# For what values of x, if any, does f(x) = 1/((2x-3)(7x-6)  have vertical asymptotes?

Dec 19, 2016

$\text{vertical asymptotes at " x=3/2" and } x = \frac{6}{7}$

#### 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 : $\left(2 x - 3\right) \left(7 x - 6\right) = 0 \Rightarrow x = \frac{3}{2} \text{or } x = \frac{6}{7}$

$\Rightarrow x = \frac{3}{2} \text{ and " x=6/7" are the asymptotes}$
graph{1/((2x-3)(7x-6) [-10.52, 10.525, -5.26, 5.26]}