# How do you find the asymptotes for f(x) = (x^2-4x+4) / (x+1)?

Oblique Asymptote is $y = x - 5$
Vertical Asymptote is $x = - 1$

#### Explanation:

from the given:

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

perform long division so that the result is

$\frac{{x}^{2} - 4 x + 4}{x + 1} = x - 5 + \frac{9}{x + 1}$

Notice the part of the quotient

$x - 5$

equate this to $y$ like as follows

$y = x - 5$ this is the line which is the Oblique Asymptote

And the divisor $x + 1$ be equated to zero and that is the Vertical asymptote

$x + 1 = 0$ or $x = - 1$

You can see the lines $x = - 1$ and $y = x - 5$ and the graph of
$f \left(x\right) = \frac{{x}^{2} - 4 x + 4}{x + 1}$
graph{(y-(x^2-4x+4)/(x+1))(y-x+5)=0[-60,60,-30,30]}

God bless...I hope the explanation is useful..