# How do you graph y=(x^2-9)/(2x^2+1) using asymptotes, intercepts, end behavior?

Feb 8, 2017

y-intercept ( x = 0 ) :$- 9$. x-intercept ( y = 0 ) : $\pm 3$
Horizontal asymptote : $\leftarrow y = \frac{1}{2} \rightarrow$
Vertical asymptote : $x = 0 \downarrow$

#### Explanation:

graph{((x^2-9)/(2x^2+1)-y)(y-1/2)x=0 [-10, 10, -5, 5]}

y-intercept ( x = 0 ) :$- 9$

x-intercept ( y = 0 ) : $\pm 3$

By actual division,

$y = \frac{1}{2} - \frac{\frac{19}{2}}{2 {x}^{2} + 1} > \frac{1}{2}$

So, y = 1/2 is the asymptote.

As $x \to \pm \infty , \to \frac{1}{2}$

Inversely, x =+-sqrt((y=9)/(1-2y)=+-sqrt((1+9/y)/(1/y-2)) to 0, as y to -oo#.

So,$x - 0 \downarrow$ is yet another asymptote.