Question

Slant asymptote help please. Check my answer

Answer 1

I get C but nobody can ever confirm if its right or not for me

Answer 2

How did you get c?

Answer 3

@JoeJoldin The "Wolf" does not concur with your answer of C.

You can look at this link to read what Wolf "thinks."

http://www.wolframalpha.com/input/?i=Asymptote+y+%3D+%28x%5E2+-2x+%2B+1%29%2F%28x-1%29

Answer 4

To my thinking, there is no slant asymptote.

The denominator (x - 1) divides out with the numerator leaving y = x - 1, where x is not 1. There is a point discontinuity as (1,0).

Answer 5

Shouldn't the quotient (which comes out to be linear after carrying out the division presented in the question) be the slant asymptote? As you tend towards \(\infty\) or \(-\infty\), the function given in the problem, \(f(x)\), tends toward the asymptote that is the resulting quotient.