As shown in the figure, the image of quadratic function y = AX2 + BX + C passes through three points a, B and C (1) Observe the image, write out the coordinates of three points a, B and C, and find the parabola analytic formula; (2) find the vertex coordinates and symmetry axis of the parabola; (3) observe the image, when x takes what value, Y & lt; 0, y = 0, Y & gt; 0

(1) A (- 1, 0), B (0, - 3), C (4, 5), let the analytic formula be y = AX2 + BX + C, substituting into a − B + C = 0C = − 316a + 4B + C = 5, the solution is a = 1b = − 2C = − 3. So the analytic formula is y = x2-2x-3; (2) y = x2-2x-3 = (x-1) 2-4, so the vertex coordinates are: (1, - 4), and the axis of symmetry is a straight line x = 1; (3) when X & lt; - 1 or X & gt; 3, Y & gt; When x = - 1 or x = 3, y = 0; when - 1 & lt; X & lt; 3, Y & lt; 0

The image of quadratic function y = ax square + BX + C (a is not equal to 0) is shown in the figure. If / ax square + BX + C / = K (k is not equal to 0) has two unequal real roots, then the value range of K is ()

You can't do it without a picture!

Using the image of quadratic function y = ax ^ 2 + BX + C (a is not equal to 0) to find the approximate root of quadratic equation AX ^ 2 + BX + C = 0 (a is not equal to 0)
(1) Draw a function_ (2) determine parabola and parabola The number of intersection points of an axis depends on which two numbers the intersection points are between , estimate the value between the two numbers, and estimate the approximate root with calculator. The approximate root is in the range of the corresponding y value_ It's a good place

(1) Draw a function_ y=ax²+bx+c_ The image of the image;
(2) Determine parabola and_ X-axis_ The number of the intersection points of the axes depends on which two numbers the intersection points are between;
（3）__________ , estimate the value between the two numbers, and estimate the approximate root with calculator. The approximate root is in the range of the corresponding y value_______ It's a good place

It is known that the image of quadratic function y = AX2 + BX + C passes through three points a (- 1,0), B (4,0) and C (0,k), where ∠ ACB = 90 ° (1) find the value of K; (2) if the opening of the function image is downward, find the value of a, B and C

Solution (1) because point C is on the y-axis, according to Pythagorean theorem: ac2 = K2 + 12 = K2 + 1, BC2 = K2 + 42 = K2 + 16 (2 points) because ∠ ACB = 90 °, so ac2 + BC2 = AB2, that is K2 + 1 + K2 + 16 = 25 The solution is k = ± 2 When k = 2, the opening of parabola is downward

It is known that the image of quadratic function y = ax & # 178; + BX + C (where a is a positive integer) passes through point a (- 1,4) and point B (2,1)
And there are two different intersections with the X axis, then the maximum value of B + C is
The analysis is as follows
Substituting AB into
a-b+c=4
4a+2b+c=1
subtract
3a+3b=-3
a+b=-1
b=-1-a
Substituting A-B + C = 4
a+1+a+c=4
c=3-2a
There are two different intersections with the X axis
So B ^ 2-4ac > 0
So (- 1-A) ^ 2-4a (3-2a) > 0
a^2+2a+1+8a^2-12a>0
9a^2-10a+1>0
(9a-1)(a-1)>0
a>1,a=2,-3a

(9a-1)(a-1)>0
a>1,a1,a>1/9
A > 1
If a is known to be a positive integer, then any positive integer greater than 1 must be greater than or equal to 2
So a ≥ 2

Given that the radius of circle C is 2, the center of circle is on the positive half axis of X axis, and the line 3x + 4Y + 4 = 0 is tangent to circle C, then the equation of circle C is ()
A. x2+y2-2x-3=0B. x2+y2+4x=0C. x2+y2+2x-3=0D. x2+y2-4x=0

Let the center of the circle be (a, 0) (a > 0), the distance from the center of the circle to the straight line 3x + 4Y + 4 = 0 d = | 3A + 4| 32 + 42 = 3A + 45 = r = 2, the solution is a = 2, so the coordinate of the center of the circle is (2, 0), then the equation of the circle C is: (X-2) 2 + y2 = 4, the simplification is x2 + y2-4x = 0, so D is selected

Given that the radius of the circle is 2, the center of the circle is on the positive half axis of the x-axis, and the circle is tangent to the line 3x + 4Y + 4 = 0, then the standard equation of the circle is______ ．

Let the center coordinate of the circle be (a, 0) and a > 0, because the circle is tangent to the straight line 3x + 4Y + 4 = 0, the distance from the center of the circle to the straight line is equal to the radius 2, that is | 3A + 4 | 32 + 42 = 2, a = 2 or a = - 143 (rounding off), so the center coordinate of the circle a = 2 is (2, 0), and the standard equation of the circle with radius 2 is: (X-2) 2 + y2 = 4, so the answer is (X-2) 2 + Y2 = 4

The equation for finding a circle with radius of 1, center of circle on X axis and tangent to the line 3x + 4y-7 = 0

Let the coordinates of the center of the circle (x, 0), (3x-7) / 5 = 1, x = 4, and write it yourself

A (- 0.5,0) B is the moving point on the square of the Circle F (x-0.5) + the square of y = 4, and the vertical bisector of the line AB intersects BF with P, then the trajectory equation of the moving point P is as follows:

x^2+4y^2/3=1
ellipse

In the circle x ^ 2 + y ^ 2 = 8, there is a point P (- 1,2). AB is the chord passing through point P
Finding the minimum chord length of point P

The center O of circle x ^ 2 + y ^ 2 = 8 is at the origin, radius = 2 √ 2
When op ⊥ AB, the chord AB gets the minimum
OP²=(-1)²+2²=5
PA=√(OA²-OP²)=√[(2√2)²-5]=√3
Minimum chord length passing through point P = AB = 2PA = 2 √ 3