How do you find the vertical, horizontal or slant asymptotes for #y=(3x^2+x-4) / (2x^2-5x) #?

1 Answer
Nov 29, 2016

Answer:

vertical asymptotes at #x=0" and " x=5/2#
horizontal asymptote at #y=3/2#

Explanation:

The denominator of y cannot be zero as this would make y 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^2-5x=0rArrx(2x-5)=0#

#rArrx=0" and " x=5/2" are the asymptotes"#

Horizontal asymptotes occur as

#lim_(xto+-oo),ytoc" ( a constant)"#

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

#y=((3x^2)/x^2+x/x^2-4/x^2)/((2x^2)/x^2-(5x)/x^2)=(3+1/x-4/x^2)/(2-5/x)#

as #xto+-oo,yto(3+0-0)/(2-0)#

#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-4)/(2x^2-5x) [-10, 10, -5, 5]}