If the right graph is a partial image of the quadratic function f (x) = x ^ 2-bx + A, then G (x) = ln x + F ′ (x) is zero

If the right graph is a partial image of the quadratic function f (x) = x ^ 2-bx + A, then G (x) = ln x + F ′ (x) is zero

The solution is as follows: because the function f (x) = x ^ 2-bx + a passes through the point (1,0), substituting into 1-B + a = 0; that is, B = a + 1; and from the image of F (x), we can know that 1 > F (0) > 0, that is, 1 > a > 0; thus, there is 2 > b = a + 1 > 1; f '(x) = 2x-b; so g (x) = LNX + F' (x) = LNX + 2x-b, it is easy to know that G (x) increases monotonically in its domain of definition, while

If the quadratic function f (x) = - X & sup2; + BX + C is a decreasing function in the interval [2, + ∞] and an increasing function in the interval [- ∞, 2], its image intersects the x-axis

F (x) = - X & # 178; + BX + C is the parabola symmetry axis with opening downward, x = 2 = - B / 2, B = 4, C ≤ 4

Given that the quadratic function y = f (x) is an increasing function on (- ∞, 2) and a decreasing function on [2, + ∞), the vertex of the image is on the straight line y = X-1, and the image passes through the point (- 1, - 8), find: (1) the analytic expression of the quadratic function y = f (x); (2) make f (x) + M

1) It is an increasing function on (- ∞, 2) and a decreasing function on [2, + ∞), that is, the axis of symmetry is x = 2 and the opening is downward
The vertex of the image is on the line y = x - 1, and its intersection with the symmetry axis is (2,1)
So let f (x) = a (X-2) ^ 2 + 1
When the image passes through the points (- 1, - 8), that is, f (- 1) = 9A + 1 = - 8, a = - 1 is obtained
So f (x) = - (X-2) ^ 2 + 1
2)m