How do you find the vertical, horizontal and slant asymptotes of #a(x)=(2x^2-1) / (3x^3-2x+1)#?

1 Answer
Aug 20, 2015

Answer:

There is a vertical asymptote at #color(red)(x=-1)#, a horizontal asymptote at #color(red)(y=0)# and #color(red)("no")# slant asymptote.

Explanation:

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

Step 1. Find the vertical asymptotes.

Set the denominator equal to zero and solve for #x#.

#3x^3-2x+1=0#

According to the rational root theorem, the rational roots of #f(x) = 0# must all be of the form #p/q# with #p# a divisor of #1# and #q# a divisor of #3#.

So the only possible rational roots are #±1,±1/3#.

We have to test all four possibilities.

The only one that works is

#-1|" "3" "color(white)(1)0" "-2" "" "1#
#" "color(white)(1)|" "color(white)(1)-3" "" "3" "-1#
#" "color(white)(1)stackrel("———————————————)#
#" "" "" "3color(white)(1)-3" "" "1" "" "color(red)(0)#

So #3x^3-2x+1=(x+1)(3x^2-3x+1)#

#x+1=0# and #3x^2-3x+1=0#

#3x^2-3x+1=0# has no real zeroes.

#x=-1# is a zero.

There is a vertical asymptote at #x=-1#.

Step 2. Find the horizontal asymptotes.

The degree of the numerator is lower than the degree of the denominator, so the #x#-axis is the horizontal asymptote.

The horizontal asymptote is at #y=0#.

Step 3. Find the slant asymptotes.

A slant asymptote occurs when the degree of the numerator is higher than the degree of the denominator.

Here, the numerator is 2nd degree, and the denominator is 3rd degree, so there are no slant asymptotes.

Graph