Potential energy curve, turning points

[J-dizzal]

Homework Statement

The figure shows a plot of potential energy U versus position x of a 0.220 kg particle that can travel only along an x axis under the influence of a conservative force. The graph has these values: UA = 9 J, UC = 20 J and UD = 24 J. The particle is released at the point where U forms a “potential hill” of “height” UB = 12 J, with kinetic energy 7.00 J. What is the speed of the particle at (a)x = 3.5 m and (b)x = 6.5 m? What is the position of the turning point on (c) the right side and (d) the left side?

http://edugen.wileyplus.com/edugen/courses/crs7165/art/qb/qu/c08/q40.jpg

Homework Equations

K=E-U

The Attempt at a Solution

I know the turning point is when K=0 but not sure how to find that from this graph.

 

[Nathanael]

I know the turning point is when K=0 but not sure how to find that from this graph.

Kinetic energy is zero when the potential energy is equal to the total energy.