[tex]~\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[/tex]
[tex]\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} \\\\ \bullet \textit{ flips it sideways if } B\textit{ is negative}\\ ~~~~~~\textit{reflection over the y-axis}[/tex]
[tex]\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]
now keeping that template for transformations above in mind, let's take a look at this one
[tex]\stackrel{\textit{original}}{f(x) = |x|}\implies f(x)=\stackrel{A}{1}|\stackrel{B}{1}x+\stackrel{C}{0}|+\stackrel{D}{0}[/tex]
now, the vertex of this type, or original that'll be the origin at (0,0), and its' graph looks like a V.
now let's take a peek at the graph, hmmm from the original V-shape it moved to the left by 2 units, with a horizontal shift of C/B = 2, we can say that +2/1 gives B = 1 and C = +2, and moved upwards by 3 units, namely D = +3.
we can also see that it flipped upside-down over the horizontal(x-axis), namely A is negative, on all choices the only value we see for A is "4", so it must be that A = -4
let's plug all those in
[tex]\begin{cases} A=-4\\ B=1\\ C=+2\\ D=+3 \end{cases}\qquad f(x)=\stackrel{A}{-4}|\stackrel{B}{1}x+\stackrel{C}{2}|+\stackrel{D}{3}\implies f(x)=-4|x+2|+3[/tex]
