How do you find the vertical, horizontal or slant asymptotes for # ((3x-2)(x+5))/((2x-1)(x+6))#?

1 Answer
Jul 14, 2016

Answer:

vertical asymptotes x = -6 , x#=1/2#
horizontal asymptote y#=3/2#

Explanation:

The denominator of the rational function 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: (2x-1)(x+6)=0#rArrx=-6,x=1/2" are the asymptotes"#

Horizontal asymptotes occur as

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

#f(x)=((3x-2)(x+5))/((2x-1)(x+6))=(3x^2+13x-10)/(2x^2+11x-6)#

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

#((3x^2)/x^2+(13x)/x^2-10/x^2)/((2x^2)/x^2+(11x)/x^2-6/x^2)=(3+13/x-10/x^2)/(2+11/x-6/x^2)#

as #xto+-oo,f(x)to(3+0-0)/(2+0-0)#

#rArry=3/2" 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{((3x-2)(x+5))/((2x-1)(x+6)) [-10, 10, -5, 5]}