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Suppose f(x) = x³ . find the graph of f(x+5)

User Ekeren
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5 votes

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


\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}


\bf ~~~~~~\textit{reflection over the y-axis} \\\\ \bullet \textit{ horizontal shift by }( C)/( B)\\ ~~~~~~if\ ( C)/( B)\textit{ is negative, to the right}\\\\ ~~~~~~if\ ( 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 }(2\pi )/( B)

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.
User Alveona
by
8.1k points
1 vote
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
f(x)=x^3, the shifted graph would be
f(x+5)=(x+5)^3 it would be the same graph except that it's shifted to the left 5 units ( passes through (-5,0) instead of (0,0) )

User Patelarpan
by
7.2k points

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