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What is the horizontal asymptote of #f(x) = (x+1) / (x^2 +3x - 4)#?

Answer 1

Caleb Alexander

#"horizontal asymptote at "y=0#

#"Horizontal asymptotes occur as"#

#lim_(xto+-oo),f(x)toc" ( a constant)"#

#"divide terms on numerator/denominator by the highest"# #"power of "x" that is "x^2#

#f(x)=(x/x^2+1/x^2)/(x^2/x^2+(3x)/x^2-4/x^2)=(1/x+1/x^2)/(1+3/x-4/x^2)#

#"as "xto+-oo,f(x)to(0+0)/(1+0-0)#

#rArry=0" is the asymptote"# graph{(x+1)/(x^2+3x-4) [-10, 10, -5, 5]}

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Answer 2

Caleb Alexander

y=0

If the polynomial in the numerator is a lower degree than the denominator, the x-axis (y = 0) is the horizontal asymptote. The degree is the power of the x variable(s).

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Answer 3

Theodore Amidon

To find the horizontal asymptote of the function (f(x) = \frac{x + 1}{x^2 + 3x - 4}), we observe the behavior of the function as (x) approaches positive or negative infinity.

First, we analyze the degrees of the numerator and denominator. Since the degree of the numerator (1) is less than the degree of the denominator (2), the horizontal asymptote occurs at (y = 0).

Therefore, the horizontal asymptote of the function (f(x) = \frac{x + 1}{x^2 + 3x - 4}) is (y = 0).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.