graph is pi, and the formula to solve for the period is then [tex]period= \frac{ \pi }{b} [/tex]. Our b value is 1, so the period of this tangent curve is pi. The formula for the phase shift is [tex]p.s.= \frac{c}{b} [/tex] and for us that is pi/1. So the period for this curve is pi, and the phase shift is pi units to the right.
