How do you find vertical, horizontal and oblique asymptotes for #(x^2+6x-9)/(x-2)#?

1 Answer
Sep 29, 2016

Answer:

vertical asymptote at x = 2
oblique asymptote is y = x + 8

Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote.

solve: #x-2=0rArrx=2" is the asymptote"#

Horizontal asymptotes occur when the degree of the numerator is ≤ degree of the denominator. This is not the case here (numerator-degree 2 , denominator-degree 1) Hence there are no horizontal asymptotes.

Oblique asymptotes occur when the degree of the numerator > degree of the denominator which is the case here.

Use #color(blue)"polynomial division"# to obtain.

#f(x)=x+8+7/(x-2)#

as #xto+-oo,f(x)tox+8+0#

#rArry=x+8" is the asymptote"#
graph{(x^2+6x-9)/(x-2) [-45, 45, -22.5, 22.5]}