It is known that the quadratic function goes through A 1,0 B 3,0 C 0, 3 vertices D 1, 4 to find the

Solution: Let the parabola pass (1,0) and (3,0) from the parabola to y=a(x-1)(x-3), and substitute y=3 when x=0.

a=1, then y=(x-1)(x-3)=x-4x+3

The analytical formula of the quadratic function is y=x -4x+3

From the analytic expression of the parabola, we can get c(0,3), and the symmetry axis of the parabola is the straight line x=2, d(2,0).

In OAC, OA=1, OC=3, AOC=90°

The coordinates of the point E are (2,m), and in DAE, da=1, de=m, ade=90°.

There are two correspondences between ADE and AOC that are similar:

If DAE OAC, there should be: de oc=da OA i.e. m 3=1 1

Solution: m = 3

Then the coordinates of point e are (2,3) or (2,-3).

If dae oca, then there should be: de oa=da oc i.e. m 1=1 3

Solution: m= 1 3

Then the coordinates of point e are (2,1 3) or (2,-1 3).

The third question, the question is not clear, to be reissued!

Are you satisfied with the above?

This is too simple, I really can't draw the coordinates, just find the four points.

It is known that the image of the quadratic function y=ax-4x+c passes through two points, a and b, and the expression of the quadratic function is obtained.

Solution: y=ax -4x+c is substituted for (-1,0) ,3,-9)0=a+4+c

9=9a-12+c

a=7/8c=36/7

y=7/8x^-4x+36/7

Bring point a b into the function.

0=a+4+c

9=9a-12+c

Solving the system of equations will do it.

a=7/8 c=-39/8

Solution: 1) Replace a(-1,-1),b(3,-9) into y=ax 2-4x+c, obtain, a+4+c=-1,9a-12+c=-9, and solve a=1,c=-6

So the analytic formula is y=x 2-4x-6

2) Replace x=m,y=m into the parabola, obtain,m=m 2-4m-6,m 2-5m-6=0,m-6)(m+1)=0

m1=6,m2=-1

Because m>0

So m=6, so p(6,6).

Because the axis of symmetry of the parabola is x=-b 2a=2

So the symmetry point q of p with respect to x=2 is (-2,6).

So the distance from q to x-axis is 6

Solution: Let the analytic formula of the function be: y=a(x

4 (vertex type).

Substituting the point b(2,-5) into the analytic formula, the solution: a= analytic formula is: y=-(x

The coordinates of the intersection of the parabola and the x-axis: y=0, x1=x2=1i.e. (-3,0) and (1,0).

Coordinates of the intersection of the parabola and the y-axis: x=0, y=3

i.e. (0,3) is not ruined.

The vertex is a(-1,-2), and the x-axis is b,c, then the opening is upward, let c(x1,0),b(x2,0).

The area of abc is equal to 4, 1 2 * bottom * height = 4, bottom edge = |x2-x1|, high = |-2|=2

Let y=a(x+1) 2-2 (vertex test) and bottom edge =|x2-x1|= root number (discriminant test) |a|, bring in to seek a, other omitted.

ATTENTION|x2-x1|=Root(x2-x1) 2=Root(x2+x1) 2-4x1x2 Vedda's theorem is brought into the bottom =|x2-x1|= root number (discriminant test) |a|

Since there is a point on the parabola that is (3,4), the parabola opening is upward, and the vertex is (1,0), so the parabola equation can be as:

y=a(x-1)^2

Substituting the coordinates of (3,4) yields 4 a(3-1) 2 a=1

The parabolic equation is: y=(x-1) 2

Substituting (3,4) into y=x+m, we get: 4 3 m m 1 ab The linear equation is: y=x+1

2) Let the coordinates of the p point be (x0,y0), which satisfies the linear equation y=x+1 y0=x0+1

The linear equation of PE is x=x0, and the coordinates of the intersection point can be obtained by substituting the parabolic equation as (x0,(x0-1) 2).

Therefore, h y0-x0-1) 2=x0+1-(x0-1) 2=-x0 2+3x0

Therefore, the relationship is as follows: h -x 2+3x x value range: 0x3

3) Parallel lines that cross point C as AB, if there is another point on the parabola, there is a parallelogram.

The slope of the parallel lines of ab at this point is 1

The c coordinate is (1,0), so the equation for parallel lines of ab is: y=x-1

Substituting the above equation into the parabolic equation has a different solution than c: x 2 y = 1

In this case, the coordinates of point E are (2,1) and the coordinates of point P are (2,3).

Substitute a(-1,0),b(3,0),c(0,-3) into y=ax square + bx+c

a=1, b=-2, c=-3 of the solution

So, y=x-square-2x-3

y=(x-1)square-4

So, d(1,-4).

Therefore, the area of the S triangle BCD = 4 times 3 times 2 times 4 = 3 triangle BCD is found by the cut-and-patch method.

From the coordinates of the three points, we know that the expression of the quadratic function is y=x 2-2x-3, then the coordinates of point c are (1,-4), and it can be found that the triangle is a right triangle with an area of 1 2*3 2*2=3

Parse the analytic formula of the function that is set once.

y=kx+b

Substituting coordinates ab

2k+b=-1 (1)

k+b=3 (2)

1) (2) Synthesis.

3k=4k=4/3

b=5 3, so the primary function y=4 3x+5 3

a(-2 -1)b(1 3)

o Distance to a straight line.

for |4x0+5+3x0|/5

1ab is (3 4).

The length of ab is 5

So area = 5x1x1 2 = 5 2

Hope it helps.

Learning progress o ( o thank you.