Precalculus, Chapter 1, Review Exercises, Section Review Exercises, Problem 14

Find the intercepts of the equation $y = x^3 - x$ and test for symmetry with respect to the $x$-axis, the $y$-axis and the origin.

$x$-intercepts


$
\begin{equation}
\begin{aligned}

y =& x^3 - x
&& \text{Given equation}
\\
0 =& x^3 - x
&& \text{To find the $x$-intercept, we let $y = 0$ and solve for $x$}
\\
0 =& (x^2 - 1)x
&&
\\
0 =& (x + 1)(x- 1)x
&&
\\
x =& -1, x = 1 \text{ and } x = 0
&&

\end{aligned}
\end{equation}
$


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

$y$-intercepts


$
\begin{equation}
\begin{aligned}

y =& x^3 - x
&& \text{Given equation}
\\
y =& (0)^3 - 0
&& \text{To find the $y$-intercept, we let $x = 0$ and solve for $y$}
\\
y =& 0
&&

\end{aligned}
\end{equation}
$


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

Test for symmetry

$x$-axis


$
\begin{equation}
\begin{aligned}

y =& x^3 - x
&& \text{Given equation}
\\
-y =& x^3 - x
&& \text{To test for $x$-axis symmetry, replace $y$ by $-y$ and see if the equation is still the same}

\end{aligned}
\end{equation}
$


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

$y$-axis


$
\begin{equation}
\begin{aligned}

y =& x^3 - x
&& \text{Given equation}
\\
y =& (-x)^3 - (-x)
&& \text{To test for $y$-axis symmetry, replace$ x$ by $-x$ and see if the equation is still the same}
\\
y =& -x^3 + x
&&

\end{aligned}
\end{equation}
$


The equation changes so the equation is not symmetric to $y$-axis.

Origin


$
\begin{equation}
\begin{aligned}

y =& x^3 - x
&& \text{Given equation}
\\
-y =& (-x)^3 - (-x)
&& \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 =& -x^3 + x
&&
\\
-y =& -(x^3 - x)
&&
\\
y =& x^3 - x

\end{aligned}
\end{equation}
$


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