It is known that the image of quadratic function y = x + bx-3 passes through the point (- 2,5)

Substitute (- 2,5) into quadratic function y = x2 + bx-3 to get: 5 = 4-2b-3, B = - 2, y = x2-2x-3 = (x-1) 2-4, the opening direction of parabola is upward, the axis of symmetry is straight line x = 1, substitute x = 1 to get: y = - 4, substitute x = 3 to get: y = 0, when 1 ＜ x ≤ 3, the value range of Y is - 4 ＜ y ≤ 0, answer: the value of B is - 2, when 1 ＜

Given the quadratic function y = x ^ 2 + 2x, when - 1
My quadratic function y = - x ^ 2 + 2x, sorry to write wrong!

According to the title: y = - (x-1) square + 1, the axis of symmetry is x = 1, y monotonically increases on the left side of x = 1, and decreases on the right side of x = 1, so - 1

As shown in the figure, it is known that the image of quadratic function y = x2 + BX + C passes through points (- 1,0), (1, - 2). When y increases with the increase of X, the value range of X is___ ．

Substituting (- 1,0), (1, - 2) into the quadratic function y = x2 + BX + C, we get 1-B + C = 01 + B + C = - 2, the solution is b = - 1C = - 2, then the analytic expression of the quadratic function is y = x2-x-2. The axis of symmetry of the function is x = 12, so when y increases with the increase of X, the value range of X is x > 12

Given that the image of quadratic function y = the square of AX + BX + C passes through (- 1,0), (1, - 2), when y increases with the increase of X, the value range of x increases

By substituting the points (- 1,0), (1, - 2) into the quadratic function respectively, we get
a-b+c=0,a+b+c=-2
B = - 1
When a > 0, the opening of the function is upward, and X is an increasing function in the region from 1 / (2a) to positive infinity,
When a

It is known that the image of quadratic function y = x + bx-3 passes through the point (- 2,5)