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

Jul 17, 2018

y=0

#### Explanation:

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).

Jul 17, 2018

$\text{horizontal asymptote at } y = 0$

#### Explanation:

$\text{Horizontal asymptotes occur as}$

${\lim}_{x \to \pm \infty} , f \left(x\right) \to c \text{ ( a constant)}$

$\text{divide terms on numerator/denominator by the highest}$
$\text{power of "x" that is } {x}^{2}$

$f \left(x\right) = \frac{\frac{x}{x} ^ 2 + \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{3 x}{x} ^ 2 - \frac{4}{x} ^ 2} = \frac{\frac{1}{x} + \frac{1}{x} ^ 2}{1 + \frac{3}{x} - \frac{4}{x} ^ 2}$

$\text{as } x \to \pm \infty , f \left(x\right) \to \frac{0 + 0}{1 + 0 - 0}$

$\Rightarrow y = 0 \text{ is the asymptote}$
graph{(x+1)/(x^2+3x-4) [-10, 10, -5, 5]}