Find the points of intersection of the curves y=2sin(x-3) and y=-4x^2 + 2

Answer 1

tough question!

I used Newton's Method
and I let
y = 2sin(x-3) + 4x^2 - 2
then dy/dx = 2cos(x-3) + 8x

x_new = x - (2sin(x-3) + 4x^2 - 2)/(2cos(x-3) + 8x)

I started with x = 1
and my first x_new was.974691
make that your current x and find the next x_new
I quickly converged to
x = 0.974237

set your calculator to radians and test my answer, it works making both equations equal.
(error was .000000418 between LS and RS

Answer 2

Reiny found a positive root but there is a negative root x=-0.55 (approximately)
We can started with x=-1