Given that the two intersections of the parabola y = (m-1) x2 + 2mx + m + 3 and the X axis are on both sides of the straight line x = 2, then the value range of M is______ .

Given that the two intersections of the parabola y = (m-1) x2 + 2mx + m + 3 and the X axis are on both sides of the straight line x = 2, then the value range of M is______ .

When M-1 > 0, then x = 2, y < 0, that is 4 (m-1) + 4m + m + 3 < 0, M has no solution; when M-1 < 0, then x = 2, Y > 0, that is 4 (m-1) + 4m + m + 3 < 0, M has no solution; when M-1 < 0, then x = 2, Y > 0, that is 4 (m-1) + 4m + m + 3 > 0, the solution is 19 < m < 1, so the value range of M is 19 < m < 1
It is known that the intersection of parabola y = x + 2mx + M-7 and x-axis is on both sides of point (1,0). Please determine the value range of M
In fact, just calculate the value of x = 1, and this value is less than zero
Given the parabola y = (m-1) x ^ 2-2mx + m + 1 (M > 1), find the intersection coordinates of the parabola and X axis
The point of intersection with X axis is (m-1) x & # 178; - 2mx + m + 1 = 0
The factor is decomposed by cross phase multiplication
m-1 -(m+1)
A kind of
1 -1
[(m-1)x-(m+1)](x-1)=0
The solution is as follows
x1=(m+1)/(m-1)
x2=1
Coordinates of the intersection point with X axis:
（(m+1)/(m-1),0）
and
（1,0）
The distance between the two intersections of the quadratic function parabola: y = x2-4x-5 and the X axis is: (), the area of the triangle formed by the intersection of the parabola and the coordinate axis
The distance between the two intersection points of quadratic function parabola: y = x2-4x-5 and X axis is: (), and the area of triangle enclosed by the intersection point of parabola and coordinate axis is ()
Let y = 0, x = 5 or - 1, so the distance between two points is 6
Let x = 0 and y = - 5, so the triangle area is 0.5 * 6 * 5 = 15
6；15
The intersection is x = 5 and - 1, so the distance is 6
Area = 6 * 5 / 2 = 15
According to the cross multiplication, we can get the focus of X axis, (- 1,0) and (5,0). The focus of area and y-axis is (0,5), and then the area formula of triangle is obtained
The distance between two intersections of quadratic function parabola: y = x2-4x-5 and X axis is (6), and the area of triangle formed by the intersection of parabola and coordinate axis is (15)
Thank you for your trust!!!!!!!
The axis of symmetry of the parabola y = (M2-2) x2 + 2mx + 1 with downward opening passes through the point (- 1,3), and the value of M is obtained
∵ the symmetric axis of the parabola y = (M2-2) x2 + 2mx + 1 with downward opening passes through the point (- 1,3), ∵ - 2M2 (M2-2) = - 1, M2-2 < 0, and the solution is: M1 = - 1, M2 = 2 (not suitable for the problem), and∵ M = - 1
The intersection point of parabola y = 1 / 3x & # 178 and straight line y = ax + B is a and B. the abscissa of these two points are 3 and - 1 respectively to find the intersection point of a and B
Title:
The intersection of parabola y = (1 / 3) x & # 178 and straight line y = ax + B is a and B, and the abscissa of intersection a and B are 3 and - 1 respectively?
Let the ordinates of a and B be y and y ', respectively, then: a (3, y), B (- 1, y')
∵ the intersection of parabola y = (1 / 3) x & # 178 and straight line y = ax + B is a and B
If a and B are on the parabola y = (1 / 3) x & #, and on the straight line y = ax + B, then
Point a (3, y) satisfies parabolic y = (1 / 3) x & # 178,
y=(1/3)x²
=(1/3)×3²
=(1/3)×9
=3
Point B (- 1, y ') satisfies parabolic y = (1 / 3) x & # 178;
y'=(1/3)x²
=(1/3)×(-1)²
=(1/3)×1
=1/3
Therefore, a (3,3), B (- 1,1 / 3)
∵ A and B are on the line y = ax + B,
Two points a and B are substituted into the straight line y = ax + B to obtain:
3=3A+B
1/3=-A+B
The results of solving the equations are as follows
A=2/3；B=1
Let two intersections be (x1, Y1), (X2, Y2)
∵ substituting X1 = 3 and X2 = - 1 into the parabola y = 1 / 3x & # 178, we get Y1 = 3 and y2 = 1 / 3
Replace (x1, Y1) (X2, Y2) with the line y = ax + B
3=3A+B
1/3=-A+B
A = 2 / 3, B = 1
If the image of the function y = ax2-ax + 3x + 1 has only one intersection point with the X axis, then the value of a and the coordinates of the intersection point are___ .
When a = 0, the function is y = 3x + 1, the image is a straight line, and there is only one intersection (- 13, 0) with X axis; when a ≠ 0, the function is y = ax2-ax + 3x + 1, the image is a parabola, △ = (3-A) 2-4 · a · 1 = a2-10a + 9; when △ = 0, the parabola has only one intersection with X axis, and a = 1 or 9
Given the set a = {x ∈ R | 12 ＜ 2x ＜ 8}, B = {x ∈ R | - 1 ＜ x ＜ m + 1}, if a sufficient and unnecessary condition for X ∈ B is x ∈ a, then the value range of real number m is______ .
A = {x ∈ R | 12 ＜ 2x ＜ 8} = {x | - 1 ＜ x ＜ 3}, because a sufficient and unnecessary condition for X ∈ B is x ∈ a, so m + 1 ＞ 3, that is m ＞ 2. So the value range of real number m is (2, + ∞). So the answer is: (2, + ∞)
Given the set a = {x ∈ Z | 2 ≤ 2 ^ (2-x) 1},
Then the number of elements in the set a ∩ (CRB) is
A．0
B．1
C．2
D．3
"2" is the subscript of log, "a ∩ (CRB)" is the intersection of a (the complement of B)
1 ≤ 2-x from a
Where is your set C? B = {x ∈ R | (log2) x > 1}
Let a = {x | - 5 ≤ x ≤ 3}, B = {x | - 2 or X ＞ 4}, find a ∩ B, (CRA) ∪ (CRB)
It is easy to get a ∩ B = {x | - 5 ≤ x ＜ - 2} CRA = (- ∞, - 5) ∪ (3, + ∞); CRB = [- 2,4] (CRA) ∪ (CRB) = (- ∞, - 5) ∪ [- 2, + ∞)