To resend the quadratic function question, please ask the quadratic function question

If p is no longer on a straight line ab, then according to the three-point formula, a parabola must be determined, and now p is no longer on any parabola across ab.

Explain that the point p is on the straight line ab, and the equation of the straight line ab is calculated, and y=x+1 substitutes the coordinates of point p, and there are m 2+1=m+1 to get m=0 or m=1, you want to be angry with me, I don't understand it after talking so carefully.

You have to draw a diagram first.

You think about it.

Now we can know that the parabola passes through two points, a and b.

Investigate point p.

1.p is not on the straight ab.

p is not on the straight line ab, then p must be outside the straight line, at this time you can determine a parabola according to the coordinates of p, a, b, three points, obviously this time does not meet the topic. (Because he has already found a parabola past the point p, which is not on topic).

2.p on the straight line ab.

In this case, because the parabola and the straight line have at most two intersection points, that is, for the points on the straight line AB, except for the two points A and B, the rest of the points are not on the parabola above the two points of AB. Now p is not on the parabola, so p is on the straight line ab.

The second sentence means that abc is no more than p(m, m 2+1) of the quadratic function y=ax 2+bx+c, regardless of any real number (not zero).

Solution: From the problem y=ax +bx+c(a<0) over the point (-1,2), then a-b+c =2;

From the intersection of the quadratic function and the x-axis x1,x2 satisfies -2 x1 -1,0 x2 1, the symmetry axis of the quadratic function must be between x =-2 and x =1;

From a <0, we can see that at x>1, this function is monotonically decreasing;

Thus f(1)=a+b+c <0;

Add a-b+c =2 and a+b+c<0 together to get (a-b+c)+(a+b+c)<2;

Thus 2(a+c)<2, we get a+c <1;The first option is wrong;

With regard to the second option, do the difference, and give b=a+c-2b +8a-4ac = a+c-2) -4ac+8a(a+c) -4(a+c)+4-4ac+8a(a-c) -4(c-a)+4

c-a-2)²≥0

Get b +8a 4ac.

Title: What does "the ordinate of the intersection with the y-axis y0 2" mean?

Solution: (1) Let the expression of the quadratic function be f1(x)=ax 2 and substitute the coordinates of the point (1,1) to obtain: a=1

f1(x)=x^2

Let f2(x)=k x, then it has two intersections with y=x, so k>0, and the coordinates are (- k, - k), k, k).

The distance between two points d=2 (2k)=8

Solution: k=8, so f2(x)=8 x

Therefore f(x)=x 2+(8 x).

2)f(a)=a^2+(8/a)

To make it: f(a)=f(x).

i.e. a 2 + (8 a) = x 2 + (8 x).

Simplification yields: (x-a)(ax 2+a 2x-8)=0 The equation has a solution: x=a, and for ax 2+a 2x-8=0, since the discriminant formula = a 4+32a=a(a 3+32)>0

Therefore ax 2+a 2x-8=0 has two different real roots, and it is not a i.e. the equation has three different real roots.

In fact, in the end, there is no need for a>3, and the equation also has f(x)=f( a) and three real number solutions.

15.Let the analytic formula be y=ax +bx+c, because the axis of symmetry x=1

So -b 2a = 1 and b = -2a

Then pass the point a(-1,0) and the point c(0,3 2) to substitute two points, and there is 0=a-b+c

3 2=c solves a=-1 2, b=1, c=3 2

So the analytic formula is y=-x 2+x+3 2

point b coordinates, (3,0).

Second question: Do PD X-axis.

s△abp=pd*ab/2

ab=3+1=4

So 4*pd 2=2*pd=2

So pd=1

When y=1.

There is 1=-x 2+x+3 2

Solve x=1 + root number 2, or x=1 - root number 2So p coordinates (1 + root number 2,1), (1 - root number 2,1).

When y=-1, there is.

1=-x²/2+x+3/2

Solve x=1 + root number 6, or x=1 - root number 6So p coordinates (1 + root number 6,1), (1 - root number 6,1).

16.The image is to move the original image to the left by 2a units, and then flip the coordinates of point A (0,a) and point B (a,0) with the x-axis as the axis of symmetry

oa=a²，ob=a

oa=ob, i.e., a=a

Solve a=1

So the function is y=(x-1).

The function after translation is y=-(x+1).

Point C coordinates (-1,0).

D point coordinates (0,-1).

ab=bc=cd=da with number 2

So CDBA circumference = 4 root number 2

17,x belongs to r

y=4x²-4x+2=4（x²-x）+2

4（x²-x+1/4-1/4）+2

4（x²-x+1/4）+2-1

4（x-1/2）²+1

When x=1 2, the function has a minimum value, and the minimum value is y=118. Let the quadratic function be y=ax +bx+c

Take the wall as the coordinate system, and know from the problem that when x=0, y=10

When x=1, y=40 3

The axis of symmetry is x=1

So there is 10=c

40/3=a+b+c

b/2a=1

Solve a=-10 3, b=20 3, c=10, so y=-10x 3+20x 3+10

When y=0, there is 0=-10x 3+20x 3+10 to solve x=3 or x=-1 (rounded).

So when the water hits the ground, it is 3 meters away from the wall.

y=x^2-2(m-1)x+m2-2m-31.Proof:

[-2(m-1)] 2 - 4 1 (m2-2m-3)= 16 > 0 Constant True.

So, regardless of the value of m, the image of this quadratic function must have two intersections with the x-axis2Remembering a(x1,0), b(x2,0), Veda's theorem: x1+x2 = 2(m-1); x1x2 = m^2-2m-3

1/x1+1/x2 = (x1+x2)/x1x2 = 2(m-1)/(m^2-2m-3) = 2/3

m = 0 or 5

y=x 2+2x-3 or y=x 2-8x+12

(1).Discriminant 4(m-1) 2-4(m 2-2m-3)=8>0, so the function image and the x-axis have two intersections.

That is, p and q are at the same time to points b and c.

Set the departure time as: t "Matching skin = 6

s=1 2(12-2*t)*4t=-4t*t+24t(t<=6) because: s=-4(t-3)*(t-3)+36s(t) increases monotonically at (0,3); (3,6] monotonically decreasing;

The maximum value is s(3)=36;The minimum value is s(6)=0, so the value range of s is [0,72].