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.