Question

Find the intercepts and graph each equation by plotting points. Be sure to label the intercepts.

$9x^2+4y=36$.

Short answer

The intercepts are :

x-intercepts are $\left(-2,0\right)$and $\left(2,0\right)$,

y-intercept is $\left(0,9\right)$.

Step-by-step solution

Given information:

Consider the equation,

$9x^2+4y=36$

To find the x-intercept:

Let $y=0$in the equation,

Then, $9x^2+4\times 0=36$

$9x^2=36$

$\frac{9x^2}{9}=\frac{36}{9}$ [Divide by 9 on both sides.]

$x^2=4$

$x^2-4=4-4$ [Add $\left(-4\right)$ on both sides.]

$x^2-4=0$

$\left(x+2\right)\left(x-2\right)=0$ [Factors]

$x+2=0$ or $x-2=0$ [Zero Product Property]

$x=-2$ or $x=2$

The equation has two solutions, $-2and2.$

$x-interceptsare\left(-2,0\right)and\left(2,0\right).$

To find the y-intercept:

Let $x=0$in the equation,

Then,$9\times 0+4y=36$

$4y=36$

$\frac{4y}{4}=\frac{36}{4}$

$y=9$

$y-interceptis(0,9).$

To draw the graph:

We can draw the graph of the equation $9x^2+4y=36$by plotting the above intercepts and the points in the table below,

$x$	$y=\frac{36-9x^2}{4}$	$\left(x,y\right)$
$-3$	$\frac{-45}{4}$	$\left(-3,\frac{-45}{4}\right)$
$-1$	$\frac{27}{4}$	$\left(-1,\frac{27}{4}\right)$
$1$	$\frac{27}{4}$	$\left(1,\frac{27}{4}\right)$
$3$	$\frac{-45}{4}$	$\left(3,\frac{-45}{4}\right)$