How do you find the vertical, horizontal or slant asymptotes for # r(x)= ((2x^2+14x-36)/(x^2+x-12))#?

1 Answer
Jul 13, 2016

Answer:

vertical asymptotes x = -4 , x = 3
horizontal asymptote y = 2

Explanation:

The denominator of r(x) cannot equal zero as this is undefined. Setting the denominator equal 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: #x^2+x-12=0rArr(x+4)(x-3)=0#

#rArrx=-4,x=3" are the asymptotes"#

Horizontal asymptotes occur as

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

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

#((2x^2)/x^2+(14x)/x^2-36/x^2)/(x^2/x^2+x/x^2-12/x^2)=(2+14/x-36/x^2)/(1+1/x-12/x^2)#

as #xto+-oo,r(x)to(2+0-0)/(1+0-0)#

#rArry=2" 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 2) Hence there are no slant asymptotes.
graph{(2x^2+14x-36)/(x^2+x-12) [-20, 20, -10, 10]}