How do you find vertical, horizontal and oblique asymptotes for #R(x) = (6x^2 + x + 12)/(3x^2 - 5x - 2)#?

1 Answer
Aug 2, 2016

Answer:

vertical asymptotes #x=-1/3,x=2#
horizontal asymptote y = 2

Explanation:

The denominator of R(x) cannot be zero as this is undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: #3x^2-5x-2=0rArr(3x+1)(x-2)=0#

#rArrx=-1/3" and " x=2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),R(x)toc" (a constant)"#

divide terms on numerator/denominator by the highest power of x, that is #x^2#

#((6x^2)/x^2+x/x^2+12/x^2)/((3x^2)/x^2-(5x)/x^2-2/x^2)=(6+1/x+12/x^2)/(3-5/x-2/x^2)#

as #xto+-oo,R(x)to(6+0+0)/(3-0-0)#

#rArry=2" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2 ) Hence there are no oblique asymptotes.
graph{(6x^2+x+12)/(3x^2-5x-2) [-10, 10, -5, 5]}