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

Question

Sadie Alexander · Accepted Answer

You evaluate the limits of the function as x approaches infinities (#-oo# and #oo#). If one of them is a real number #k#, than #y=k# is a horizontal asymptote.

There are no horizontal asymptotes in this function.

Horizontal asymptote means that it approches a certain value of #y# and tends to make a horizontal line as it approaches (limit involvement) infinite values of x (whether positive or negative). Therefore, the two possible asymptotes of the function must be constant functions and can be calculated with the following way: Possible horizontals #lim_(x->oo)f(x)=k_1# #lim_(x->-oo)f(x)=k_2# If #k_1# or #k_2# are not real numbers then there are no asymptotes. Solution #f(x)=(3x^2+2x-1)/(x+1)# Solving the numerator: #Δ=b^2-4*a*c=2^2-4*3*(-1)=16# #x=(-b+-sqrt(Δ))/(2*a)=(-2+-sqrt(16))/(2*3)=(-2+-4)/6# The two solutions are #x=1/3# and #x=-1# So the numerator can be written as follows: #3x^2+2x-1=3(x-1/3)(x+1)# The two limits are now easier to find: #k_1=lim_(x->oo)f(x)=lim_(x->oo)(3x^2+2x-1)/(x+1)=# #=lim_(x->oo)(3(x-1/3)(x+1))/(x+1)=lim_(x->oo)3(x-1/3)=oo# #k_2=lim_(x->-oo)f(x)=lim_(x->-oo)(3x^2+2x-1)/(x+1)=# #=lim_(x->-oo)(3(x-1/3)(x+1))/(x+1)=lim_(x->-oo)3(x-1/3)=-oo# Since neither #k_1# nor #k_2# are real numbers, there are no horizontal asymptotes.

Sadie Alexander · Answer

undefined To find the horizontal asymptote of the function $ f(x) = \frac{3x^2 + 2x - 1}{x + 1} $, you can look at the behavior of the function as $ x $ approaches positive infinity or negative infinity.

1. As $ x $ approaches positive infinity:
$$ f(x) = \frac{3x^2 + 2x - 1}{x + 1} $$
$$ \lim_{{x \to +\infty}} f(x) = \lim_{{x \to +\infty}} \frac{3x^2 + 2x - 1}{x + 1} $$
$$ \lim_{{x \to +\infty}} f(x) = \lim_{{x \to +\infty}} \frac{3x^2}{x} = \lim_{{x \to +\infty}} 3x = +\infty $$

2. As $ x $ approaches negative infinity:
$$ f(x) = \frac{3x^2 + 2x - 1}{x + 1} $$
$$ \lim_{{x \to -\infty}} f(x) = \lim_{{x \to -\infty}} \frac{3x^2 + 2x - 1}{x + 1} $$
$$ \lim_{{x \to -\infty}} f(x) = \lim_{{x \to -\infty}} \frac{3x^2}{x} = \lim_{{x \to -\infty}} 3x = -\infty $$

Since the limit as $ x $ approaches both positive and negative infinity is either positive or negative infinity, there is no horizontal asymptote for this function.