College Algebra, Chapter 2, 2.2, Section 2.2, Problem 36

Make a table of values and sketch the graph of the equation $y = 16 - x^4$. Find the $x$ and $y$ intercepts.

$
\begin{array}{|c|c|}
\hline\\
\text{Let } x & y = 16 - x^4 \\
\hline\\
-3 & -65 \\
\hline\\
-2 & 0 \\
\hline\\
-1 & 15 \\
\hline\\
1 & 15 \\
\hline\\
2 & 0 \\
\hline\\
3 & -65\\
\hline

\end{array}
$


$
\begin{equation}
\begin{aligned}

y =& 16 - x^4
&& \text{Given}
\\
\\
y =& (4 - x^2)(4 + x^2)
&& \text{Difference of squares}

\end{aligned}
\end{equation}
$


To solve for $y$ intercept, we set $x = 0$


$
\begin{equation}
\begin{aligned}

y =& 16 - (0)^4
\\
\\
y =& 16

\end{aligned}
\end{equation}
$


The $y$ intercept is at $(0,16)$

To solve for the $x$ intercept, we set $y = 0$


$
\begin{equation}
\begin{aligned}

0 =& (4 - x^2)(4 + x^2)
&& \text{Zero Product Property}
\\
\\
0 =& 4 - x^2 \text{ and } 0 = 4 + x^2
&& \text{Solve for } x
\\
\\
x =& \pm 2 \text{ and } x = \pm \sqrt{-4}
&& \text{Absurd}

\end{aligned}
\end{equation}
$


Thus, the $x$ intercepts are at $(2,0)$ and $(-2,0)$