# How do you find the vertical, horizontal or slant asymptotes for y = e^x (1-x^2)?

Jan 27, 2018

Horizontal asymptote at $y = 0$

#### Explanation:

Firstly, there are no singularities in this function (there is nowhere where we would have to "divide by 0"). As such there are no vertical asymptotic.

Lets look at the case where:

$x \to + \infty$ The function then becomes:

${e}^{x} \left(1 - {x}^{2}\right) \to - {e}^{x} {x}^{2}$

as the ${x}^{2}$ term dominates.

This increases non-linearly and as such will not be asymptotic. So there are no asymptotes going when $x \to + \infty$.

Now let's look at when $x \to - \infty$. The function becomes:

${e}^{x} \left(1 - {x}^{2}\right) \to {e}^{x} {x}^{2}$

again.

However in this case, $x$ is negative so the exponential product: ${e}^{x}$ will get closer and closer to $0$, and will do much faster than the ${x}^{2}$ product can "pull the function up."

As such there will be a horizontal asymptote at $y = 0$:

graph{e^x(1-x^2) [-10, 10, -5, 5]}