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!