# For what values of x, if any, does f(x) = 1/((5x+8)(x-6)  have vertical asymptotes?

Jan 29, 2016

$x = - \frac{8}{5}$ and $x = 6$

#### Explanation:

For a rational function such as the one presented, vertical asymptotes occur whenever the denominator of the function is equal to $0$. These are the "holes" in the function's domain since it's impossible to divide by $0$.

Now, we must find the times when

$\left(5 x + 8\right) \left(x - 6\right) = 0$

What we have here are two terms, $5 x + 8$ and $x - 6$, that are being multiplied and equalling $0$. This means that either one of the terms must be equal to $0$, but it doesn't matter which.

Thus, to find the times when the whole expression equals $0$, we can set either term equal to $0$.

$m a t h b f \left(\left(1\right)\right)$

$5 x + 8 = 0$

$5 x = - 8$

$x = - \frac{8}{5}$

There is a vertical asymptote at $x = - \frac{8}{5}$.

$m a t h b f \left(\left(2\right)\right)$

$x - 6 = 0$

$x = 6$

There is a vertical asymptote at $x = 6$.

We can check a graph of the original function:

graph{1/((5x+8)(x-6)) [-7.75, 12.25, -4.915, 5.085]}