How do you find the vertical, horizontal or slant asymptotes for #f(x) = (2x^2 + x + 2) /( x + 1)#?

2 Answers
May 12, 2016

Answer:

Vertical asymptote at #x=-1#
No horizontal asymptote
Slant asymptote: #f(x)=2x-1#

Explanation:

Given: #f(x)=(2x^2+2x+2)/(x+1)#

#f(x)# is undefined when #(x+1)=0#
giving us the vertical asymptote of #x=-1#

#lim_(xrarroo) f(x) rarr oo#
and
#lim_(xrarr-oo) f(x) rarr -oo#
so there is no horizontal asymptote.

Since the degree of the numerator is greater than the degree of the denominator,
we can divide the denominator into the numerator to get a slant asysmptote:
#f(x)=(2x^2+x+2)div(x+1)=(2x-1) +3/(x+1)#
So the slant asymptote is #f(x)=2x-1#

Answer:

Slant Asymptote: #y=2x-1#

Vertical Asymptote: #x=-1#

Horizontal Asymptote: None

Explanation:

The given function is #y=(2x^2+x+2)/(x+1)#

To find the slant asymptote, divide numerator by the denominator of the given rational function.
#" " " " " "underline(" "2x-1" " " " " ")#
#x+1""|~" "2x^2+x+2#
#" " " " " "underline(2x^2+2x" " " " " ")#
#" " " " " " " " " "-x+2#
#" " " " " " " " " "underline(-x-1)#
#" " " " " " " " " " " " " " "+3#

The result of the division is

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

The whole number part of the quotient which is #2x-1# becomes the right side part of the linear equation

#color(red)(y=2x-1)#

which is the #color(red)("Slant Asymptote")#

To solve for the Vertical Asymptote, use the divisor and equate to zero

#x+1=0#

and the Vertical Asymptote is

#color(red)(x=-1 " is the Vertical Asymptote")#

There is #color(red)("No Horizontal Asymptote")#

Kindly see the graph of the function #y=(2x^2+x+2)/(x+1)# colored blue, the slant asymptote #y=2x-1# colored red, vertical asymptote #x=-1# colored green below.

desmos

God bless....I hope the explanation is useful.