How to find a horizontal asymptote #(x^2 - 5x + 6)/( x - 3)#?

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Answer 1 · 2016-02-29T16:15:25

There is no horizontal asymptote

There is no horizontal asymptote as degree of numerator #2# is greater than that of denominator #1# by one.

In such case, there is a possibility of a slant asymptote, but before concluding that let us factorize #(x^2−5x+6)# as follows:

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

Hence #(x^2−5x+6)/(x−3)# can be simplified as follows:

#((x-2)(x-3))/(x-3)# or

#(x-2)#

Hence #(x^2−5x+6)/(x−3)# is the equation of just the line #y=(x-2)#
(I do not think tis can be considered as slanting asymptote).

But as #x-3# appears in denominator the domain of#x# in #y# does not include #3#.