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 with that template in mind,

notice from a parent function of   f(x) = x³,
a derived function with f(x+5) = (x + 5)³

has a C component of 5, C = 5, which means is the as the parent, just shifted to the left by 5 units.
to move a function to the left by c units, add c to every xso f(x) shifted to the left by 5 units is f(x+5)
for [tex]f(x)=x^3[/tex], the shifted graph would be [tex]f(x+5)=(x+5)^3[/tex] it would be the same graph except that it's shifted to the left 5 units ( passes through (-5,0) instead of (0,0) )