Answer:
A and C
Step-by-step explanation:
so the final g(x) would be y=2^x+5
A because of the +5
C because there are no parenthesis and therefor cant be a horizontal shift
[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}[/tex]
[tex]\bf ~~~~~~\textit{reflection over the x-axis} \\\\ \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}\\[/tex]
[tex]\bf \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]
now, with that template in mind, let's take a look
[tex]\bf \begin{cases} f(x)=&2^x\\ g(x)=&f(x+k)\\ & f(x-5)\\ &2^{x-5} \end{cases}~\hspace{10em}g(x)=2^{\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{-5})}+\stackrel{D}{0} \\\\\\ \textit{horizontal shift by }\cfrac{C}{B}\implies \cfrac{-5}{1}\implies -5\qquad \qquad \textit{ of 5 units to the right}[/tex]
