The quadratic function y x 1 2 2 10 is known

Substituting x=0 into y=(x-1)2+2 gives y=3

then a(0,3).

m1（1,2）

Substituting x=0 into y=-(x+1)2-2 gives y=-3, then b(0,-3).

m2（-1,-2）

then the s quadrilateral.

S-triangle. AM1B+S triangle BM2A=3+3=6A: The area is 6

Solution: Let the coordinates of point a be (x1,0), the coordinates of b are (x2,0), and the coordinates of c are (0,-1), because the two roots of the equation m2x2 2(n-1)x-1=0 are x1,x2, that is, the abscissa of the two points of points a and b, so x1 x2=-2(n-1) m2,x1*x2=-1 m2, because the image symmetry axis is x=-1, so -2(n-1) 2m2=-1, so -2(n-1) m2=-2, so x1 x2=-2 because the area of the triangle abc is 2, so 1 2 1 ab=2, and ab=x2-x1 so x2-x1=4 because x1 x2=-2, so x2=1, x1=-3, so 1 (-3)=-1 m2, so m2=3, because -2(n-1) m2=-2, so n=4, so the function image is y=3x2 6x-1. Satisfied.

1. Match y=-1 2x 2+x+3 2 = 1 2(x-1) 2 + 2 2, the vertex is located at (1,2), the axis of symmetry x=1 3, the world is (1,2) (0,3 2), (2,3 2) (1,0), (3,0) (2,-5 2),(4,-5 2) 4, (-1)< x (3) y>0; (1) > x or > (3) y

When y=1, x2

2x-2=1, i.e. x2

2x-3=0, the solution is x1

3, x2-1, i.e., when x=3 or -1, y=1

Therefore, the answer is 3 or Jingzu -1 group sales.

y=x²-2x-1

When x=0 y=-1

When y=0 x -2x-1=0

x-1)²=2

x=1 2 under the root number

Therefore, the intersection point with the y-axis is (0,-1) and the intersection point with the x-axis of the collapse width is (1 + root number is good under 2,0) (1-root number is 2,0).

2 y=x²-2x-1=(x-1)²+2

From the left to the right and the right to subtract from the top, we can see that x-1 is to move y=x to the right and move it by one unit, and +2 is to move y=x up by 2 units.

So if you move y=x one unit to the right and two units up, you get y=(x-1) +2

Don't ask for everything, it's better to try it yourself.

This question is relatively basic. Take the numbers 0, 1, 2, 3 and so on of the X generation and calculate the y. You can draw a picture. Establish a Cartesian coordinate system.

Quadratic function y=-12x 2-x+32 or y=-1 2x 2-x+3 2

The quadratic function y=-1 2x 2-x+3 2 is known

y=-1/2x^2-x+3/2=(-1/2)(x+1)²+2==(-1/2)(x+3)(x-1)

1) Axis of symmetry: x=-1, vertex m(-1,2), and the intersection point of x-axis a(-3,0)b(1,0), opening upward; You can draw an image of this function;

2) When y 0, the value range of x is ((3) (1,+.)

3) Translate the image 3 units to the right along the x-axis, and the functional relationship corresponding to the translated image is:

y=(-1/2)(x-3+1)²+2==(-1/2)(x-2)²+2==(-1/2)x²+2x