How do you find vertical, horizontal and oblique asymptotes for $\frac{x + 3}{x^{2} + 8 x + 15}$ $(x+3 )/ (x^2 + 8x + 15)$ ?

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Answer 1 · 2016-04-04T23:05:16

Vertical Asymptotes: $x = - 5$ $x=-5$
Horizontal Asymptote: $y = 0$ $y=0$
No Oblique Asymptote

To obtain the Horizontal Asymptote, take the limit of the function

$lim_(x rarr oo) y=lim_(x rarr oo) (x+3)/(x^2+8x+15)=zero$

therefore, $y=0$ is a Horizontal Asymptote

To obtain the Vertical Asymptote, equate the factors of the denominator to zero the solve for x.

$x+5=0$

$x=-5$

Kindly see the graph of $y=(x+3)/(x^2+8x+15)$ and the location of the imaginary asymptotes.

graph{y=(x+3)/(x^2+8x+15)[-20,20,-10,10]}

God bless....I hope the explanation is useful.

How do you find vertical, horizontal and oblique asymptotes for (x+3 )/ (x^2 + 8x + 15)x+3x2+8x+15(x+3 )/ (x^2 + 8x + 15)?