# How do you identify all horizontal and slant asymptote for f(x)=(3x^2+1)/(x^2+x+9)?

Dec 18, 2016

Horizontal: $\leftarrow y = 3 \rightarrow$

#### Explanation:

By actual division,

$f \left(x\right) = 3 - \frac{3 x + 26}{{x}^{2} + x + 9}$

y = quotient = 3 gives the asymptote.

For x <= -26/3, y>=3.

On the left side, (-10, 3.04040404..) is on the curve.

Yet, as $x \to - \infty , y \to 3$, turning back to the asymptote

graph{y(x^2+x+9)-3x^2-1 = 0 [-10, 10, -5.21, 5.21]}

Dec 18, 2016

horizontal asymptote at y = 3

#### Explanation:

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 ${x}^{2}$

$f \left(x\right) = \frac{\frac{3 {x}^{2}}{x} ^ 2 + \frac{1}{x} ^ 2}{{x}^{2} / {x}^{2} + \frac{x}{x} ^ 2 + \frac{9}{x} ^ 2} = \frac{3 + \frac{1}{x} ^ 2}{1 + \frac{1}{x} + \frac{9}{x} ^ 2}$

as $x \to \pm \infty , f \left(x\right) \to \frac{3 + 0}{1 + 0 + 0}$

$\Rightarrow y = 3 \text{ is the asymptote}$

Slant 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 slant asymptotes.
graph{(3x^2+1)/(x^2+x+9) [-10, 10, -5, 5]}