Given that the image of a quadratic function passes through points (4, - 3), and when x = 3, there is a maximum value of 4, find the analytic expression of the quadratic function

Given that the image of a quadratic function passes through points (4, - 3), and when x = 3, there is a maximum value of 4, find the analytic expression of the quadratic function

The vertex coordinates are (3,4)
Let the analytic formula be y = a (x-3) ^ 2 + 4
(4. - 3) substitute - 3 = a * 1 + 4, a = - 7
That is, the analytic formula is y = - 7 (x-3) ^ 2 + 4

The parabola y = a (X-H) is known. When x = 2 has the maximum value, find the parabola analytical formula: and point out that when x is the value, y increases with the increase of X!

The quadratic function y = a (X-H) ^ 2 can formally determine that the vertex has the maximum value at (h, 0) when the opening is downward, so a

The analytical formula of parabola is y = - x square + 6x, the edge BC of rectangle is on the x-axis, a and D are on the parabola (in the first quadrant), and the maximum perimeter of rectangle is obtained

It is known that the parabola intersects with X axis at (0,0) and (6,0), and a and D are required to be in the first quadrant, then the abscissa of the four vertices of the rectangle is within (0,6)
Let B (x, 0), then C (6-x, 0), a (x, - X & sup2; + 6x)
The perimeter of the rectangle
=2(AB+BC)
=2[(-x²+6x)+(6-x-x)]
=-2(x-2)²+20
Therefore, the maximum perimeter of the rectangle is 20