1.S= 1/ 16C^2

2.B

3.( 1) The opening is upward, the vertex coordinate is (2, -7), and the symmetry axis is a straight line x = 2.

(2) The opening is downward, the vertex coordinates are (1,-1), and the symmetry axis is a straight line X = 1.

4. Difference: the opening direction is different; The former passes through the second quadrant and the latter does not; When x≤3, y decreases with the increase of x, while when x≤3, y increases with the increase of x. ...

The same point: the symmetry axes are all straight lines x = 3；; All pass through the first quadrant; The vertex is in the first quadrant ...

5.( 1)y = 1/2x 2-2x- 1。 The image is omitted.

(2) When x≥2, y increases with the increase of x; When x≤2, y decreases with the increase of x.

6. There is a solution. x 1≈5.2，x2≈0.8。

7.D

8. If {m 2+2m=-4 8 = 0 and m-2≠0 is equal to m=-4, then y =-6x2-4x =-6 (x+1/3) 2+2/3. This parabola can be defined as parabola y =-6x.

9.( 1)y=(- 1/90)(x-60)^2+60

(2) from (-1/90) (x-60) 2+60 = 0, x = 60+30 √ 6 < 150 will not exceed the green belt.

10.( 1) A (1, 0), B (3 3,0), C (0 0,3), D(2,-1), the area of quadrilateral ACBD is 4.

(2) From 3s △ ABC = s △ ABP, the distance from point P to X axis is 9. Substitute Y = 9 for Y = X 2-4x+3, and X = 2 √ 10. So there is the point p, whose coordinates are (2+√10,9).

1 1.( 1) Points A (0 0,0) and B (2 2,0) are symmetrical about the axis of symmetry of parabola X = 1, so △ABD is an isosceles right triangle.

(2)∫△BOC is an isosceles triangle, ∴ OB = OC. And point c (0, 1-m 2) is on the negative semi-axis, ∴ m 2- 1 = m+ 1, and the solution is M65438.

12.( 1)y = 1/2√2x √2/2( 1-x)=- 1/2x^2+ 1/2x,0