Answer
a) $d(x)=\sqrt{x^4-13x^2+49}$
b) $d=7$ units
c) $d=\sqrt{37}$ units
d) See the attached graph below.
e) $x=\pm 2.55$
Work Step by Step
(a) The distance $d$ from $P(x,y)$ to the point $(0,-1)$ is:
$$\begin{align*}
d&=\sqrt{(x-x_1)^2+(y-y_1)^2}\\
&=\sqrt{(x-0)^2+(y-(-1))^2}\\
&=\sqrt{x^2+(y+1)^2}
\end{align*}$$
Since $P$ is a point on the graph of $y=x^2-8$ , substitute $x^2-8$ for $y$.
Then,
$$\begin{align*}
d&=\sqrt{x^2+[(x^2-8)+1]^2}\\
&=\sqrt{x^2+(x^2-7)^2}\\
&=\sqrt{x^2+x^4-14x^2+49}\\
&=\sqrt{x^4-13x^2+49}
\end{align*}$$
Therefore, the distance from $P$ to the origin expressed as a function of $x$ is:
$$d(x)=\sqrt{x^4-13x^2+49}$$
(b) If $x=0$ , $d$ can be computed by substituting $0$ to $x$ in $d(x)$ to obtain:
$$d(0)=\sqrt{0^4-13(0^2)+49}=\sqrt{49}=7$$
Hence, if $x=0$, $d=7$ units.
(c) If $x=-1$, $d$ can be computed by substituting $-1$ to $x$ to obtain:
$$d(-1)=\sqrt{(-1)^4-13(-1)^2+49}=\sqrt{1-13+49}=\sqrt{37}$$
Hence, if $x=-1$, $d=\sqrt{37}$ units.
(d) Use a graphing utility to the graph $d(x)=\sqrt{x^4-13x^2+49}$. Refer to the attached graph below.
(e) From the graph it is clear that when the value of $x$ is $\pm2.55$ , $d$ is smallest.
