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

Question

Liam Brennan · Accepted Answer

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

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]}

Elijah Allen · Answer

undefined To find the vertical asymptote(s) of a rational function, set the denominator equal to zero and solve for x. In this case, the vertical asymptote occurs when x - 4 = 0, yielding x = 4.

To find the horizontal asymptote, compare the degrees of the numerator and denominator. If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is y = 0. If the degree of the numerator is equal to the degree of the denominator, divide the leading coefficients of both polynomials. In this case, since both the numerator and denominator have the same degree (1), the horizontal asymptote is y = coefficient of x in numerator / coefficient of x in denominator, which is 1/1 = 1.

To find oblique asymptotes, divide the numerator by the denominator using long division or synthetic division. The quotient obtained will represent the equation of the oblique asymptote, if it exists.