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.