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Question #52abf

1 Answer

Mar 10, 2017

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!

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