How do you find the vertical, horizontal or slant asymptotes for #f(x) = (2x^2-5x-12)/(3x^2-11x-4 )#?

Question

Sadie Alexander · Accepted Answer

vertical asymptote at #x=-1/3#
horizontal asymptote at #y=2/3# The first step is to factorise and simplify f(x). #f(x)=((2x+3)cancel((x-4)))/((3x+1)cancel((x-4)))=(2x+3)/(3x+1)# The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the value that x cannot be and if the numerator is non-zero for this value then it is a vertical asymptote. solve : #3x+1=0rArrx=-1/3" is the asymptote"# Horizontal asymptotes occur as #lim_(xto+-oo),f(x)to c" (a constant)"# divide numerator/denominator by x #f(x)=((2x)/x+3/x)/((3x)/x+1/x)=(2+3/x)/(3+1/x)# as #xto+-oo,f(x)to(2+0)/(3+0)# #rArry=2/3" is the asymptote"# Slant asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 1 ) Hence there are no slant asymptotes. graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}

Anna Allen · Answer

undefined To find the vertical asymptotes, set the denominator equal to zero and solve for $x$. The values of $x$ obtained are the vertical asymptotes if they are not canceled out by factors in the numerator.

To find the horizontal asymptotes, 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 degrees are equal, divide the leading coefficients of the numerator and denominator to find the horizontal asymptote. If the degree of the numerator is greater, there is no horizontal asymptote.

To find the slant asymptote, perform polynomial long division on the numerator and denominator. The quotient obtained is the equation of the slant asymptote.