# How do you find the Vertical, Horizontal, and Oblique Asymptote given f(x) = ((x – 10)(x + 5) )/( (13x + 10)(10–x))?

Nov 26, 2017

The vertical asymptote is $x = - \frac{10}{13}$. The horizontal asymptote is $y = - \frac{1}{13}$. (There's a removable discontinuity at $x = 10$.)

#### Explanation:

First notice that $10 - x = - \left(x - 10\right)$. So we can simplify the given function:

f(x) = ((x – 10)(x + 5) )/( (13x + 10)(10–x)) = -frac(x+5)(13x+10),x!=10.

(Since that factor reduces out we know that the function has a removable discontinuity--or hole--at x = 10, but that wasn't asked.)

Now, vertical asymptotes are the zeros of the factors of the denominator that do not cancel with factors of the numerator, so in this case a vertical asymptote is at $x = - \frac{10}{13}$.

Horizontal asymptotes are the ratio of the leading coefficients of the numerator and denominator if both the numerator and denominator have the same degree (which they do in this case).

The horizontal asymptote in this case is $y = - \frac{1}{13}$ for this function because the leading coefficient in the numerator is 1, the leading coefficient in the denominator is 13, and there's a negative just hanging out from the cancelled factor.