So it is expressed as: y =1/2 (x+4) (x-1) =1/2x? +3 times /2-2
Substitute X=0 and Y-=-2. So the c coordinate is (0, -2)
(2)EF‖AC, simple triangle BAC is similar to triangle BEF.
BE/BA=BF/BC
△CEF and △BEF are equal in height. When CF=2BF, the area of △CEF is twice that of △BEF.
At this time BF/BC= 1/3, then BE/BA= 1/3,
AB=5, then BE=5/3. So the coordinate of point E is (-2/3,0).
(3) Let the expression of AC function be y = kx+b.
Replace (-4,0) (0,2)
B=-2, K=- 1/2. So the function is Y=-X/2-2.
Since P is on a parabola, let the coordinate of P be (x, 1/2X? +3X/2-2)
PQ is parallel to Y axis and Q is on AC, so let Q coordinate be (x, -X/2-2).
Because point Q is above p, PQ=-X/2-2-( 1/2X? +3X/2-2)=- 1/2X？ -2 times
When x =-(-2)/[2× (- 1/2)] =-2, the PQ value is the largest.
At this time, the abscissa of point P is -2, and X=-2 is substituted into the parabolic analytical formula.
Y=-3
The coordinates of point P are (-2, -3).
The above is a complete process.
I hope it helps you.