How do you find the asymptotes for $h(x) = (x^3-8)/(x^2-5x+6)$ ?

Question

1457 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-15T13:07:27

Slant asymptote { #y=x+7.
Vertical asymptote : x = 3.

graph{((x^2+4x+4)/(x-3)-y)(y-x-7.9)(x-2+.01y)=0 [-45, 45, -20, 25]}

The graph is a hyperbola

$(y-x-7)(x-3)=15$ having asymptotes given by

$(y-x-7)(x-3)=0$ , with a hole at (2, -16)

$h =((x-2)/(x-2))((x^2+4x+4)/(x-3))$ .

Sans the hole at x = 2,

$h=(x^2+4x+4)/(x-3)$

$=x+7+25/(x-3)$ .

So, $y = quotient=x+7$ gives the slant asymptote and

$x-3 = 0$ gives the vertical asymptote.

How do you find the asymptotes for h(x) = (x^3-8)/(x^2-5x+6)h(x) = (x^3-8)/(x^2-5x+6)?