## Thomas' Calculus 13th Edition

$y=1-8x^3$
Given the graph of $y=f(x)$ and a real number $c>1$, we obtain the graph of $y=f(cx)$ by horizontally stretching the graph of $y=f(x)$ by a factor of $c$. Hence to horizontally stretch the graph of $y=1-x^3$ by a factor of 2, we replace each occurrence of $x$ in this equation by $2x$, and, thus, obtaining the equation $$y=1-(2x)^3,$$ or equivalently, $$y=1-8x^3.$$