Question

Question #52abf

Answer 1

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

Explanation:

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

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

`0^3(e^x) - x^2 + 5e^0 = 1`

`-x^2 + 5(1) = 1`

`-x^2 + 4= 0`

`4 = x^2`

`x = +-2`

Hopefully this helps!