## College Algebra (6th Edition)

a. $5\neq 3$, so we calculate $h(5)$ using the $\displaystyle \frac{x^{2}-9}{x-3}$ part $h(5)=\displaystyle \frac{5^{2}-9}{5-3}=\frac{25-9}{2}=\frac{16}{2}=8$ b. $0\neq 3$, so we calculate $h(0)$ using the $\displaystyle \frac{x^{2}-9}{x-3}$ part $h(5)=\displaystyle \frac{0^{2}-9}{0-3}=\frac{-9}{-3}=3$ c. for $x=3, h(3)=6$