## Thomas' Calculus 13th Edition

$y=1-\dfrac{1}{8}x^3$
Given the graph of $y=f(x)$ and a real number $c>1$, we obtain the graph of $y=f\left(\dfrac{1}{c}x\right)$ by horizontally compressing the graph of $y=f(x)$ by a factor of $c$. Hence to horizontally compress the graph of $y=1-x^3$ by a factor of 2, we replace each occurrence of $x$ in this equation by $\dfrac{1}{2}x$ obtaining the equation $$y=1-\left(\dfrac{1}{2}x\right)^3,$$ or equivalently, $$y=1-\dfrac{1}{8}x^3.$$