Find the intercepts of the equation $x^2 + 4x + y^2 - 2y = 0$ and test for symmetry with respect to the $x$-axis, the $y$-axis and the origin.
$x$-intercepts
$
\begin{equation}
\begin{aligned}
& x^2 + 4x + y^2 - 2y = 0
&& \text{Given equation}
\\
& x^2 + 4x + (0)^2 - 2(0) = 0
&& \text{To find the $x$-intercept, we let $y = 0$ and solve for $x$}
\\
& x^2 + 4x = 0
&&
\\
& x (x + 4) = 0
&&
\\
& x = 0 \text{ and } x + 4 = 0
&&
\\
& x = 0 \text{ and } x = -4
&&
\end{aligned}
\end{equation}
$
The $x$-intercepts are $(0,0)$ and $(-4,0)$
$y$-intercepts
$
\begin{equation}
\begin{aligned}
& x^2 + 4x + y^2 - 2y = 0
&& \text{Given equation}
\\
& (0)^2 + 4(0) + y^2 - 2y = 0
&& \text{To find the $y$-intercept, we let $x = 0$ and solve for $y$}
\\
& y^2 - 2y = 0
&&
\\
& y(y-2) = 0
&&
\\
& y = 0 \text{ and } y - 2 = 0
&&
\\
& y = 0 \text{ and } y = 2
\end{aligned}
\end{equation}
$
The $y$-intercepts are $(0,0)$ and $(0,2)$.
Test for symmetry
$x$-axis
$
\begin{equation}
\begin{aligned}
x^2 + 4x + y^2 - 2y =& 0
&& \text{Given equation}
\\
x^2 + 4x + (-y)^2 - 2(-y) =& 0
&& \text{To test for $x$-axis symmetry, replace $y$ by $-y$ and see if the equation is still the same}
\\
x^2 + 4x + y^2 + 2y =& 0
&&
\end{aligned}
\end{equation}
$
The equation changes so the equation is not symmetric to $x$-axis.
$y$-axis
$
\begin{equation}
\begin{aligned}
x^2 + 4x + y^2 - 2y =& 0
&& \text{Given equation}
\\
(-x)^2 + 4(-x) + y^2 - 2y =& 0
&& \text{To test for $y$-axis symmetry, replace$ x$ by $-x$ and see if the equation is still the same}
\\
x^2 - 4x + y^2 - 2y =& 0
&&
\end{aligned}
\end{equation}
$
The equation changes so the equation is not symmetric to $y$-axis.
Origin
$
\begin{equation}
\begin{aligned}
& x^2 + 4x + y^2 - 2y = 0
&& \text{Given equation}
\\
& (-x)^2 + 4(-x) + (-y)^2 - 2(-y) = 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 - 4x + y^2 + 2y = 0
&&
\end{aligned}
\end{equation}
$
The equation changes so the equation is not symmetric to the origin.
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