Answer
$-\frac{15}{4}$
Work Step by Step
We are given that $f(x)=x^{2}-9$, $g(x)=2x$, and $h(x)=x-3$.
We are asked to find the value of $(g+h)(-\frac{1}{4})$. We know that $(g+h)(x)=g(x)+h(x)$.
Therefore, $(g+h)(-\frac{1}{4})=(2(-\frac{1}{4}))+(-\frac{1}{4}-
\frac{12}{4})=(-\frac{2}{4})+(\frac{-1-12}{4})=(-\frac{2}{4})+(\frac{-13}{2})=-\frac{2}{4}-\frac{13}{4}=\frac{-2-13}{4}=\frac{-15}{4}=-\frac{15}{4}$.
