Precalculus, Chapter 1, 1.2, Section 1.2, Problem 56

Find the intercepts of the equation $y^2 = x + 9$ and test for symmetry.

$x$-intercepts:


$
\begin{equation}
\begin{aligned}

y^2 =& x + 9
&& \text{Given equation}
\\
0^2 =& x + 9
&& \text{To find the $x$-intercept, we let } y = 0
\\
0 =& x + 9
&&
\\
-9 =& x
&&

\end{aligned}
\end{equation}
$



The $x$-intercept is $(-9,0)$

$y$-intercepts:


$
\begin{equation}
\begin{aligned}

y^2 =& x + 9
&& \text{Given equation}
\\
y^2 =& 0 + 9
&& \text{To find the $y$-intercept, we let } x = 0

\\
y^2 =& 9
&&
\\
y =& \pm 3
&&

\end{aligned}
\end{equation}
$


The $y$-intercepts are $(0,3)$ and $(0,-3)$

Test for symmetry

$x$-axis:


$
\begin{equation}
\begin{aligned}

y^2 =& x + 9
&& \text{Given equation}
\\
(-y)^2 =& x + 9
&& \text{To test for $x$-axis symmetry, replace $y$ by $-y$ and see if the equation is still the same}
\\
y^2 =& x + 9
&&

\end{aligned}
\end{equation}
$


The equation is still the same, therefore the equation is symmetric to the $x$-axis.

$y$-axis:


$
\begin{equation}
\begin{aligned}

y^2 =& x + 9
&& \text{Given equation}
\\
y^2 =& -x + 9
&& \text{To test for $y$-axis symmetry, replace$ x$ by $-x$ and see if the equation is still the same}

\end{aligned}
\end{equation}
$


The equation changes so the equation is not symmetric to the $y$-axis

Origin:


$
\begin{equation}
\begin{aligned}

y^2 =& x + 9
&& \text{Given equation}
\\
(-y)^2 =& -x + 9
&& \text{To test for origin symmetry, replace both $x$ by $-x$ and y by $-y$ and see if the equation is still the same}
\\
y^2 =& -x + 9
&&

\end{aligned}
\end{equation}
$


The equation changes so the equation is not symmetric to the origin.

Therefore, the equation $y^2 = x + 9$ has an intercepts $(-9,0), (0,3)$ and $(0,-3)$ and it is symmetric to the $x$-axis.