91Ó°ÊÓ

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

role="math" localid="1646797031387" $\begin{array}{rcl}16{\left(0\right)}^2& =& 120x-225\\ 120x& =& 225\\ x& =& \frac{225}{120}\\ & =& \frac{15}{8}\end{array}$

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

$$\begin{gathered} 16y^2=120\left(0\right)-225 \\ 16y^2=-225 \\ y=Â\pm \sqrt{\frac{-225}{16}} \end{gathered}$$

The value of $y$is imaginary because inside the square root the value is negative.

Perform the test for symmetry with respect to the $x$-axis. Put $y=-y$

$\begin{array}{rcl}16(-y{)}^2& =& 120x-225\\ 16y^2& =& 120x-225\end{array}$

Since after substituting $y=-y$into the given equation, the resulting equation is equivalent to the original equation hence the graph of the given equation is symmetrical about the $x$-axis.

Perform the test for symmetry with respect to the $y$-axis. Put $x=-x$

$\begin{array}{rcl}16y^2& =& 120(-x)-225\\ 16y^2& =& -120x-225\\ & ≠& 120x-225\end{array}$

Perform the test for symmetry with respect to the origin. Put $x=-x;y=-y$

$\begin{array}{rcl}16(-y{)}^2& =& 120(-x)-225\\ 16y^2& =& -120x-225\\ & ≠& 120x-225\end{array}$

a) There is no $y$-intercept and $x$-intercept is $\left(\frac{15}{8},0\right)$.

b) The graph of the given equation is symmetrical about the $x$-axis but not about the $y$-axis hence it will not have about the origin.