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

May 27, 2017

Vertical asymptote at $x = - 1$ , horizontal asymptote is absent.
$f \left(x\right) = x - 5$ is the oblique asymptote.

#### Explanation:

$f \left(x\right) = \frac{{x}^{2} - 4 x + 4}{x + 1}$

Vertical asymptote: Solving denominator for zero we get $x + 1 = 0 \mathmr{and} x = - 1$ 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.

$\frac{{x}^{2} - 4 x + 4}{x + 1} = \left(x - 5\right) \left(\frac{9}{x + 1}\right)$

So straight line $f \left(x\right) = x - 5$ is the oblique asymptote.

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