How do you find the vertical, horizontal and slant asymptotes of: #f(x)= ( x-3) /( x^2-3x+2)#?

1 Answer
Oct 11, 2016

Answer:

#H.A => y = 0#

#V.A =>x=1# and #x=2#

Explanation:

Remember: You cannot have three asymptotes at the same time. If the Horizontal Asymptote exists, the Oblique/Slant 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)#

For this problem, #f(x)=(x-3)/(x^2-3x+2)#

#color (red)n color (green)< color (blue) m#, #H.A => y = 0#

#V.A=>x^2-3x+2=0#

Find the answer by using the tools that you already know. As for me, I always use #Delta=b^2-4ac#, with #a=1#, #b=-3# and #c=2#

#Delta=(-3)^2-4(1)(2)=1=>sqrt Delta=+-1#

#x_1=(-b+sqrt Delta)/(2a)# and #x_2=(-b-sqrt Delta)/(2a)#

#x_1=(3+1)/(2)=2# and #x_2=(3-1)/(2)=1#

So, the #V.A# are #x=1# and #x=2#

Hope this helps :)