A question about a junior high school function 10 points first, then extra points .

1.Since point A is on the y-axis, its abscissa is 0, i.e., 4-2a=0a=2 and point b is positioned at (4,3).

2.If point B is on the right side of the y-axis and point A is on the right-hand side of the y-axis.

Then 4-2a>0 and 2+a>0 i.e., -2, then 4-2a=-(2+a) i.e., a=6

b-5=3 i.e.: b=8

So a-b=-2

4.If the highlights of A and B are connected parallel to the Y axis.

Then 4-2a=2+a, i.e., a=2 3

b-5 is not equal to 3, i.e., b cannot be equal to 8

-2a=0 gives a=,3).

a>0,4-2a>0 to get -23, (4-2a)+(2+a)=0, b-5=3 to get a=6,b=8,a-b=-2

2a=2+a, b-5 is not equal to 3 and a=2 3, and b is not equal to 8

1) Known: 4-2a=0

then a=2b(4,3) is in the first quadrant.

2) Known: 2+a>0,4-2a>0

then a>-2, a<2

i.e. -23) is known: 4-2a=-(2+a), b-5=3, then a=6, b=8

a-b=-2

4) It is known that 4-2a=2+a, b-5 is not equal to 3a=2, and 3b is not equal to 8

1 a on the y-axis 4-2a = 0 a = 2 b (4, 3) will only be able to give points if this can be given.

is a quadratic function about x.

m^2-m=2,(m-2)(m+1)=0,m1=2,m2=-1.

When m=-1, its image is a parabola with the opening pointing downward.

2.Under the condition of 1, when x satisfies the condition, y decreases with the increase of x.

y=-x 2, when x 0, y decreases as x increases.

2.It is known that the parabola y=ax 2 passes through the point p(1,4)4=a*1,a=4

When x=3, the value of y is 3*4=12

3.Determine whether the function image passes through the point (-1,2).

y=4x2, when x=, y=4≠2, the function image does not pass through the point (-1,2)3The side length of the square is 3, if the side length increases x, then the area increases y, and the functional relationship between y and x is clearly attacked.

According to the formula for the area of the square, s=3*3=9, y=(3+x) 2, (x 0).

The functional relation between y and x is:

y=(3+x)^2,(x≥0)

1) If the image of the inverse proportional function y=k of x passes through the point (-3, -4), then its image is in quadrant (1, 3).

2) Translate the parabola y=-2x in the horizontal direction by three units, and the analytic formula of the parabola is (y=-2x +3) Translate the parabola y=3x downward by 1 unit in the horizontal direction, and the analytic formula of the parabola is (y=-2x-1).

3) The parabola y=-3x square + 2 openings downward, the axis of symmetry is (x=1/3) and the vertex coordinates are (1/3, 1).

4) The vertex coordinates of the parabola y=-one-half (x-2) square + 5 are (2,5) and the axis of symmetry is (x=2),

(1) One three quadrants.

2）y=-2x^2+3

y=3x^2-1

3) The axis of symmetry x=0, i.e. the y-axis (0,2)(4)(2,5) x=2

1) Third Quadrant Because the points (-3, -4) are in the third quadrant 2) y-3=-2x i.e. y=-2x +3 because the up-and-down movement is translational on the y-axis y-1=3x i.e. y=3x +1

3)y=-3x +2 axis of symmetry is x=0 fixed-point coordinates (0,2)4)y=-1 2(x-2) +5 vertex coordinates are (2,5) axis of symmetry x=2