Question

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

Answer 1

Vertical asymptote at `x= -1` , horizontal asymptote is absent.
`f(x)=x-5` is the oblique asymptote.

Explanation:

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

Vertical asymptote: Solving denominator for zero we get `x+1=0 or x = -1` So V.A is at `x= -1`

Since degree of numerator(2) is greater than denominator(1) horizontal asymptote is absent.

Since degree of numerator(2) is greater than denominator(1) by a margin of 1 there is a slant/oblique asymptote , which can be found by long division.

`(x^2-4x+4)/(x+1) =( x-5) (9/(x+1) )`

So straight line `f(x)=x-5` is the oblique asymptote.

graph{(x^2-4x+4)/(x+1) [-80, 80, -40, 40]} [Ans]