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}[/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}[/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{ period of }\frac{2\pi }{ B}[/tex]
now with that template in mind
[tex]\bf h(x)=\stackrel{A}{2}|\stackrel{B}{1}x+\stackrel{C}{3}|\stackrel{D}{-1}\qquad \qquad \stackrel{\textit{C=C-2 a translation to the right}}{h(x)=2|x\boxed{+3-2}|-1} \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ~\hfill h(x)=2|x+1|-1~\hfill[/tex]
Using shifting concepts, it is found that the equation after a translation of 2 units to the right is:
d. f(x)=2|x+1|−1
The parent function is:
[tex]h(x) = 2|x + 3| - 1[/tex]
Shifting a function 2 units to the right, the equivalent function is:
[tex]f(x) = h(x - 2)[/tex]
Then:
[tex]h(x - 2) = 2|x - 2 + 3| - 1[/tex]
[tex]f(x) = 2|x + 1| - 1[/tex]
Thus, the function is:
d. f(x)=2|x+1|−1
A similar problem is given at https://brainly.com/question/24465194
