The image of quadratic function can be solved by (1.6) (- 1. - 18) (2.9)

The image of quadratic function can be solved by (1.6) (- 1. - 18) (2.9)

Let the analytic expression of quadratic function be y = ax ^ 2 + BX + C. substituting the coordinates of known points, we can get (1) a + B + C = 6; (2) A-B + C = - 18; (3) 4A + 2B + C = 9; (1) - (2) 2B = 24, so B = 12; (3) - (1) 3A + B = 3, so 3A = 3-B = 3-12 = - 9

Given that the image of quadratic function passes through points (1, - 2), (0,3), (1,6), find the analytic expression of quadratic function

Let the analytic expression of quadratic function be y = ax ^ 2 + BX + C (a is not equal to 0). Substitute three points (1, - 2), (0,3), (1,6) into the quadratic equation respectively, and get the equation system a * 1 ^ 2 + b * 1 + C = - 2
a*0^2+b*0+c=3
a*1^2+b*1+c=6
Three equations are combined into a system of equations
A C B can be obtained
There is something wrong with the title

Given that the image of quadratic function passes through points (1,0), (2,0) and (3,4), the vertex coordinates of its analytical symmetry axis are obtained

Suppose that the quadratic function is y = ax & # 178; + BX + C, and the points (1,0), (2,0), (3,4) are substituted into the system of equations: a + B + C = 0, ① 4A + 2B + C = 0, ② 9A + 3B + C = 4, ③ a = 2, B = - 6, C = 4, then y = 2x & # 178; - 6x + 4, ④ suppose that its vertex is a (m, n), then its vertex equation can be set as y = 2 (x-m)

When the image of a quadratic function passes through (1,0) (2,0) (0,2), the analytic expression of the function is obtained

Let y = ax, square + BX + C, and bring the three points in
0=a+b+c
0=4a+2b+c
2=c
A = 1, B = - 3
So the analytic formula is y = x square - 3x + 2

If the image of a quadratic function passes through (1,0), (2,0) and (0,2), the analytic expression of the function is

y=x^2-3x+2

The analytic expression of quadratic function is obtained according to the following conditions: (1) the image of quadratic function is known to pass (0,11 △ 3) (0,3) (2,0) three points; (2) when x = - 1, y = 4; (3) the image of quadratic function intersects with X axis at point (2,0) (- 3,0) and passes (1,5)

(1) Using the undetermined coefficient method, the data can be brought into y = AX2 + BX + C to solve the problem. (2) 1 point can't be solved. There are other conditions you haven't found, such as several solutions, openings, axes of symmetry of the function. If it's a fixed point, it's brought to the vertex coordinates to calculate. (3) brought to the 3-point formula, y = a (x-x1) (x-x2) x1, X2

Given that the image of a quadratic function passes through (3,0), (0, - 3), (1, - 4), the analytic expression of the quadratic function is obtained

Let the analytic expression of quadratic function be y = AX2 + BX + C. according to the meaning of the title, we get 9A + 3B + C = 0C = - 3A + B + C = - 4, and the solution is a = 1b = - 2C = - 3, so the analytic expression of quadratic function is y = x2-2x-3

Given that the image of quadratic function passes through three points a (- 3,0) B (1.5) C (0.0), the analytic expression of quadratic function is obtained

The solution is a (- 3,0), C (0.0)
So let y = a (x-x1) (x-x2), a ≠ 0
y=ax（x+3）
Bring B (1,5) into the presence of
5=4a
a=5/4
So y = (5 / 4) x (x + 3)

According to the coordinates of three points on the image of quadratic function, the analytic expression of function is obtained
1.（-1,0）,（3,0）,（1,-5）
2.（1,2）,（3,0）,（-2,20）

1.（-1,0）,（3,0）,（1,-5）
Because (- 1,0), (3,0) is the intersection of x-axis
Let y = a (x + 1) (x-3)
(1, - 5) is substituted by:
a*2*(-2)=-5
a=5/4
Analytic formula of function: y = 5 / 4 (x + 1) (x-3)
y=5/4x²-5/2x-15/4
2.（1,2）,（3,0）,（-2,20）
Let the analytic expression of the function be y = ax & # 178; + BX + C
Three points are substituted
a+b+c=2
9a+3b+c=0
4a-2b+c=20
The solution is as follows
a=1；b=-5；c=6
The analytic formula of the function is: y = x & # 178; - 5x + 6

According to the coordinates of three points on the image of quadratic function, the analytic expression of function is obtained: (- 1.0) (3,0) (1, - 5)
Substituting y = ax ^ 2 + BX + C

Let the analytic expression of quadratic function be y = a (x + 1) (x-3), and substitute x = 1, y = - 5 into y = a (x + 1) (x-3)
-5=a(1+1)(1-3)
-5=-4a
a=5/4
The analytic expression of quadratic function is: y = 5 / 4 (x + 1) (x-3)
Supplement;
Y = 5 / 4 (x + 1) (x-3) is reduced to a general formula
y=5/4x²-5/2x-15/4