How do you find the asymptotes for #(2x^2 + x + 2) / (x + 1)#?

Question

Elijah Allen · Accepted Answer

The vertical asymptote is #x=-1#
The slant asymptote is #y=2x-1#
No horizontal asymptote

Let #f(x)=(2x^2+x+2)/(x+1)# The domain of #f(x)# is #D_f(x)=RR-{-1}# As you cannot divide by #0#, #x!=-1# The vertical asymptote is #x=-1# As we the degree of the numerator is #># than the degree of the denominator, there is a slant asymptote. Let's do a long division #color(white)(aaaa)##2x^2+x+2##color(white)(aaaa)##∣##x+1# #color(white)(aaaa)##2x^2+2x##color(white)(aaaaaaa)##∣##2x-1# #color(white)(aaaaa)##0-x+2# #color(white)(aaaaaaa)##-x-1# #color(white)(aaaaaaaaa)##0+3# Therefore, #f(x)=(2x-1)+(3)/(x+1)# So, #lim_(x->-oo)(f(x)-(2x-1))=lim_(x->-oo)3/(x+1)=0^-# #lim_(x->+oo)(f(x)-(2x-1))=lim_(x->+oo)3/(x+1)=0^+# The slant asymptote is #y=2x-1# graph{(y-(2x^2+x+2)/(x+1))(y-2x+1)(y-50x-50)=0 [-25.64, 25.67, -12.83, 12.84]}

Lincoln Anderson · Answer

undefined To find the asymptotes of the rational function $ \frac{{2x^2 + x + 2}}{{x + 1}} $, you need to consider two types of asymptotes: vertical asymptotes and horizontal asymptotes.

1. Vertical Asymptotes:
   Vertical asymptotes occur where the denominator of the rational function equals zero, but the numerator does not. In this case, set the denominator, $ x + 1 $, equal to zero and solve for $ x $. The solution gives you the equation of the vertical asymptote(s).

$ x + 1 = 0 $
   $ x = -1 $

Therefore, there is a vertical asymptote at $ x = -1 $.

2. Horizontal Asymptotes:
   Horizontal asymptotes can be found by analyzing the behavior of the function as $ x $ approaches positive or negative infinity. To find horizontal asymptotes, compare the degrees of the numerator and denominator of the function.

Since the degree of the numerator (2) is less than the degree of the denominator (1), there is a horizontal asymptote at $ y = 0 $, which is the $ x $-axis.

In summary:
- Vertical asymptote: $ x = -1 $
- Horizontal asymptote: $ y = 0 $