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

An asymptote is defined as a line or curve that another line or curve approaches, but never meets. As such, we simply find the values of #x# and #y# that the y can never cross.

To find the vertical asymptote, you simply look at the denominator. The denominator, as you already know, cannot equal 0, else the curve would be undefined.

So, the #x-3# in the denominator can never equal 0

#x-3!=0#
#x!=3#

Therefore, #x=3# is the asymptote as #x# can never equal #3#, or else the denominator would become 0 and undefined.

To find the horizontal asymptote, divide every symbol/number in both the numerator and denominator by the highest #x# power in the denominator.

For example, in this equation, the highest #x# that can be found is #x^2#, but that is in the denominator. The highest power of #x# in the denominator, however, is simply #x#, and so that is what we will use.

So,

#(x^2/x-(5x)/x+6/x)/(x/x-3/x)#
#(x-5+6/x)/(1-3/x)#

For all values that are any value over x, replace that with a 0 as, as #x# in the denominator gets infinitely larger, the overall number will get smaller and approach 0.

#y=x-5# is the horizontal/oblique asymptote.

graph{(x^2-5x+6)/x-3 [-3.095, 16.905, -4.44, 5.56]}