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

1 Answer
Jul 5, 2018

Answer:

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

Explanation:

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]}