We know the quadratic function y = - x square + 2x + 3, when - 2

We know the quadratic function y = - x square + 2x + 3, when - 2

Because the symmetry axis of quadratic function is x = 1, the maximum value of Y is 4 when x = 1, and the minimum value is - 5 when x = - 2. So the value range of Y is (- 5,4]

It is known that for quadratic function y = x2-2x-a (1), if there are two intersections between this function and X axis, the value range of a is obtained
(2) If the abscissa of the two intersections of this function and X axis is x1, X2, and 1 / X1 + 1 / x2 = - 2 / 3, the value of a is obtained

Analysis: (1) consider the relationship between the function and the equation. If there are two intersections between the function and the X axis, then the equation x ^ 2-2x-a = 0 △ > 0, that is B ^ 2-4ac > 0. The range of a can be obtained

Given that the vertex of quadratic function y = 2x2 MX + 2 is on the X axis, then M=______ ．

The vertex of the quadratic function y = 2x2 MX + 2 is on the x-axis, ∧ = b2-4ac = M2-4 × 2 × 2 = 0, ∧ M2 = 4 × 2 × 2 = 16, ∧ M = ± 4

If the vertex of the image of the quadratic function y = 2x ^ 2-x + C is on the X axis, then the value of C is

If the vertex is on the X axis, the discriminant is 0,
That is (- 1) ^ 2-4 * 2 * C = 0, 1-8c = 0,
So C = 1 / 8

As shown in the figure, the image of the quadratic function y = AX2 + BX + C (a ≠ 0) passes through the point (- 1,2), and the abscissa of the intersection point with the X axis is X1 and X2 respectively, where - 2
As shown in the figure, the image of the quadratic function y = ax ^ 2 + BX + C (a ≠ 0) passes through the point (- 1,2), and the abscissa of the intersection point with the X axis are X1 and X2 respectively, where - 2 ＜ x1 ＜ - 1,0 ＜ x2 ＜ 1. The following conclusion: ① b > 3a ② B ^ 2-4ac > 0 is correct? Please write the proof process
Wrong number. There is no picture. I have to draw it myself

Substituting (- 1,2) into the function, we get A-B + C = 2, sorting out C = 2-A + B, and the abscissa of the intersection point with the X axis are X1 and X2 respectively, where - 2 < X1 < 1,0 < x2 < 1, so the opening of the image is downward. When x = - 2, Y & lt; 0 is 4a-2b + C & lt; 0. Substituting C = 2-A + B, we get 4a-2b + (2-A + b) & lt; 0 sorting out: 3a-b {0}

It is known that the abscissa of the intersection of the image of quadratic function f (x) = ax ^ 2 + X + 1 (a > 0) and X axis are X1 and X2 respectively
(1)  prove that: (1 + x1) (1 + x2) = 1
(2)  proof: x1 ＜ - 1, X2 ＜ - 1;
(3) If X1 and X2 satisfy the inequality LG X1 / x2 ≤ 1, try to find the value range of A
Want to (2) (3) ask the detailed process

The axis of symmetry 0 is a

Given that the value of quadratic function y = x - (2m + 6) x + m + 3 is always nonnegative, the value range of real number m is obtained

Because a = 1 ＞ 0, the image opening of function y is upward. Because y is always nonnegative, that is, y is always greater than 0, so there is no intersection between Y and X axis. That is, X & # 178; - (2m + 6) x + M & # 178; + 3 = 0 has no real number solution. The discriminant △ = B & # 178; - 4ac ＜ 0 substitutes the values of a, B, C into it

If y = (k-1) x ^ 2 + kx-2 is a quadratic function, then the value range of K is

K is not equal to 1

It is known that the parabola y = - x2 + 2 (M + 1) x + m + 3 has two intersections with the x-axis, a, B and y-axis intersect at point C, where point a is on the negative half axis of x-axis, point B is on the positive half axis of x-axis, and OA: OB = 3:1. (1) find the value of M; (2) if P is a point on the parabola and satisfies s Δ PAB = 2S Δ ABC, find the coordinates of point P

(1) From the parabola y = - X & # 178; + 2 (M + 1) x + m + 3 and X-axis have two intersections a, B can be obtained: Δ = (2m + 2) &# 178; + 4 (M + 3) > 0, that is M & # 178; + 3M + 4 > 0, it is easy to know that for any real number m, the above formula is constant, and point a is on the negative half axis of x-axis, point B is on the positive half axis of x-axis, then point a and B coordinates are (x1, Y1)

25. It is known that the parabola y = - x2 + 2 (M + 1) x + m + 3 has two intersections a with the x-axis, B and y-axis intersect at point C, where a
It is known that the parabola y = - x2 + 2 (M + 1) x + m + 3 has two intersections with the X axis, a, B and Y axis intersect at point C, where point a is on the negative half axis of X axis, point B is on the positive half axis of X axis, and OA: OB = 3: (1) find the value of M; (2) if P is a point on the parabola and satisfies s Δ PAB = 2S Δ ABC, find the coordinates of point P.

Sister, the title? Incomplete!
got it.
Do this: - x2 + 2 (M + 1) x + m + 3 = 0
Let ob = a > 0, then OA = 3A, B (a, 0) a (- 3a, 0) is the two zeros of the function
a-3a=2(m+1) -3a^2=-m-3
The solution is m = 0 or M = - 5 / 3 and a > 0, so m = 0, M = - 5 / 3
（2）SΔPAB=2SΔABC
So the distance from P to X axis should be twice that of OC
If M = - 5 / 3 generation, the function = - x ^ 2-4 / 3x + 4 / 3, so OC = 4 / 3
Let P (m, 8 / 3) or (m, - 8 / 3)
When p (m, 8 / 3), substituting into no real number root
When p (m, - 8 / 3), substitute M = (- 2 + 2 radical 10) / 3 or M = M = (- 2-2 radical 10) / 3
That is, P ((- 2 + 2 radical 10) / 3, - 8 / 3) or = (- 2-2 radical 10) / 3, - 8 / 3)