Respuesta :

[tex]\bf ~~~~~~~~~~~~\textit{function transformations} \\\\\\ % templates f(x)= A( Bx+ C)+ D \\\\ ~~~~y= A( Bx+ C)+ D \\\\ f(x)= A\sqrt{ Bx+ C}+ D \\\\ f(x)= A(\mathbb{R})^{ Bx+ C}+ D \\\\ f(x)= A sin\left( B x+ C \right)+ D \\\\ --------------------[/tex]

[tex]\bf \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}[/tex]

[tex]\bf ~~~~~~\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]

and now with that template in mind,

bearing in mind the parent function for  both is just y = x², or y = 1(1x + 0)² + 0.

[tex]\bf y=\stackrel{A}{\cfrac{1}{2}}(x\stackrel{C}{+4})^2[/tex]

is x² but expanded by twice as much vertically, and has a horizontal shift of +4, namely 4 units to the left.

y=(x-1)² - 3

is just x² but with a horizontal shfit of -1, namely 1 unit to the right, and a vertical shift of -3, namely 3 units down.