(1) Substitute point A(- 1 0) and point D (0,2) into the functional relationship:

b = 8/3，c = 2

Therefore, the function of parabola is:

y = -2/3x^2 + 4/3x + 2

(2) The key of this topic is to draw a picture, and draw an image of a parabola that is symmetrical about the X axis, which is clear at a glance.

As shown in the above figure, we can know that it is enough to translate the distance between two D points to X, that is, the ordinate of D point.

So the number of units required for translation is: 2×2 = 4.

(3) Find the coordinates of the parabola vertex.

Solution 1:

Imagine: if we know the abscissa of the vertex, is it equal to knowing the coordinates of the vertex?

So now the question becomes: how to find the abscissa of the vertex?

After careful observation, it is not difficult to find that the parabola is an axisymmetric figure, the vertex is located on the axis of symmetry (the axis of symmetry has been drawn in the above figure), and the parabola has two intersections with the X axis. So is the midpoint of these two intersections also on this axis of symmetry?

The answer is yes, so the first question now is to find the coordinates of another intersection point B between parabola and X axis.

It can be determined that the ordinate of point B is 0, so:

Let-2/3x2+4/3x2 = 0, then x =-1, or x = 3.

So the coordinates of the two points intersecting the X axis are: A (- 1 0) and B (3 3,0).

So the abscissa of the midpoint of AB is: (-1+3)/2 = 1.

Substitute x = 2 into the parabola function, and you get y = 8/3.

So the vertex coordinates are: (1, 8/3)

Solution 2:

As can be seen from the figure, there are two abscissas corresponding to the same ordinate Y except the vertex.

In other words, the equation has two solutions -2/3x 2+4/3x+2-y = 0.

The equation can be simplified as: -(x 2-2x-(6-3y)/2) = 0, that is, x 2-2x+(3y-6)/2 = 0.

In this case, the equation must be in this form: (x -a)(x-b) = 0.

When it is a vertex, there is only one solution.

In this case, the equation must be in this form: (x-a) 2 = 0 (complete square form).

So: (3y-6)/2 = 1, the solution is: y = 8/3.

Substitute y = 8/3 into the functional relationship: x = 1.

So the vertex coordinates are: (1, 8/3)

(4)

Question: Are the two points of MN symmetrical about the straight line y =2 (the straight line y =2 is actually a straight line passing through the D point and parallel to the X axis)?

The answer is yes, so the length of line segment MN is twice the distance from point m or point n to line y=2.

The coordinates of point M are: (n, m) (0

Then: put x? = n into the functional relationship:

m = -2/3n^2 + 4/3n + 2

So the distance from point M to line y = 2 is: m-2 =-2/3N2+4/3N2+2-2 =? -2/3n^2 + 4/3n

So the length of line segment MN is d = 2x (m-2) =-4/3n 2+8/3n.

So the functional relationship between d and n is:

d = -4/3n^2+8/3n(0 & lt； N<2) (don't forget the range of n)