Given the quadratic function y = AX2 + BX + C, when x = 3, the maximum value of the function is 10, and the length of the line cut by the image on the x-axis is 4, try to find the expression of the quadratic function

Given the quadratic function y = AX2 + BX + C, when x = 3, the maximum value of the function is 10, and the length of the line cut by the image on the x-axis is 4, try to find the expression of the quadratic function

Let the abscissa of the intersection of the parabola and the x-axis be x1, X2,... X1 + x2 = - BA, x1 ·x2 = Ca, | x1-x2 · = (x1 + x2) 2 − 4x1x2 = B2 − 4aca2 = 4, and when x = 3, the maximum value is 10, | - B2A = 3, ② 4ac − b24a = 10, ③ simultaneous solution is obtained: a = - 52, B = 15, C = - 252

The image of quadratic function y = AX2 + BX + C (a > 0) intersects the coordinate axis at (- 1,0) (0, - 1), and its vertex is in the fourth quadrant
Then the value range of a + B + C is?

Substitute 2 points to get
a-b+c=0
c=-1
∴a-b=1,b=a-1
The vertex is in the fourth quadrant, so the abscissa is positive and the ordinate is negative
Abscissa - B / 2A > 0, so B < 0, so A-1 < 0
So a + B + C = a + a-1-1 = 2 (A-1) < 0

The image of the quadratic function f (x) = ax ^ 2 + BX + C (a > 0) intersects the coordinate axis at (- 1,0) and (0, - 1) respectively, and the vertex is in the fourth quadrant. The value range of a + B + C is obtained

Because A-B + C = 0, C = - 1
(4ac-b*b)/4a0
So {- 4 (B + 1) - B * B} / 4 (B + 1) 0
So - 1

If the triangle formed by the intersection of the quadratic function y = - x2 + 2x + C and the coordinate axis is a right triangle, find the value of C

This parabola and the three intersections a, B, C of the coordinate axis form a right triangle, then: [a, B on the X axis]
OC²=OA×OB
Because C (0, c), a (x1,0), B (x2,0)
And X1 and X2 are the two roots of the equation - X & # 178; + 2x + C = 0, then:
c²=|x1x2|=2
C = ± √ 2 (negative value rounded off)
Then: C = √ 2

It is known that the vertex coordinates of the image with quadratic function y = ax ^ 2 + BX + C are C (1,4), and the coordinates of the intersection point with y axis are (0,3)
It is known that the vertex coordinates of the image of the quadratic function y = ax ^ 2 + BX + C are C (1,4), and the coordinates of the intersection with the Y axis are (0,3). If the intersection of the image of the quadratic function and the X axis is a, B, the coordinates of the secondary two points are obtained, and a, B, C are the area of the vertex △ ABC

If the vertex coordinates are C (1,4), then y = a (x-1) ^ 2 + 4
Substitute (0,3) to get: 3 = a + 4, get a = - 1
So y = - (x-1) ^ 2 + 4 = - x ^ 2 + 2x + 3
From y = 0, x = 3, - 1,
So the point of intersection with X axis is a (3,0), B (- 1,0)
Area of ABC = 1 / 2 * AB * OC = 1 / 2 * 4 * 3 = 6

Quadratic function y = ax ^ 2 + BX + C (a ≠ 0). The vertex of the image is D, and the abscissa of the intersection a and B of the image and X axis are - 1, respectively
Quadratic function y = ax ^ 2 + BX + C (a ≠ 0). The vertex of the image is d (the fourth quadrant). The abscissa of the intersection a and B of the image and X-axis are - 1 and - 3 respectively, and they intersect with the negative half axis of y-axis at C
1. Only when a = 1 / 2, △ abd is an isosceles right triangle; 2. If △ ACB is an isosceles triangle, there can be three values of A
It's an isosceles right triangle!

From a (- 1,0), B (3,0), we can get: B = - 2A, C = - 3a, so, vertex coordinates (1, - 4A) so: 1 correct: when a = 1 / 2, we can see that vertex coordinates (1, - 2), △ abd is isosceles right triangle; 2 incorrect: because point C coordinates (0, c), that is, (0, - 3a), and in △ ACB, AC cannot be equal to BC

On the sign of quadratic function?
④ To judge the sign of 2A + B, 2a-b, we need to compare the size of symmetry axis - B / 2a with 1 and - 1, and then simplify to get the size relationship of 2A + B, 2a-b and 0

It's hard. I haven't reached that level yet

The story of quadratic function symbol

We may use all the commonly used mathematical symbols in one exam, but their invention has gone through many years. Let's talk about the four symbols + - × △ first. More than 500 years ago, the driving mathematician Vader may invented "+", which vividly pointed out that it was adding a vertical to a horizontal line. Later, he thought that removing the vertical line would reduce the number, so he invented "-". 300 years later, In the 18th century, when Hannah, a Swiss, split the watermelon into two parts, he invented the △, which is to separate the two points with a line. The ＝ was invented by lekold, an English scholar in the 16th century, In 1631, a man named harriott opened the "=" to both sides and invented the greater than sign and the less than sign

How to find the value of ABC for quadratic function
The image of the quadratic function y = ax & sup2; + BX + C (a ≠ 0) is shown in the figure (not on the graph)
Find the values of a, B and C
The graph is like this. There are two points, a (- 2,0), B (4,0), the axis of symmetry is 1, and the intersection of parabola and Y axis is (0,4)

Substituting a (- 2,0), B (4,0) and (0,4) into y = ax & # 178; + BX + C, we get
0=4a-2b+c
0=16a+4b+c
4=c
The solution is b = 1, a = - 1 / 2

How to determine the positive and negative of C in the general expression of quadratic function

The intersection point of quadratic function on Y axis is positive on C
In this case, C is negative