What are the vertical and horizontal asymptotes of $y = \frac{x + 3}{x^{2} - 9}$ $y = (x+3)/(x^2-9)$ ?

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Answer 1 · 2018-06-20T15:15:54

vertical asymptote at $x = 3$ $x=3$

horizontal asymptote at $y = 0$ $y=0$

hole at $x = - 3$ $x=-3$

$y = \frac{x + 3}{x^{2} - 9}$ $y = (x+3)/(x^2-9)$

First factor:

$y = \frac{(x + 3)}{(x + 3) (x - 3)}$ $y = ((x+3))/((x+3)(x-3))$

Since the factor $x + 3$ $x+3$ cancels that is a discontinuity or hole, the factor $x - 3$ $x-3$ does not cancel so it is a asymptote:

$x - 3 = 0$ $x-3=0$

vertical asymptote at $x = 3$ $x=3$

Now let's cancel out the factors and see what the functions does as x gets really big in the positive or negative:

$x \to \pm \infty, y \to ?$ $x -> +-oo, y ->?$

$y = cancel((x+3))/(cancel((x+3))(x-3))=1/(x-3)$

As you can see the reduced form is just $1$ over some number $x$ , we can ignore the $-3$ because when $x$ is huge it is insignificant.

We know that: $x ->+-oo, 1/x -> 0$ hence, our original function has the same behavior:

$x ->+-oo, ((x+3))/((x+3)(x-3)) -> 0$

Therefore, the function has a horizontal asymptote at $y=0$

graph{y = (x+3)/(x^2-9) [-10, 10, -5, 5]}

What are the vertical and horizontal asymptotes of y = (x+3)/(x^2-9)y=x+3x2−9y = (x+3)/(x^2-9)?