What is the equation of the oblique asymptote $f (x) = \frac{x^{2} - x - 2}{x + 1}$ $f(x)=(x^2-x-2)/(x+1)$ ?

Question

2050 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-07-18T10:22:20

$f (x) = \frac{x^{2} - x - 2}{x + 1} = \frac{(x - 2) (x + 1)}{x + 1} = x - 2$ $f(x) = (x^2-x-2)/(x+1) = ((x-2)(x+1))/(x+1) = x-2$

So $f (x)$ $f(x)$ is a straight line (with excluded point $(- 1, - 3)$ $(-1,-3)$ ).

It has no asymptote.

$\frac{x^{2} - x - 2}{x + 1}$ $(x^2-x-2)/(x+1)$

$= \frac{x^{2} + x - 2 x - 2}{x + 1}$ $= (x^2+x-2x-2)/(x+1)$

$= \frac{x (x + 1) - 2 (x + 1)}{x + 1}$ $= (x(x+1)-2(x+1))/(x+1)$

$= \frac{(x - 2) (x + 1)}{x + 1} = x - 2$ $= ((x-2)(x+1))/(x+1) = x-2$

with exclusion $x \neq - 1$ $x != -1$

What is the equation of the oblique asymptote f(x)=(x^2-x-2)/(x+1)f(x)=x2−x−2x+1f(x)=(x^2-x-2)/(x+1)?