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We know that the image of quadratic function y = AXX + BX + C passes through a (0, a), B (1,2), @ # Given the image of quadratic function y = AXX + BX + C through a (0, a), B (1,2), @ #, prove: the symmetry axis of this quadratic function image is a straight line x = 2. (in the title, @ #, it is a piece of contaminated and unrecognizable text) (1) according to the existing information, can we find out the quadratic function analytic formula of the title? If yes, please write out the solving process; if not, please write down the solving process, Please explain the reason. (2) add a condition to the original subject according to all the information
(1) according to the meaning of the question, we get the equations
-b/2a=2
a+b+c=2,
c=a
The solution is a = - 1, B = 4, C = - 1,
∴Y=-X^2+4X-1
(2) the parabola intersects the Y axis at C (0, - 1)
The quadratic function f (x) = ax ^ 2-bx + C and the two roots of F (x) = 0 are in the interval (0,1), it is proved that f (0) * f (1) ≤ a ^ 2 / 16
Certification:
Let f (x) = 0 be p, Q
Because the two roots of F (x) = 0 are in the interval (0,1)
Then: 0
Let f (x) = ax & # 178; + BX + C (a, B, C ∈ R), and f (1) = - 2 / A, a > 2B > B. (1)?
(2) It is proved that f (x) = 0 has at least one real root in the interval (0,2)
a> Because 2B > b, b > 0
(2) Because a > 0, b > 0, the axis of symmetry should be on the left side of the y-axis with the opening upward. In addition, f (1) = - 2 / A, so C = - 2 / a-a-b0 is OK. By substituting x = 2, y = 5 / 2A + b > 0 is obtained. Therefore, at least one real root is in the interval (0,2)
In addition, f (1) should be equal to - A / 2, that is, equal to - a divided by 2, otherwise it cannot be proved
RELATED INFORMATIONS
- 1. Let f (x) = ax ^ 2 + BX + C (a > 0) and f (1) = - A / 2 (1) prove that f (x) has two zeros
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- 9. If the function f (x) defined on R satisfies f (x) + 2F (- x) = 2x + 1, then f (2)=________
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- 12. It is known that the two zeros of the function f (x) = ax & # 178; + BX + C are - 1 and 2, and f (5)
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