How do you find the vertical, horizontal or slant asymptotes for $f(x) = (x^2 - 8)/(x+3)$ ?

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Answer 1 · 2017-10-14T12:23:11

Vertical asymptote at $x = -3$ , no horizontal asymptote and
slant asymptote $y = x-3$

$f(x)=(x^2-8)/(x+3)$ .The vertical asymptotes will occur at those

values of x for which the denominator is equal to zero.

$:. x+3=0 or x=-3$ .Thus, the graph will have vertical

asymptote at $x = -3$ .

To find the horizontal asymptote, here the degree of the numerator

is $2$ and the degree of the denominator is $1$ . Since the larger

degree occurs in the numerator, the graph will have no horizontal

asymptote.

If the numerator's degree is greater (by a margin of 1), then

we have a slant asymptote which can be found by doing long

division of $(x^2-8)/(x+3)$ of which quotient is $y=x-3 :.$

Slant asymptote is $y=x-3$

graph{(x^2-8)/(x+3) [-10, 10, -5, 5]} [Ans]

How do you find the vertical, horizontal or slant asymptotes for f(x) = (x^2 - 8)/(x+3)f(x) = (x^2 - 8)/(x+3)?