Verification on Intersection of Tangent Lines.

Question

Given a function $\gamma (t) = (-1,4)(2-3t)^2 + (1,0)(3t - 1)^2 $ figure out the control point $P1$ which exist at the intersection of the tangent lines of $P0$ = $\gamma (0) $ and $P2$ = $\gamma (1) $.

I'm a little rusty with intersections using the tangent lines, so any clarification/ correction would be appreciated.

I got $P1 = (1/3, -112)$ which seems odd cause -112 is really far off for a control point for such a parabola.

EDIT: figured it out I'm an idiot $P1=(0, -8)$

Answer 1

Hint: convert the equation to Bézier form. Then the middle control point is the intersection of the two end tangents.