Respuesta :
[tex]\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ f(x)=Asin(Bx+C)+D \qquad \qquad f(x)=Acos(Bx+C)+D \\\\ f(x)=Atan(Bx+C)+D \qquad \qquad f(x)=Asec(Bx+C)+D \\\\---------------------------------\\\\ \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}[/tex]
[tex]\bf \bullet \textit{ flips it sideways if }B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }\frac{C}{B}\\ ~~~~~~if\ \frac{C}{B}\textit{ is negative, to the right}\\\\[/tex]
[tex]\bf ~~~~~~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)[/tex]
so, with that template in mind,
[tex]\bf f(x)=\stackrel{A}{-3}cos(\stackrel{B}{2}x\stackrel{C}{-\pi })\stackrel{D}{+2}~~ \begin{cases} C=-\pi \\ B=2\\ \frac{C}{B}=-\frac{\pi }{2}&\qquad \stackrel{\textit{horizontal shift}}{\frac{\pi }{2}\textit{ to the right}}\\[-0.5em] \hrulefill\\ D=+2&\qquad \stackrel{\textit{vertical shift}}{2\textit{ upwards}} \end{cases}[/tex]