How do you find vertical, horizontal and oblique asymptotes for #(x^2-5x+6)/(x-4)#?

1 Answer
Mar 7, 2017

Answer:

The vertical asymptote is #x=4#
The oblique asymptote is #y=x-1#
No horizontal asymptote

Explanation:

As you cannot divide by #0#, #=>#, #x!=4#

The vertical asymptote is #x=4#

The degree of the numerator is #># than the degree of the denominator, there is an oblique asymptote.

Let #f(x)=(x^2-5x+6)/(x-4)#

Let's do a long division

#color(white)(aaaa)##x^2-5x+6##color(white)(aaaa)##|##x-4#

#color(white)(aaaa)##x^2-4x##color(white)(aaaaaaaa)##|##x-1#

#color(white)(aaaaa)##0-x+6#

#color(white)(aaaaaaa)##-x+4#

#color(white)(aaaaaaa)##-0+2#

Therefore,

#f(x)=(x-1)+2/(x-4)#

#lim_(x->-oo)(f(x)-(x-1))=lim_(x->-oo)2/(x-4)=0^-#

#lim_(x->+oo)(f(x)-(x-1))=lim_(x->+oo)2/(x-4)=0^+#

The oblique asymptote is #y=x-1#

graph{(y-(x^2-5x+6)/(x-4))(y-x+1)(y-100(x-4))=0 [-18.3, 17.74, -6.74, 11.28]}