College Algebra, Chapter 2, Review Exercises, Section Review Exercises, Problem 26

a.) Test the equation $6x + y^2 = 36$ 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}
6(-x) + y^2 &= 36\\
\\
-6x + y^2 &= 36
\end{aligned}
\end{equation}
$

Since the new equations is not equal to the original equation, then $6x + y^2 = 36$ is not symmetric to $y$-axis.



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

$
\begin{equation}
\begin{aligned}
6x + (-y)^2 &= 36\\
\\
6x + y^2 &= 36
\end{aligned}
\end{equation}
$

Since the new equation is equal to the original equation, then $6x + y^2 = 36$ is symmetric to the $x$-axis.



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

$
\begin{equation}
\begin{aligned}
6(-x) + (-y)^2 &= 36\\
\\
-6x + y^2 &= 36
\end{aligned}
\end{equation}
$

Since the new equation is not equal to the original equation. Then $6x + y^2 = 36$ is not symmetric to the origin.

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

$
\begin{equation}
\begin{aligned}
6x + (0)^2 &= 36\\
\\
6x &= 36\\
\\
x &= 6
\end{aligned}
\end{equation}
$


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

$
\begin{equation}
\begin{aligned}
6(0) + y^2 &= 36\\
\\
y^2 &= 36\\
\\
y &= \pm 6
\end{aligned}
\end{equation}
$

The $x$-intercept is at $(6,0)$ and the $y$-intercepts are at $(0,6)$ and $(0,-6)$