## Intermediate Algebra for College Students (7th Edition)

$f(x)=|x-1|\sqrt{5}$
Factor out 5 in the trinomial to obtain: $f(x)=\sqrt{5(x^2-2x+1)}$ The trinomial is a perfect square whose factored form is $(x-1)^2$. Thus, $f(x)=\sqrt{5(x-1)^2}$ The principal square root of any number/expression is always non-negative. Since $x$ can be any real number, then an absolute value must be applied to the principal square root of $(x-1)^2$ to make it non-negative for all values of $x$. Thus, simplifying the function gives: $f(x)=\sqrt{5(x-1)^2} \\f(x)=|x-1|\sqrt{5}$