How do you find vertical, horizontal and oblique asymptotes for $f(x) =( x^3 - 4x^2 + 5x + 5)/(x - 1)$ ?

Question

1659 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-12-21T19:44:15

$f(x)$ has one vertical asymptote $x=1$ and no other linear asymptotes.

$f(x) = (x^3-4x^2+5x+5)/(x-1)$

$color(white)(f(x)) = (x^3-x^2-3x^2+3x+2x-2+7)/(x-1)$

$color(white)(f(x)) = x^2-3x+2 + 7/(x-1)$

This function has a vertical asymptote $x=1$ .

As $x->+-oo$ we have $7/(x-1)->0$ , so $f(x)$ is asymptotic to $x^2-3x+2$ , which is not a line.

So there are no other linear asymptotes.

graph{(x^3-4x^2+5x+5)/(x-1) [-10, 10, -200, 200]}

How do you find vertical, horizontal and oblique asymptotes for f(x) =( x^3 - 4x^2 + 5x + 5)/(x - 1)f(x) =( x^3 - 4x^2 + 5x + 5)/(x - 1)?