Given that the quadratic function y = f (x), when x = 2, it has a maximum value of 16, and the length of the line segment obtained by its image cutting x-axis is 8. We try to prove that the equation f (x) = 0 has two unequal real values (1) It is proved that the equation f (x) = 0 has two unequal real roots, and the two roots are in the interval (- 3, - 1) and (5,7) respectively (2) The zero point of the function is obtained

Given that the quadratic function y = f (x), when x = 2, it has a maximum value of 16, and the length of the line segment obtained by its image cutting x-axis is 8. We try to prove that the equation f (x) = 0 has two unequal real values (1) It is proved that the equation f (x) = 0 has two unequal real roots, and the two roots are in the interval (- 3, - 1) and (5,7) respectively (2) The zero point of the function is obtained

The two zeros are symmetric about the axis of symmetry,
The axis of symmetry is x = 2,
So a zero at 2 + 4 = 6
The other is at 2-4 = - 2

It is known that the maximum value of quadratic function y = f (x) is f (0.5) = 6.25, and the expression of F (x) can be obtained through (2,4) points

Let y = a (x-1 / 2) ^ 2 + 25 / 4 (a)

When x = 1, y has a maximum value of 4, and when x = 2, y = 1, we can find the expression of quadratic function

Let the analytic formula of parabola be y = a (x-1) * 2 + 4 ∵ when x = 2, y = 1 ∵ 1 = a (2-1) * 2 + 4, the solution is a = - 3, so y = - 3 (x-1) * 2 + 4

If the quadratic function y = ax & # 178; + BX + C passes through the tangent graph of (1,0) and is symmetrical with respect to the straight line x = half, which point must the image pass through?

Over (0,0)
Because a + B + C = 0
And - B / 2A = 1 / 2
So a + B = 0, so C = 0
So it's past (0,0)