Question

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

Answer 1

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

Explanation:

from the given:

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

perform long division so that the result is

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