Precalculus, Chapter 1, 1.2, Section 1.2, Problem 60

Find the intercepts of the equation $x^2 - y - 4 = 0$ and test for symmetry.

$x$-intercepts:


$\begin{equation} \begin{aligned} x^2 - y - 4 =& 0 && \text{Given equation} \\ x^2 - (0) - 4 =& 0 && \text{To find the $x$-intercept, we let } y = 0 \\ x^2 - 4 =& 0 && \\ x^2 =& 4 && \\ x =& \pm 2 && \end{aligned} \end{equation}$



The $x$-intercepts are $(-2,0)$ and $(2,0)$

$y$-intercepts:


$\begin{equation} \begin{aligned} x^2 - y - 4 =& 0 && \text{Given equation} \\ (0)^2 - y - 4 =& 0 && \text{To find the $y$-intercept, we let } x = 0 \\ -y-4 =& 0 && \\ -4 =& y && \end{aligned} \end{equation}$


The $y$-intercept is $(0,-4)$

Test for symmetry

$x$-axis:


$\begin{equation} \begin{aligned} x^2 - y - 4 =& 0 && \text{Given equation} \\ x^2 - (-y) - 4 =& 0 && \text{To test for $x$-axis symmetry, replace $y$ by $-y$ and see if the equation is still the same} \\ x^2 + y - 4 =& 0 && \end{aligned} \end{equation}$


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

$y$-axis:


$\begin{equation} \begin{aligned} x^2 - y - 4 =& 0 && \text{Given equation} \\ (-x)^2 - y - 4 =& 0 && \text{To test for $y$-axis symmetry, replace$ x$ by $-x$ and see if the equation is still the same} \\ x^2 - y - 4 =& 0 && \end{aligned} \end{equation}$


The equation is still the same so it is symmetric to the $y$-axis

Origin:


$\begin{equation} \begin{aligned} x^2 - y - 4 =& 0 && \text{Given equation} \\ (-x)^2 - (-y) - 4 =& 0 && \text{To test for origin symmetry, replace both $x$ by $-x$ and y by $-y$ and see if the equation is still the same} \\ x^2 + y - 4 =& 0 && \end{aligned} \end{equation}$


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

Therefore, the equation $x^2 - y - 4 = 0$ has an intercepts $(0,-4), (-2,0)$ and $(2,0)$ and it is symmetric to the $y$-axis.