ABC problem of quadratic function The teacher said ABC is a constant Y = ax ^ 2 + BX + C (a, B, C are constants, a ≠ 0, and a determines the opening direction of the function. When a > 0, the opening direction is upward, a ≠ 0

ABC problem of quadratic function The teacher said ABC is a constant Y = ax ^ 2 + BX + C (a, B, C are constants, a ≠ 0, and a determines the opening direction of the function. When a > 0, the opening direction is upward, a ≠ 0

There are many kinds of analytic expressions for quadratic functions,
For example: y = x ^ 2 + 3x-1, where a = 1, B = 3, C = - 1;
Y = 2x ^ 2 + 1, where a = 2, B = 0, C = 1;
Y = 7x ^ 2 + 3x, where a = 7, B = 3, C = 0;
In quadratic function, only x and y are variables, others are constants or parameters. Different quadratic functions have different constants or parameters
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The meaning of quadratic function ABC
This is the knowledge of the second year of junior high school. For example, in the quadratic function y = AX2 + BX + C, how can we see in the image that when ABC is greater than or less than zero

[don't be bothered. It's better to memorize these properties!]
1. The parabola is an axisymmetric figure. The axis of symmetry is a straight line x = - B / 2A
The only point of intersection between the axis of symmetry and the parabola is the vertex P of the parabola
In particular, when B = 0, the symmetry axis of the parabola is Y-axis (that is, the line x = 0)
2. The parabola has a vertex P whose coordinates are p (- B / 2a, (4ac-b ^ 2) / 4A)
When - B / 2A = 0, P is on the y-axis; when Δ = B ^ 2-4ac = 0, P is on the x-axis
3. The quadratic coefficient a determines the opening direction and size of the parabola
When a > 0, the parabola opens up; when a < 0, the parabola opens down
|The larger a | is, the smaller the opening of the parabola is
4. The position of the axis of symmetry is determined by the coefficient b of the first term and the coefficient a of the second term
When a and B have the same sign (AB > 0), the symmetry axis is on the left of Y axis;
When a is different from B (AB < 0), the axis of symmetry is on the right of Y axis
5. The constant term C determines the intersection of the parabola and the y-axis
The intersection of parabola and y-axis at (0, c)
6. The number of intersections of parabola and x-axis
When Δ = B ^ 2-4ac ＞ 0, there are two intersections between the parabola and X axis
When Δ = B ^ 2-4ac = 0, there is one intersection point between the parabola and X axis
When Δ = B ^ 2-4ac ＜ 0, there is no intersection between the parabola and X axis. The value of X is an imaginary number (x = - B ± √ the opposite of the value of B ^ 2-4ac, multiplied by the imaginary number I, the whole formula divided by 2a)
When a > 0, the minimum value f (- B / 2a) = 4ac-b ^ 2 / 4A is obtained at x = - B / 2A; it is an increasing function on {x | X-B / 2A}; the opening of parabola is upward; the value range of function is {y | y ≥ 4ac-b ^ 2 / 4A} on the contrary
When B = 0, the axis of symmetry of the parabola is y-axis, then the function is even, and the analytic expression is y = ax ^ 2 + C (a ≠ 0)
7. Domain of definition: R
Range: (corresponding to the analytic expression, and only discuss the case that a is greater than 0, and the case that a is less than 0, please infer by yourself) ① [(4ac-b ^ 2) / 4A, positive infinity); ② [T, positive infinity]
Parity: even function
Periodicity: None
Analytical formula:
① Y = ax ^ 2 + BX + C [general formula]
⑴a≠0
(2) when a > 0, the opening of parabola is upward; when a < 0, the opening of parabola is downward;
(3) extreme point: (- B / 2a, (4ac-b ^ 2) / 4A);
⑷Δ=b^2-4ac,
The image intersects the x-axis at two points
([- B + √ Δ] / 2a, 0) and ([- B + √ Δ] / 2a, 0);
Δ = 0, the image intersects the x-axis at one point
（-b/2a,0）；
There is no intersection between the image and the x-axis;
② Y = a (X-H) ^ 2 + T
In this case, the corresponding extreme point is (h, t), where h = - B / 2a, t = (4ac-b ^ 2) / 4A);

How to determine ABC in quadratic function

A: Image up, greater than 0, down less than 0
B: If x = 0, then the value of Y is the value of C, and a looks at the opening direction, upward is greater than 0, downward is less than 0
C: Let x = 0, then y is the value of C

The quadratic function f (x) satisfies f (4 + x) = f (- x), and f (2) = 1, f (0) = 3. If f (x) has minimum value 1 and maximum value 3 on [0, M], then the value range of real number m is ()
A. [2，4]B. （0，2]C. （0，+∞）D. [2，+∞）

From F (4 + x) = f (- x), we can see that f (4) = f (0) = 3 is the maximum, while f (2) = 1 is the minimum, and f (x) has the minimum value 1 and the maximum value 3 on [0, M], then M must have 2, and f (4) = f (0) = 3, so m can also be equal to 4, so the answer is a

How to determine the value range of X in mathematical quadratic function?

1. First look at the clear scope of the topic
2. Second, look at the implied conditions. For example, the denominator is not 0, the radical is greater than 0, etc

Mathematical problems - quadratic function for parameter values
It is known that f (x) = 2x ^ 2-4ax + A + B
1) For X ∈ [0,1], is there a real number a such that when x = a, y has the minimum value 2A? If so, find out the value of A. if not, explain the reason
2) For X ∈ [0,1], Let f (x) be a monotone function, the minimum value of which is nonnegative, and the maximum value of which is no more than 11, then the value range of real number a is obtained

1) When f (x) = 2x ^ 2-4ax + A + B = 2 (x ^ 2-2ax + A ^ 2) - 2A ^ 2 + A + B = 2 (x-a) ^ 2-2a ^ 2 + A + BX = a, y = 2A = - 2A ^ 2 + A + B, i.e. 2A ^ 2 + A-B = 0, B = a (2a + 1) 2) f (x) = 2 (x-a) ^ 2 + 2AF (x) monotone function, when a = 1, i.e. a = 11ax = 0, f (x) = 2 (0-A) ^ 2 + 2A = 2 (a + 1 / 2) ^ 2

Quadratic function y = AX2 + BX + C the coordinates of some points on the image satisfy the following table: X -3 -2 -1 0 1 … y … -3 -2 -3 -6 -11 … Then the vertex coordinates of the function image are______ ．

When x = - 3 and x = - 1, the function values are all - 3, which are equal. The symmetry axis of the function image is a straight line x = - 2, and the vertex coordinates are (- 2, - 2)

If the vertex of the image of quadratic function y = x ^ 2 + 1 / 2 and y = - 2x ^ 2 + K coincides, the following conclusion is incorrect
A. This function has the same axis of symmetry
B. The opening direction of the two function images is opposite
C. The maximum value of quadratic function y = - 2x ^ + k is 1 / 2
D. The equation - 2x ^ 2 + k = 0 has no real roots

Choose D
If the vertices are on the axis of symmetry, if the vertices coincide, the axis of symmetry will coincide
The coefficient of quadratic term is positive and negative, the opening is up and down, and B is right
The vertex of y = x ^ 2 + 1 / 2 is (0,1 / 2), and the vertices coincide, so the maximum value of y = - 2x ^ 2 + k is 1 / 2
The vertex of the corresponding function image of the equation is above the x-axis, the opening is down, and there is an intersection point with the x-axis, so it has a real root, D is wrong

The image vertex m (- 1, - 8) of quadratic function is known, and its intersection with y axis is (0,6)
1. Find the analytic expression of quadratic function f (x) 2. When x takes what value, f (x)

(1) ∵ image vertex m (- 1, - 8) of quadratic function,
Let f (x) = a (x + 1) & # 178; - 8 be the analytic expression of quadratic function,
Substituting the point (0,6) into, we get
a(0+1)²-8=6
The solution is: a = 14
The analytic expression of quadratic function is f (x) = 14 (x + 1) &# 178; - 8 = 14x & # 178; + 28x + 6
(2) Let f (x) = 0, we get 14x & # 178; + 28x + 6 = 0
The solution is: X1 = - 1 + [(2 / 7) √ 7], X2 = - 1 - [(2 / 7) √ 7],
The parabolic opening is upward,
When - 1 - [(2 / 7) √ 7] < x ＜ - 1 + [(2 / 7) √ 7], f (x)

It is known that the vertex coordinates of the image with quadratic function y = ax ^ 2 + BX + C are (1, - 6), and there are two intersections between the image and the X axis, the cubic sum of the abscissa of the two intersections
Is 6, then the analytic expression of this quadratic function is__________ .

Let y = a (x-1) ^ 2-6
y=ax^2-2ax+a-6
x1+x2=2,x1*x2=(a-6)/a
x1^3+x2^3=(x1+x2)(x1^2-x1x2+x2^2)=6
x1^2-x1x2+x2^2=3
X1 ^ 2 + x2 ^ 2 = (2a + 12) / A, the solution is a = 3
∴y=3x^2-6x-3