How do you find vertical, horizontal and oblique asymptotes for # Q(x)= (3x^2-2x-8)/(2x^3+x^2-3x)#?

1 Answer
Jul 15, 2016

Answer:

vertical asymptotes x#=-3/2# , x = 0 , x = 1
horizontal asymptote y = 0

Explanation:

The denominator of Q(x) 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^3+x^2-3x=0rArrx(2x+3)(x-1)=0#

#rArrx=-3/2,x=0,x=1" are the asymptotes"#

Horizontal asymptotes occur as

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

divide terms on numerator/denominator by the highest power of x , that is #x^3#

#((3x^2)/x^3-(2x)/x^3-8/x^3)/((2x^3)/x^3+x^2/x^3-(3x)/x^3)=(3/x-2/x^2-8/x^3)/(2+1/x-3/x^2)#

as #xto+-oo,Q(x)to(0-0-0)/(2+0-0)to0/2#

#rArry=0" is the asymptote"#

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here (numerator-degree 2 ,denominator-degree 3) Hence there are no oblique asymptotes.
graph{(3x^2-2x-8)/(2x^3+x^2-3x) [-10, 10, -5, 5]}