How do you find vertical, horizontal and oblique asymptotes for #f(x) = (2x^2 +x -1) /( x-1)#?

1 Answer
Sep 29, 2016

Answer:

vertical asymptote at x = 1
oblique asymptote is y = 2x + 3

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-1=0rArrx=1" is the asymptote"#

Horizontal asymptotes occur when the degree of the numerator ≤ 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)=2x+3+2/(x-1)#

as #xto+-oo,(f(x)to2x+3+0#

#rArry=2x+3" is the oblique asymptote"#
graph{(2x^2+x-1)/(x-1) [-40, 40, -20, 20]}