The absolute value problem of quadratic function Given the function y = ax ^ 2 + BX + C, when - 1 The original function y = f (x) = ax ^ 2 + BX + C

The absolute value problem of quadratic function Given the function y = ax ^ 2 + BX + C, when - 1 The original function y = f (x) = ax ^ 2 + BX + C

1. When x = 1 - 1

What is the meaning of quadratic function 2A + B and how to deduce it

For example, if the opening of the parabola is downward and the axis of symmetry is less than 1, then - B / 2A

The relationship between the image of quadratic function and b value
for instance.
When B is greater than zero, it is
When B is less than zero, it is
Note: I don't want x = - B / 2A

It is impossible to determine the relationship between B and quadratic function image only by B. as you have written, the symmetry axis = - B / 2a, a and B can judge the position of the symmetry axis of quadratic function image
If AB has the same sign, then the axis of symmetry is on the left side of the Y axis; if AB has different signs, then the axis of symmetry is on the right side of the Y axis. This is what the formula tells us. B has so many functions

In the rectangular coordinate system as shown in the figure, a and B are two points on the x-axis. A circle with diameter AB intersects the y-axis at C. The analytical formula of the parabola passing through three points a, B and C is
Y = xx-mx + N, the sum of the two reciprocal of the equation xx-mx + n = 0 is - 2
(1) Find the value of n
(2) Find the analytical formula of this parabola
(3) Let a straight line parallel to the x-axis intersect the parabola at two points E and F. ask if there is a circle with the diameter of line segments E and f just tangent to the x-axis? If so, find out the radius of the circle; if not, explain the reason

(1) Parabola y = x & sup2; - MX + n
Point C (0, n) let a (x1,0) B (x2,0) and x10
1+1+r>0
r>-2
Solution equation: x = [2 ± √ 4 (R + 2)] / 2 = 1 ± √ (R + 2)
According to the meaning of the title,
1-[1-√(r+2)]=｜r｜
√(r+2)]=｜r｜
r²-r-2=0
(r-2)(r+1)=0
R = 2 or - 1
So there is a circle whose diameter is e and F, and its radius is 1 or 2
The center of the circle is (1, - 1) or (1,2)

Find the equation of the circle passing through point a (3,2) whose center is on the line y = 2x and tangent to the line y = 2x + 5

Let the coordinates of the center of the circle be (a, 2a), then according to the meaning of the title, we get | 2A − 2A + 5 | 5 = (a − 3) 2 + (2a − 2) 2 = R, and the solution is: a = 2, r = 5 or a = 45, r = 5, the equation of the circle is: (X-2) 2 + (y-4) 2 = 5 or (x-45) 2 + (y-85) 2 = 5

Given that the circle passes through two points a (5,2) and B (3, - 2), and the center of the circle is on the straight line 2x-y-3 = 0, the equation of the circle is solved

Since the center of the circle is on the straight line 2x-y-3 = 0, the center coordinates of the circle C can be set as C (a, 2a-3). Then from the circle C passing through a (5, 2) and B (3, - 2), we can get | Ca | = | CB |, | Ca | - 2 = | CB |, and | (a-5) 2 + (2a-3-2) 2 = (A-3) 2 + (2a-3 + 2) 2

Given that the center of circle C is on y = 2x and passes through point m (3,1), the equation of circle C is obtained

If the equation of the line passing through the dot m is y = ax + B and perpendicular to y = 2x, then a = - 1 / 2
According to m (3,1), B = 5 / 2
Y = 2x and y = - 1 / 2x + 5 / 2 intersect to get x = 1, y = 2
The center C coordinates are (1,2)
The radius is (3-1) ^ 2 + (1-2) ^ 2 = R ^ 2, and the radius is √ 5
The equation of circle C is (Y-2) ^ 2 + (x-1) ^ 2 = 5

Find the standard equation of a circle passing through two points a (5,2) B (3, - 2) on the line 2x-y = 3

The center of the circle is the intersection of AB vertical bisector L and 2x-y = 3
AB：y=2x-8,|AB|^2=20
AB midpoint (4,0), KL = - 1 / 2
L：x+2y=2
Center (- 4 / 5,7 / 5), distance d to ab
D^2=121/5
Radius r, R ^ 2 = D ^ 2 + (AB / 2) ^ 2 = 221 / 5
Equation: (x + 4 / 5) ^ 2 + (Y-7 / 5) ^ 2 = 221 / 5

It is known that the equation of circle C is x ^ 2 + y ^ 2 + (m-2) x + (M + 1) y + m-2 = 0. Determine the value of m according to the following conditions, and write the center coordinates and radius of the circle
(1) (2) the center of the circle is closest to the origin of the coordinate

[x-(m-2)/2]^2+[y-(m+1)/2]^2=[(m-2)/2]^2+[(m+1)/2]^2-m+2
Center of circle [(m-2) / 2, (M + 1) / 2]
1、
The smallest area is the smallest radius
So R ^ 2 = [(m-2) / 2] ^ 2 + [(M + 1) / 2] ^ 2-m + 2 is the smallest
[(m-2)/2]^2+[(m+1)/2]^2-m+2
=(m^2-4m+4+m^2+2m+1-4m+8)/4
=(2m^2-6m+13)/4
=[(m-3/2)^2+17/2]/4
So when m = 3 / 2, the minimum of R ^ 2 is 17 / 8
That is, M = 3 / 2
The center of the circle (- 1 / 4,5 / 4)
Radius √ 34 / 4
2、
Square of distance from center to origin = [(m-2) / 2] ^ 2 + [(M + 1) / 2]
=(2m^2-2m+5)/2
=[2(m-1/2)^2+9/2]/2
So m = 1 / 2
The center of the circle (- 3 / 4,3 / 4)
Radius 25 / 8

Given that the circle C passes through the point m (4, - 2) n (1,1) and the center of the circle is on the straight line x + y + 1 = 0 (1), we can find the equation of circle C
How to find the center of 2x + 3Y + 1 = 0?

Let the center of the circle be C (x0, Y0),
According to the definition of circle, | cm | = | cn |, so
(x0-4)^2+(y0+2)^2=(x0-1)^2+(y0-1)^2,
It is reduced to: x0-y0 = 3, ①
From the center of the circle on the straight line: x0 + Y0 + 1 = 0, ②
The simultaneous equations (1) and (2) are solved as follows: x0 = 1, Y0 = - 2,
Radius r = | cn | = 3,
The equation of circle C point is: (x-1) ^ 2 + (y + 2) ^ 2 = 9