Respuesta :
[tex]\bf ~\hspace{10em}\textit{function transformations} \\\\\\ \begin{array}{llll} f(x)= A( Bx+ C)^2+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \end{array}\qquad \qquad \begin{array}{llll} f(x)=\cfrac{1}{A(Bx+C)}+D \\\\\\ f(x)= A sin\left( B x+ C \right)+ D \end{array} \\\\[-0.35em] ~\dotfill\\\\ \bullet \textit{ stretches or shrinks horizontally by } 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}\\\\ ~~~~~~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{ period of }\frac{2\pi }{ B}[/tex]
with that template in mind, let's check
C = -3, three units to the right
D = -2, two units down.
[tex]\bf f(x)=2^x~\hspace{10em}\stackrel{\stackrel{C=-3\qquad D=-2}{\cfrac{}{}}}{g(x)=2^{x-3}-2}[/tex]
Answer:
g(x) = 2^(x - 3) - 2.
Step-by-step explanation:
Translating 2^x 3 units to the right . This is done by changing it to 2^(x - 3). Then 2 units down is given by 2^(x -3) - 2.
Answer is 2^(x - 3) - 2.
