# How do you find the horizontal asymptote for f(x)= (x-3)/ (x^2-3x-10)?

Jun 27, 2018

No horizontal asymptote in given case.

#### Explanation:

There is no horizontal asymptote for $f \left(x\right) = \frac{x - 3}{{x}^{2} - 3 x - 10}$.

graph{(x-3)/(x^2-3x-10) [-10, 10, -5, 5]}

Horizontal asymptote is there only if in such expressions, degree of numerator is equal to that of denominator. If degree of numerator is one more than that of denominator, we have an oblique or slanting asymptote. The example of latter is $f \left(x\right) = \frac{{x}^{2} - 3 x - 10}{x - 3}$ as

${\lim}_{x \to \infty} \frac{{x}^{2} - 3 x - 10}{x - 3} = {\lim}_{x \to \infty} \frac{x - 3 - \frac{10}{x}}{1 - \frac{3}{x}}$

= $x - 3$

and hence oblique or slanting asymptote of $f \left(x\right) = \frac{{x}^{2} - 3 x - 10}{x - 3}$ is $y = x - 3$.

If $f \left(x\right) = \frac{{x}^{2} - 3 x - 10}{2 {x}^{2} - 3}$ we have

${\lim}_{x \to \infty} \frac{{x}^{2} - 3 x - 10}{2 {x}^{2} - 3}$

= ${\lim}_{x \to \infty o =} \frac{1 - \frac{3}{x} - \frac{10}{x} ^ 2}{2 - \frac{3}{x} ^ 2}$

= $\frac{1}{2}$

graph{(x^2-3x-10)/(2x^2-3) [-10, 10, -5, 5]}