Question

In Problems 57, list the intercepts and test for symmetry.

$y=\frac{3x}{x^2+9}$

Short answer

The graph of the equation is symmetric with respect to the origin, and the point of the equation $(0,0)$is both a $x$- and a $y$-intercept.

Step-by-step solution

Given information.

The given equation is$y=\frac{3x}{x^2+9}$.

Determine the x-intercept.

For $x$-intercept, substitute 0 for yand then solve for $x$.

$\begin{array}{rcl}0& =& \frac{3x}{x^2+9}\\ 3x& =& 0\\ x& =& 0\end{array}$

Determine the y-intercept.

For $y$-intercept, substitute 0 for $x$and then solve for $y$.

$\begin{array}{rcl}y& =& \frac{3\left(0\right)}{{\left(0\right)}^2+9}\\ & =& 0\end{array}$

Determine the symmetry.

Substitute $y$for $-y$in the equation to check for symmetry with respect to $x$axis:

$\begin{array}{rcl}-y& =& \frac{3x}{x^2+9}\\ y& =& -\frac{3x}{x^2+9}\end{array}$

Notice that the right side of the equation after substitution is not equal to the right side of the initial equation.

Since the equations are not equivalent, the graph is not symmetric with respect to the $x$-axis.

Substitute $x$with $-x$in the equation to check for the symmetry with respect to $y$-axis:

$\begin{array}{rcl}y& =& \frac{3(-x)}{{(-x)}^2+9}\\ & =& -\frac{3x}{x^2+9}\end{array}$

Notice that the right side of the equation after substitution is not equal to the right side of the initial equation.

Since the equations are not equivalent, the equation is not symmetric with respect to the $y$-axis.

Substitute $x$with $-x$and $y$with $-y$in the equation to check for the symmetry with respect to the origin:

$\begin{array}{rcl}-y& =& \frac{3(-1)}{{(-x)}^2+9}\\ -y& =& -\frac{3x}{x^2+9}\\ y& =& \frac{3x}{x^2+9}\end{array}$

Notice that the right side of the equation after substitution is equal to the right side of the initial equation.

Since the equations are equivalent, the equation is symmetric with respect to the origin.

Write the conclusion

The $x$-intercept and $y$-intercept is $\left(0,0\right)$.

The given equation is symmetric about the origin.