Question

Question #29ac7

Answer 1

`x=1`
`y = x+1`

Explanation:

One obvious asymptote is `x=1` as `f(1) = 1^2/(1-1) = 1/0` which is impossible because of the division by `0`.

But there is another one, to find it you first have to divide `x^2` by `x-1` :

`x^2/(x-1) = x+1` with a remainder of `1`.

So we know that `x^2 = (x-1)(x+1) + 1`. Dividing both sides by `x-1` we get:

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

As `x ->oo : 1/(x-1) = 0`

So for infinitely large values of `x, f(x)` approches `y= x+1` but never reaches it.

So `y=x+1` is an asymptote.

graph{x^2/(x-1) [-13.16, 30.92, -2.24, 19.8]}