How do you find vertical, horizontal and oblique asymptotes for #(4x^2-x+2)/(x+1)#?

Question

1322 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-10T12:09:25

Vertical asymptote is at # x = -1#
Slant asymptote is #y=4x-5#

#f(x) = (4x^2-x +2)/(x+1)#

Vertical asymptote is fomed when deniminator is zero.

# :. x +1= 0 or x = -1# So vertical asymptote is at # x = -1#

Here degree in numerator is #1# greater than degree in

denominator . So we have a slant asymptote which is found by

doing long division . By long division we find Quotient as

#y=4x-5# . So slant asymptote is #y=4x-5#

graph{(4x^2-x+2)/(x+1) [-101.25, 101.2, -50.6, 50.7]} [Ans]