# For what values of x, if any, does f(x) = 1/((x-3)(x+4)(e^x-x))  have vertical asymptotes?

Dec 3, 2016

The graph depicts the answer: x = 3 and $x = - 4$.

#### Explanation:

${e}^{x} - x$ has the minimum 1, at x = 0.

So, $y \to \pm \infty$, as $x \to 3 \mathmr{and} x \to - 4$.

And so, the vertical asymptotes are given by x = 3 and $x = - 4$.

The y-intercept of the graph is $- \frac{1}{12}$ and it is the local minimum.

graph{y(x-3)(x+4)(e^x-x)-1=0 [-5, 5, -2.5, 2.5]}