How do you find the vertical, horizontal or slant asymptotes for #(x^2 - 5x + 6)/( x - 3)#?

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Answer 1 · 2016-11-22T17:53:23

We have a hole when #x=2#

Let's factorise the numerator.

#x^2-5x+6=(x-3)(x-2)#

Therefore,

#(x^2-5x+6)/(x-3)=(cancel(x-3)(x-2))/cancel(x-3)#

#=(x-2)#

We have a hole at #x=2#

#y=(x-2)# is the equation of a straight line.

graph{x-2 [-10, 10, -5, 5]}