College Algebra, Chapter 2, Review Exercises, Section Review Exercises, Problem 28

a.) Test the equation $x^4 = 16 + y $ for symmetry with respect to the $x$-axis, $y$-axis and the origin.
Testing the symmetry with respect to $y$-axis, set $x = -x$

$
\begin{equation}
\begin{aligned}
(-x)^4 &= 16 + y\\
\\
x^4 &= 16 + y
\end{aligned}
\end{equation}
$

Since the new equations is equal to the original equation, then $x^4 = 16 + y$ is symmetric to $y$-axis.



Testing the symmetry with respect to $x-$axis, set $y = -y$

$
\begin{equation}
\begin{aligned}
x^4 &= 16 + (-y)\\
\\
x^4 &= 16 - y
\end{aligned}
\end{equation}
$

Since the new equation is not equal to the original equation. Then $x^4 = 16 + y$ is not symmetric to the $x$-axis.



Testing the symmetry with respect to origin. Set, $x = -x$ and $y = -y$

$
\begin{equation}
\begin{aligned}
(-x)^4 &= 16 + (-y)\\
\\
x^4 &= 16 - y
\end{aligned}
\end{equation}
$

Since the new equation is not equal to the original equation. Then $x^4 = 16 + y$ is not symmetric to the origin.

b.) Find the $x$ and $y$-intercepts of the equation.
To find for $y$-intercept, set $x = 0$,

$
\begin{equation}
\begin{aligned}
(0)^4 &= 16 + y\\
\\
y &= -16
\end{aligned}
\end{equation}
$


To find for $x$-intercept, set $y = 0$

$
\begin{equation}
\begin{aligned}
x^4 &= 16 + 0\\
\\
x^4 &= 16\\
\\
x &= \pm \sqrt[4]{16} = \pm 2
\end{aligned}
\end{equation}
$

The $x$-intercepts are at $(2,0)$ and $(-2,0)$, while the $y$-intercept is at $(0,-16)$