If the image of the quadratic function y = - x2 + MX-1 has two different intersections with the line AB whose two ends are a (0,3), B (3,0), then the value range of M is______ .

If the image of the quadratic function y = - x2 + MX-1 has two different intersections with the line AB whose two ends are a (0,3), B (3,0), then the value range of M is______ .

It is known that the equation of line segment AB is y = - x + 3 (0 ≤ x ≤ 3). Because there are two different intersections between quadratic function image and line segment AB, the equations y = − x2 + MX − 1y = − x + 3, 0 ≤ x ≤ 3 have two different real solutions. The elimination result is: X2 - (M + 1) x + 4 = 0 (0 ≤ x ≤ 3), Let f (x) = X2 - (M + 1) x +
In the bivariate linear equation 7x-8y = 1, X and y are opposite to each other
X and y are opposite numbers, so x = - Y
So 7x-8y = 1 can be changed to - 7y-8y = 1, so y = - 1 / 15
So x = 1 / 15
Y = - x is substituted into the equation
7x-8(-x)=1
15x=1
x=1/15
y=-x=-1/15
x=1/15 y=-1/15
Substituting y = - x, 7x-8 (- x) = 1, that is, 15x = 1, so x = 1 / 15, then y = - 1 / 15
∵-x=y
The original equation can be reduced to 7x-8 (- x) = 1
The solution is x = 1 / 15
The coordinates of points a and B are (1,0), (2,0) respectively. If there is only one intersection point between the image of quadratic function y = x2 + (A-3) x + 3 and line AB, then the value range of a is______ .
If the vertex of quadratic function is below the x-axis, if YX = 1 ＜ 0 and YX = 2 ≥ 0, that is 1 + (a − 3) + 3 ＜ 04 + 2 (a − 3) + 3 ≥ 0, the solution of the inequality is - 1 ≤ a ＜ - 12; if YX = 2 ＜ 0 and YX = 1 ≥ 0, that is 1 + (a − 3) + 3 ≥ 04 + 2 (a − 3) + 3 ＜ 0, the solution is - 1 ≤ a ＜ - 12
Solving binary linear equations y = x-3,7x-5y = 9
y=x-3
Substituting 7x-5y = 9
7x-5x+15=9
2x=-6
therefore
x=-3
y=x-3=-6
As shown in the figure, it is known that the vertex coordinates of the image of the quadratic function are C (1,0), and the line y = x + m intersects the image of the quadratic function at two points a and B, where the coordinates of point a are (3,4), and point B is on the Y axis
(1) Find the relationship between the value of M and the quadratic function
(2) Point E is a moving point on the parabola. Let the abscissa of point e be n (0 ＜ n ＜ 3), and let the area of △ Abe be s, then does s have a maximum? If so, ask for its maximum and find out the coordinates of point e at this time
PS. the parabolic opening is upward
(1) From the meaning of the question, a (3,4) and B (0, m) are both on the straight line and on the parabola (because of the editing problem, X2 represents the square of x). Let the parabola y = AX2 + BX + m (a > 0) bring a (3,4) into y = x + m, that is: 4 = 3 + m; we get: M = 1, so
Let the vertex formula of quadratic function be y = a (x-1) ^ 2
Because a (3,4), on quadratic function 4 = a (3-1) ^ 2, so a = 1
So it's a quadratic function y = (x-1) ^ 2
Since a (3,4) is y = x + m, 4 = 3 + m leads to M = 1
So y = x + 1
2 A（3，4) B（0，1) E（n，(n-1)^2)
The distance of line AB is the root sign (3-0) ^ 2 + (4-1) ^ 2 = 3 * root sign 2
Binary linear equations Y-X = 3,7x + 5Y = 6
Y-X = 3 1
7x + 5Y = 62
1 formula X7 is 7y-7x = 21 3 formula
Formula 2-3 12x = - 15
x=-5/4
y=3-5/4=7/4
Classmate, don't be lazy to ask: if I can solve it myself, can I still use the Internet to ask?
The intersection of the image of quadratic function y = - x2 + (m-1) x + m and X axis and the vertex C of AB two points to find the value of m on the line y = 4x
Second, if the y-axis image in the graph is folded to the left along the y-axis, C coincides with C1, and B coincides with B1, then the analytic expression of the quadratic function on the left side of the y-axis after folding is obtained
To complete the process
Take C into C2, find 20, and then if K and H are known
Then the topic will tell you that the function passes through a point beyond the vertex, and you can substitute this point
y=a(x-x1)(x-x2)
It will also tell you a point that is not on the x-axis, so we can get a
If not, there should be other conditions
The solution of binary linear equations y = x + 37x + 5Y = 9 is______ .
Y = x + 3, 1 7x + 5Y = 9, 2, 1 substituting into 2, 7x + 5 (x + 3) = 9, the solution is x = - 12, substituting x = - 12 into 1, y = - 12 + 3 = 52, so the solution of the equations is x = - 12Y = 52. So the answer is: x = - 12Y = 52
The quadratic function f (x) = AX2 + BX + C is known. (1) if a ＞ B ＞ C and f (1) = 0, it is proved that f (x) must have two zeros; (2) if for x1, X2 ∈ R and x1 ＜ X2, f (x1) ≠ f (x2), the equation f (x) = 12 [f (x1) + F (x2)] has two unequal real roots, it is proved that there must be a real root belonging to (x1, x2)
It is proved that: (1) f (1) = 0, ∵ a + B + C = 0, and ∵ a ＞ B ＞ C, ∵ a ＞ 0, C ＜ 0, that is, AC ＜ 0. And ∵ △ = b2-4ac ≥ - 4ac ＞ 0, the equation AX2 + BX + C = 0 has two unequal real roots. Therefore, the function f (x) must have two zeros. (2) Let G (x) = f (x) - 12 [f (x1) + F (x2)], then
What is the solution of the system of equations {x + Y-1 = 0,2x & # 178; + X + Y-3 = 0,
｛x+y-1=0,
y=1-x;
2x²+x+y-3=0
2x²+x+1-x-3=0;
2x²-2=0;
x²-1=0;
x=±1;
x=1;y=0;
x=-1;y=2;
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A:
x+y-1=0
2x^2+x+y-3=0
The result of subtracting the two formulas is as follows:
2x^2-2=0
x^2=1
X = 1 or x = - 1
Substituting x + Y-1 = 0, i.e. y = - x + 1, we get:
x=1，y=0
x=-1，y=2
y=1-x （1）
2*x^2+1-x+x=0 （2）
2 * x ^ 2 + 1 = 0 (3) from (2)
x^2=-0.5
x=sqrt(0.5)*i