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

1 Answer
Mar 12, 2016

Answer:

Vertical asymptote at #x=4# and slanting asymptote at #y=2x#

Explanation:

In #y=(2x^3-5x+3)/(x^2-5x+4)#, the factors of denominator are #(x-4)(x-1)# and numerator is also divisible by #x-1#, as such

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

Hence, as denominator will be zero for #x=4#, we will have vertical asymptote at #x=4#.

Further as degree of numerator in #(2x^2+2x-3)/(x-4)# is greater than that of denominator and ratio of highest degrees by one in the two is #2x^2/x=2x#, we will have a slanting asymptote at #y=2x#

graph{(2x^3-5x+3)/(x^2-5x+4) [-15, 15, -5, 5]}