# How do you find the Vertical, Horizontal, and Oblique Asymptote given #f(x)=((3x-2)(x+5))/((2x-1)(x+6))#?

##### 1 Answer

vertical asymptotes x = -6 , x

horizontal asymptote

#### Explanation:

For this rational function the denominator cannot be zero. This would lead to division by zero which is undefined.By setting the denominator equal to zero and solving for x we can find the values that x cannot be and if the numerator is also non-zero for these values of x then they must be vertical asymptotes.

solve : (2x-1)(x +6 ) =0

#rArrx=-6,x=1/2#

#rArrx=-6" and " x=1/2" are the asymptotes"# Horizontal asymptotes occur as

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

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