How do you find vertical, horizontal and oblique asymptotes for #h(x) = (2x^2-5x-12)/(3x^2-11x-4 )#?

1 Answer
Aug 13, 2018

Answer:

#"vertical asymptote at "x=-1/3#
#"horizontal asymptote at "y=2/3#

Explanation:

#"factorise numerator/denominator and simplify"#

#h(x)=((2x+3)cancel((x-4)))/(cancel((x-4))(3x+1))=(2x+3)/(3x+1)#

The denominator of h(x) cannot be zero as this would make h(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),h(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by " x#

#h(x)=((2x)/x+3/x)/((3x)/x+1/x)=(2+3/x)/(3+1/x)#

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

#y=2/3" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator is greater than the degree of the denominator. This is not the case here hence there is no oblique asymptote.
graph{(2x+3)/(3x+1) [-10, 10, -5, 5]}