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

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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.