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

Question

Sadie Alexander · Accepted Answer

Slant Asymptote: #y=2x-1#

Vertical Asymptote: #x=-1#

Horizontal Asymptote: None

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.

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

Chloe Alexander · Answer

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

Madelyn Anderson · Answer

undefined To find the vertical asymptote(s), set the denominator equal to zero and solve for $ x $. In this case, set $ x + 1 = 0 $, so $ x = -1 $. Thus, there's a vertical asymptote at $ x = -1 $.

To find the horizontal asymptote, compare the degrees of the numerator and the denominator. Since the degree of the numerator (2) is less than the degree of the denominator (1), there is no horizontal asymptote.

To check for slant asymptotes, perform polynomial long division of the numerator by the denominator. After dividing $ (2x^2 + x + 2) $ by $ (x + 1) $, if there's a remainder, there is a slant asymptote. If the remainder is not zero, it indicates a slant asymptote. If the remainder is zero, there is no slant asymptote.