Answer
a) $-1$
b) $x=3$ or $x=-3$
c) $0$
d) $-1$
e) $\pm3$
Work Step by Step
(a) Substituting $x=0$ gives $f(0)=\sqrt{0^2+16}-5=\sqrt{16}-5=4-5=-1$.
(b) We want to find $x$ such that $f(x)=\sqrt{x^2+16}-5=0$. Thus, we have
$$
\begin{aligned}
\sqrt{x^2+16}-5 & =0 \\
\sqrt{x^2+16} & =5 \\
x^2+16 & =25 \\
x^2 & =9 \\
x & = \pm 3 .
\end{aligned}
$$
Thus, $f(x)=0$ for $x=3$ or $x=-3$.
(c) $f(3)=\sqrt{3^2+16}-5=0$
(d) Let $x=0$. The vertical intercept is:
$\sqrt{0^2+16}-5=-1$
(e) The graph crosses the $x$-axis when $f(x)=0$:
$\sqrt{x^2+16}-5=0$
$\sqrt{x^2+16}=5$
$x^2+16=5^2$
$x^2=25-16$
$x^2=9$
$x=\pm 3$
