Given the quadratic function f (x), when x = 2, there is a maximum value of 16. The length of the line segment obtained by cutting x-axis of its image is 8 If the quadratic function f (x) is known, when x = 2, it has a maximum value of 16. The length of the line segment obtained by its image cutting the x-axis is 8. The expression of F (x) can be obtained

Given the quadratic function f (x), when x = 2, there is a maximum value of 16. The length of the line segment obtained by cutting x-axis of its image is 8 If the quadratic function f (x) is known, when x = 2, it has a maximum value of 16. The length of the line segment obtained by its image cutting the x-axis is 8. The expression of F (x) can be obtained

When x = 2, there is a maximum of 16
That is, the axis of symmetry is a straight line, x = 2, and the vertex is (2,16). Let y = a (X-2) &# 178; + 16
Because the axis of symmetry is x = 2, the length of the line cut on the X axis is 8,
The two intersections with X axis are (- 2,0) and (6,0) respectively
Substituting the point (6,0) into the analytic expression, we get: 0 = 16A + 16, we get: a = - 1
Therefore, f (x) = - (X-2) &# 178; + 16
If you don't understand, please hi me,

It is known that the length of the line segment obtained from the image of the quadratic function y = f (x) is 8. When x = 2, f (x) has a maximum value of 16
1. Find the analytic expression of the function. 2. Try to prove that the equation f (x) = 0 has two unequal real roots, and the two roots are in the interval (- 3, - 1) and (5,7). 3. Find the zero point of the function

Vertex (2,16) y = a (X-2) & sup2; + 16 = ax & sup2; - 4ax + 4A + 16x1 + x2 = 4, x1x2 = (4a + 16) / a | x1-x2 | = 8 (x1-x2) & sup2; = (x1 + x2) & sup2; - 4x1x2 = 6416-4 (4a + 16) / a = 64a = - 1F (x) = - X & sup2; + 4x + 12 discriminant = 16 + 48 > 0 has two unequal real roots f (- 3) * f (- 1)

When x = 2, the quadratic function f (x) has a maximum value of 16, and the length of the line segment obtained by cutting x-axis of the image is 8

If f (x) has a maximum value of 16 when x = 2 and the length of the line segment obtained by cutting x-axis of the image is 8, then the abscissa of the intersection of the function and x-axis is 2 + 8 / 2 = 6 and 2-8 / 2 = - 2 respectively
Then let f (x) = a (x + 2) (X-6)
The vertex coordinates (2,16) are substituted to get 16 = a * (2 + 2) (2-6), a = - 1
That is, f (x) = - (x + 2) (X-6) = - x ^ 2 + 4x + 12