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

Question

1155 views around the world

You can reuse this answer
Creative Commons License

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]