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

1 Answer
Clara M.
Oct 7, 2016

Remember: You cannot have three asymptotes at the same time. If the Horizontal Asymptote exists, the Oblique Asymptote doesn't exist. Also, #color (red) (H.A)# #color (red) (follow)# #color (red) (three)# #color (red) (procedures).# Let's say #color (red)n# = highest degree of the numerator and #color (blue)m# = highest degree of the denominator,#color (violet) (if)#:
#color (red)n color (green)< color (blue) m#, #color (red) (H.A => y = 0)#
#color (red)n color (green)= color (blue) m#, #color (red) (H.A => y = a/b)#
#color (red)n color (green)> color (blue) m#, #color (red) (H.A) # #color (red) (doesn't)# #color (red) (EE)#

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

#V.A: x-3=0 => x = 3#
#O.A: y=x-2#

Please, take a look at the picture.

The oblique/slant asymptote is found by dividing the numerator by the denominator (long division.)

Notice that I did not do the long division in the way some people excepted me to. I always use the "French" way because I've never understood the English way, also I'm a francophone :) but it is the same answer.

Hope this helps :)
enter image source here

Related questions
See all questions in Asymptotes
Impact of this question
1663 views around the world
You can reuse this answer
Creative Commons License