How do you find the asymptotes for #f(x)=(2x^3+3x^2+x)/(6x^2+x-1)#?

Question

Sophie Adams · Accepted Answer

The vertical asymptote is #x=1/3#. No horizontal asymptote. The slant asymptote is #y=1/3x+4/9#

The numerator is

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

#=x(2x+1)(x+1)#

The denominator is

#6x^2+x-1=(3x-1)(2x+1)#

The function is

#f(x)=(2x^3+3x^2+x)/(6x^2+x-1)#

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

#=(x(x+1))/(3x-1)#

As the denominator must de #!=0#, therefore

#3x-1!=0#, #=>#, #x!=1/3#

The vertical asymptote is #x=1/3#

There is no horizontal asymptote, as

#lim_(x->oo)f(x)=lim_(x->oo)(x^2)/(3x)=lim_(x->oo)x/3=+oo#

#lim_(x->-oo)f(x)=lim_(x->-oo)(x^2)/(3x)=lim_(x->-oo)x/3=-oo#

As the degree of the numerator is #># the degree of the denominator, there is a slant asymptote.

Perform a long division

#color(white)(aaaa)##x^2+x##color(white)(aaaa)##|##3x-1#

#color(white)(aaaa)##x^2-1/3x##color(white)(aaaa)##|##1/3x+4/9#

#color(white)(aaaaa)##0+4/3x#

#color(white)(aaaaaaa)##+4/3x+0#

#color(white)(aaaaaaa)##+4/3x-4/9#

#color(white)(aaaaaaaaa)##+0+4/9#

Therefore,

#f(x)=(x^2+x)/(3x-1)=(1/3x+4/9)+(4/9)/(3x-1)#

#lim_(x->+oo)(f(x)-(1/3x+4/9))=lim_(x->+oo)(4/9)/(3x-1)=0^+#

#lim_(x->-oo)(f(x)-(1/3x+4/9))=lim_(x->-oo)(4/9)/(3x-1)=0^-#

The slant asymptote is

#y=1/3x+4/9#

graph{(y-(x(x+1))/(3x-1))(y-1/3x-4/9)=0 [-4.382, 4.386, -2.19, 2.193]}

Sophie Adams · Answer

undefined To find the asymptotes of the rational function $ f(x) = \frac{2x^3 + 3x^2 + x}{6x^2 + x - 1} $, you need to determine both vertical and horizontal asymptotes.

Vertical asymptotes occur where the denominator equals zero and the numerator doesn't. 
To find vertical asymptotes, solve $6x^2 + x - 1 = 0$ for $x$.

Horizontal asymptotes can be found by comparing the degrees of the numerator and denominator. 
If the degree of the numerator is greater than the degree of the denominator, there is no horizontal asymptote. 
If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is at $y = 0$.
If the degrees are equal, divide the leading coefficients of both terms to find the horizontal asymptote.

So, to summarize:
1. Find vertical asymptotes by solving $6x^2 + x - 1 = 0$ for $x$.
2. Determine horizontal asymptotes by comparing the degrees of the numerator and denominator.