Question #52abf

Mar 10, 2017

Yes, you are correct. The curve does indeed have x-intercepts at $\left(2 , 0\right)$ and $\left(- 2 , 0\right)$.

Explanation:

A point that lies on the x-axis will be of the form $\left(x , 0\right)$ where $a$ is any real number.

To find the value of $x$, we must set $y$ to $0$.

${0}^{3} \left({e}^{x}\right) - {x}^{2} + 5 {e}^{0} = 1$

$- {x}^{2} + 5 \left(1\right) = 1$

$- {x}^{2} + 4 = 0$

$4 = {x}^{2}$

$x = \pm 2$

Hopefully this helps!