How do you find all the asymptotes for function # f(x)= (4x^2-x+2)/(x+1)#?

Question

Sadie Ackerman · Accepted Answer

If you have also studied slant asymptotes, you will include: #y=4x-5# as a slant asymptote.

By either long division of synthetic division, get: #(4x^2-x+2)/(x+1) = 4x-5+ 7/(x+1)# So the line #y=4x-5# is a slant asymptote.

Sophie Adams · Answer

undefined To find the asymptotes of the function $ f(x) = \frac{{4x^2 - x + 2}}{{x + 1}} $, we need to identify any vertical, horizontal, or slant asymptotes.

Vertical asymptotes occur where the denominator equals zero, but the numerator does not. 
Horizontal asymptotes are found by analyzing the behavior of the function as $ x $ approaches positive or negative infinity.
Slant asymptotes may occur when the degree of the numerator is one more than the degree of the denominator.

1. Vertical asymptotes:
Set the denominator equal to zero and solve for $ x $:

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

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

2. Horizontal asymptote:
Check the limit of the function as $ x $ approaches positive or negative infinity:

$$ \lim_{{x \to \infty}} \frac{{4x^2 - x + 2}}{{x + 1}} = \infty $$
$$ \lim_{{x \to -\infty}} \frac{{4x^2 - x + 2}}{{x + 1}} = \infty $$

Since the limit approaches infinity for both positive and negative infinity, there is no horizontal asymptote.

3. Slant asymptote:
If the degree of the numerator is one more than the degree of the denominator, there may be a slant asymptote. Here, the degree of the numerator (2) is equal to the degree of the denominator (1), so there is no slant asymptote.

Therefore, the only asymptote for the function $ f(x) = \frac{{4x^2 - x + 2}}{{x + 1}} $ is the vertical asymptote at $ x = -1 $.