Precalculus, Chapter 1, 1.2, Section 1.2, Problem 62

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

$x$-intercepts:


$
\begin{equation}
\begin{aligned}

4x^2 + y^2 =& 4
&& \text{Given equation}
\\
4x^2 + (0)^2 =& 4
&& \text{To find the $x$-intercept, we let } y = 0
\\
4x^2 =& 4
&&
\\
x^2 =& 1
&&
\\
x =& \pm 1

\end{aligned}
\end{equation}
$



The $x$-intercepts are $(-1,0)$ and $(1,0)$

$y$-intercepts:


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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

Test for symmetry

$x$-axis:


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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

$y$-axis:


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


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

Origin:


$
\begin{equation}
\begin{aligned}

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

\end{aligned}
\end{equation}
$


The equation is still the same so it is symmetric to the origin.

Therefore, the equation $4x^2 + y^2 = 4$ has an intercepts $(-1,0), (1,0), (0,-2)$ and $(0,2)$ and it is symmetric to the $x$-axis, to the $y$-axis and to the origin.