## Precalculus (6th Edition)

When we see $($... x ...$)^{2}$ in the function expression, our first thought should be "this is probably not one-to-one", because, for example, $1^{2}=(-1)^{2}=1$... We can get 1 in the parentheses if $x=7$ and $-1$ if $x=5.$ So, for two different values from the domain, $-7$ and $5$... $f(-7)=-3(1)^{2}+8=5$ $f(5)=-3(-1)^{2}+8=5$ ... we have the same function value.