When x = 3, the minimum value of y = - 1, and the image is over (0,7), the analytic expression of quadratic function about X is obtained

When x = 3, the minimum value of y = - 1, and the image is over (0,7), the analytic expression of quadratic function about X is obtained

When x = 3, the minimum value of y = - 1
So y = a (x-3) ^ 2-1
There is a minimum, so a > 0
Substituting (0,7) into
7=a*(0-3)^2-1
A = 8 / 9, a > 0
So y = (8 / 9) (x-3) ^ 2-1
=(8/9)x^2-(16/3)x+7