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}\\ ~~~~~~\textit{reflection over the y-axis}[/tex]

[tex]\bf \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,

down 4 units, D = -4

flipped over the y-axis, B = -1

[tex]\bf y=x^2\implies y=\stackrel{A}{1}(\stackrel{B}{1}x\stackrel{C}{+0})^2\stackrel{D}{+0}\qquad \qquad \stackrel{shifted}{y=\stackrel{A}{1}(\stackrel{B}{-1}x\stackrel{C}{+0})^2\stackrel{D}{-4}} \\\\\\ y=-x^2-4[/tex]
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