How to find the asymptotes of $f (x) = \frac{x + 3}{x^{2} + 8 x + 15}$ $f(x) = (x+3) /( x^2 + 8x + 15)$ ?

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How to find the asymptotes of $f (x) = \frac{x + 3}{x^{2} + 8 x + 15}$ $f(x) = (x+3) /( x^2 + 8x + 15)$ ?

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Answer 1 · 2016-02-28T09:56:22

Simplify $f (x)$ $f(x)$ to see that there is a horizontal asymptote $y = 0$ $y=0$ , vertical asymptote $x = - 5$ $x=-5$ and hole (removable singularity) at $(- 3, \frac{1}{2})$ $(-3, 1/2)$

Explanation:

$f (x) = \frac{x + 3}{x^{2} + 8 x + 15}$ $f(x) = (x+3)/(x^2+8x+15)$

$= \frac{x + 3}{(x + 3) (x + 5)}$ $=(x+3)/((x+3)(x+5))$

$= \frac{1}{x + 5}$ $=1/(x+5)$

with exclusion $x \neq - 3$ $x != -3$

This function has horizontal asymptote $y = 0$ $y=0$ since $\frac{1}{x + 5} \to 0$ $1/(x+5)->0$ as x->+-oo#

It has a vertical asymptote $x = - 5$ $x=-5$ , where the denominator is zero, but the numerator non-zero.

It has a hole (removable singularity) at $(- 3, \frac{1}{2})$ $(-3, 1/2)$ since the left and right limits exist at $(- 3, \frac{1}{2})$ $(-3, 1/2)$ and are equal.

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How to find the asymptotes of $f (x) = \frac{x + 3}{x^{2} + 8 x + 15}$ $f(x) = (x+3) /( x^2 + 8x + 15)$ ?

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