# How do you find vertical, horizontal and oblique asymptotes for f(x) = (x^2 - 3x)/(x^2 + 1)?

Aug 16, 2016

horizontal asymptote at y = 1

#### Explanation:

The denominator of f(x) cannot be zero as this would make f(x) undefined. Equating the denominator to zero and solving gives the values that x cannot be and if the numerator is non-zero for these values then they are vertical asymptotes.

solve: ${x}^{2} + 1 = 0 \Rightarrow {x}^{2} = - 1$ which has no real solutions hence there are no vertical asymptotes.

Horizontal asymptotes occur as

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

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

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

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

$\Rightarrow y = 1 \text{ is the asymptote}$

Oblique asymptotes occur when the degree of the numerator > degree of the denominator. This is not the case here ( both of degree 2) Hence there are no oblique asymptotes.
graph{(x^2-3x)/(x^2+1) [-10, 10, -5, 5]}