The parabola y = AX2 + BX + C passes through point a (3,0). B (2, - 3) with x = 1 as the symmetry axis Is there a point P on the symmetry axis X = 1 so that PA = Pb in the triangle PAB? If so, find out the coordinates of point P. if not, explain the reason

The parabola y = AX2 + BX + C passes through point a (3,0). B (2, - 3) with x = 1 as the symmetry axis Is there a point P on the symmetry axis X = 1 so that PA = Pb in the triangle PAB? If so, find out the coordinates of point P. if not, explain the reason

Because the parabola y = ax ^ 2 + BX + C passes through the point a (3,0). B (2, - 3), two points are brought into the equation, 9A + 3B + C = 0, 4A + 2B + C = - 3, the symmetry axis is x = - B / 2A = 1. From the above three formulas, a = 1, B = - 2, C = - 3 parabola y = x ^ 2-2x-3 suppose that there is P point so that PA = Pb, then P point is on the vertical bisector of line AB, and the line

Given that the coordinates of the vertex of the parabola are (1, - 3), and intersect with the Y-axis and (0,1), the analytical expression of the parabola is obtained

Let the analytic formula of parabola be y = ax & sup2; + BX + C, so the vertex coordinates are [- B / 2a, (4ac-b & sup2;) / 4A]. We know that the vertex coordinates are (1, - 3), that is - B / 2A = 1, (4ac-b & sup2;) / 4A = - 3, and because the parabola passes through the point (0,1), then C = 1, then a = 4, B = - 8