The set of intersection points between the image of function y = x + 1 and the image of function y = x * 2 + a (constant a ∈ R) is represented by enumeration Quick return

The set of intersection points between the image of function y = x + 1 and the image of function y = x * 2 + a (constant a ∈ R) is represented by enumeration Quick return

Function y = x + 1 and function y = x * 2 + a (x ^ 2-x + A-1 = 0)
Δ=b^2-4ac=1-4(a-1)
Let Δ = 0, a = 5 / 4
Then when a = 5 / 4, there is only one intersection point, x = 1 / 2, y = 3 / 2
When a > 5 / 4, there is no intersection between the two images
When a

For quadratic function y = ax (square) + BX + C (a is not equal to 0), the real number with function value 0 is called the zero point of this function, then the number of zeros of quadratic function y = x (square) - MX + m-2 is ()
A,1 B.2 C.3 D.4
Why?

Because △ = M & sup2; - 4 (m-2) = M & sup2; - 4m + 8
=(m-2)²+4>0
So the equation x (square) - MX + m-2 = 0 has two unequal real roots
So choose B

Given that the quadratic function f (x) = ax square + BX = C, f (- 2) = f (0) = 0, the minimum value of F (x) is - 1
(1) Finding the analytic expression of function f (x)

F (- 2) = 4a-2b = C, f (0) = C = 0, so 4a-2b = 0, f (x) minimum is - 1, so 4ac-b square = - 1, B = 1
4a=2 a=1/2

The quadratic function y = 1 / 2x ^ 2 + X-5 / 2 is known
(1) Find its vertex coordinates and axis of symmetry
(2) If the two intersection points of parabola and X-axis are a and B, the length of line AB is calculated

(1) When a = 1 / 2, B = 1, C = - 5 / 2, vertex coordinate is (- B / 2a, (4ac-b ^ 2) / 4A), that is (- 1, - 3), or change the function into vertex formula: y = 1 / 2 (x + 1) ^ 2-3, vertex coordinate is (- 1, - 3), symmetry axis is x = - 1 (2) y = 0, there are two intersection points between function and X axis, which is the common root of equation 1 / 2x ^ 2 + X-5 / 2 = 0

Given the quadratic function y = - x + X-2, when the independent variable x is m, the corresponding value is greater than 0, when the independent variable x is M-1, M + 1, respectively
Given the quadratic function y = - x + X-2, when the independent variable x is m, the corresponding value is greater than 0, when the independent variable x is M-1, M + 1, the corresponding function values are Y1, Y2, then Y1, Y2 satisfy a.y1 > 0, Y2 > 0, b.y1,0, y2,0, c.y10, d.y1 > 0, Y2

Δ﹦‐7

It is known that the quadratic function y = - X & # 178; + 2x + M-1 with X as the independent variable intersects with y axis at (0,3) point (1) to find the value of M (2) to find its intersection with X axis
It is known that the quadratic function y = - X & # 178; + 2x + M-1 with X as the independent variable intersects the Y axis at (0,3)
(1) Finding the value of M
(2) Find out the intersection of it and X axis and the coordinates of the parabola vertex
(3) What is the value of X when the parabola is above the x-axis?
(4) What is the value of X axis, y decreases with the increase of X?

(1) Bring in the coordinates (0,3) to get: 3 = M-1, so m = 4
(2) The intersection of X axis: let y = 0 get: - x ^ 2 + 2x + 3 = 0 simplify (- x + 3) (x + 1) = 0, the intersection is (3,0) (- 1,0)
(3) For the function formula method: - (x-1) ^ 2 + 4 = y, draw the function image directly above the X axis, that is, Y > = 0,
Get {x | - 1

The function independent variable, known quadratic function y = x & sup2; - x + a [a > 0], when the independent variable x is m, its corresponding function value is less than 0, then whether the function value of M-1 is large
Function independent variable, known quadratic function y = x square - x + a [a > 0], when the independent variable x takes m, its corresponding function value is less than 0, then whether the function value of M-1 is greater than 0, why?

If M & # 178; - M + A is less than 0, then M is between (1-radical 1-4a) / 2 and (1 + radical 1-4a) / 2, then: (m-1) &# 178; - (m-1) + a = M & # 178; - M + A-2 (m-1), while - 2 (m-1) is between (1-radical 1-4a) and (1 + radical 1-4a), M & # 178; - M + A is less than 0, so it can not guarantee that the function value is greater than 0

If a > 0, the equation AX ^ 2 + BX + C = 0 has two unequal real roots, and the vertex of the parabola y = ax ^ 2 + BX + C may be in the quadrant

If B ^ 2-4ac > 0, y = ax ^ 2 + BX + C vertex abscissa = - B / 2A ordinate = (4ac-b ^ 2) / 4A is less than 0, it may be in the third and fourth quadrant

The quadratic function f (x) = 2x ^ 2-4 (A-1) - A ^ 2 + 2A + 9 is known
If f (m) > 0 for all real numbers m in the interval [- 1,1], find the value range of real number a

F (x) = 2x ^ 2-4 (A-1) x-a ^ 2 + 2A + 9
Then f (x) = 2x ^ 2-4 (A-1) x-a ^ 2 + 2A + 9 = 2 (x - (A-1)) ^ 2-3a ^ 2 + 6A + 7
That is, the axis of symmetry is x = A-1
When A-1

What is the constant hold of quadratic function? And when x2 + ax + B is greater than or equal to 0, why is △ less than or equal to 0?

X2 + ax + B is equal to or greater than 0, which means that in the domain R, all real numbers can satisfy the condition that the value of this polynomial is equal to or greater than 0
That is to say, y = x2 + ax + B with the opening upward, there is either no or only one intersection point with the X axis, otherwise there will be X so that it is less than 0
It is not always greater than or equal to 0. At most, there is only one intersection, △ ≤ 0
You draw an image to see~