A quadratic function, please enter! A quadratic function, urgent!

First of all, the general form of the quadratic function: y ax 2 bx c know.

Take the yax2bxc recipe, extract a, add and subtract one-half squared of the term once, and get:

y a(x b 2a) 2+(4ac-b 2) 4a, because the axis of symmetry of y (x-h) 2 k is x h

So the axis of symmetry of y a(x b 2a) 2+(4ac-b 2) 4a is x -b 2a

It is known that the quadratic function y=mx of the quadratic + 2(m+1)x+m+1 image opening is downward, and there is an opening with the x-axis downward, so m<0;There are two different intersection points with the x-axis, delta>0, and m>-1; is solvedSo d

The general formula of the quadratic function y=ax 2+bx+c formulates y=a(x+b 2a) 2+(4ac-b 2) 4a, which is the vertex formula of the quadratic function, and the axis of symmetry is x=-b 2a

The axis of symmetry formula for the quadratic function y=ax 2+bx+c is x=-b 2a

Here a=-1, b=m-2, c=m+1

The axis of symmetry is x=-b 2a=(m-2) 2

That's the theorem, it's in the math textbook.

Substituting (1,0) into y=ax 2+bx-2 gives a+b-2=0; a=2-b;

That is, the quadratic function is y=(2-b)x +bx-2

Substituting (1,-b) into the primary function y=kx (past the origin) gives -b=k, that is, the primary function is y=-bx

Synact y=(2-b)x +bx-2

y=-bx.

2-b)x²+bx-2=-bx

2-b)x²+2bx-2=0

From the meaning of the title, x1, x2 is (2-b)x +2bx-2=0, then x1+x2=2b (b-2); x1x2=-2/(2-b)(x1-x2)²=(x1+x2)²-4x1x2=4b²/(b-2）²+8/((2-b)

4（b²-2b+4)/(b-2)²

ix1-x2i=2/ib-2i√[（b-1)²+3]∵a>b>0

2-b＞b0＜b＜1

ix1-x2i=2/ib-2i√[（b-1)²+3]=2/(2-b)√[b-1)²+3]

i.e. 2 ix1-x2i 2 3

a+b-2=0

y=-bxx1-x2|= root number (x1 +x2 -2x1x2) = root number [(x1+x2) -4x1x2] = root number (b a +8 a).

Substituting a+b-2=0 into the root number (b a +8 a) gives |x1-x2|=|(a+2)/a|=|1+1/a|So |x1-x2|>1

The image of y=ax 2+bx-2 passes through the dot (1,0), 0=a+b-2

A function image passes through (1,-b) and the origin.

Let the analytical formula of the primary function be: y=kx

b=ky=-bx

bx=(2-b)x²+bx-2

b-2)x²-2bx+2=0

x1+x2=2b/(b-2)

x1x2=2/(b-2)

x1-x2)²=(x1+x2)²-4x1x2=4b²/(b-2)²-8/(b-2)=4(b²-2b+2)/(b-2)²

x1-x2| =2√(b²-2b+2)/(b-2)

After (3,b), substituting the straight line to get b=2*3-7=-1 so the point (3,-1), and then substituting the parabola, we get -1=a*9, and getting a=-1 9 points (-b,-ab) is (1,-1 9).

When x=1, the parabola y=-1 9*1*1=-1 9, so the point (-b, -ab) is on the parabola.

Substituting (3,b into a straight line, finding b=-1, and substituting (3,-1) into a parabola, finding a=-1 9.

The point to be judged is (-1, -1 9), and the parabola is true, so the point is on the curve.

(3,b) Substituting y=2x-7 yields b=-1

Then substitute (3,-1) with y=ax

a=y/x²=-1/9

So y=-1 9x

b,-ab) i.e. (1,-1 9) substitution, the equation holds, so on that parabola.

Substituting (3,b) into y=2x-7 gives b=-1, substituting the point (3,-1) into y=ax gives a=-1 9, so (-b,-ab) is (1,-1 9), substituting this point into y=ax gives -1 9=1x-1 9, so on top.

Solution: (1) x1+x2 = b x1 x2 = c-1+2 = b = 1 -1 2 = c = 2b = 1 c = 2

y=ax²-x-2

The axis of symmetry of this parabola is -b 2a =x

i.e. -(1) 2a = x

If a 0 then -(1) 2a 0

If a 0 then -(1) 2a 0

Regardless of any real number a, the intersection of the parabolic image with the x-axis is on either side of the origin.

a, b on the x-axis.

0=ax²+bx+c

x1= -c/a x2= -b-c/a

tan∠cbo-tan∠cao=1

i.e. bo=2ao

2l-c/al=lb-c/al

b=3c substitute x1= -c a x2= -b-c a.

c=b-a a=2c

The solution gives a=8 15 b=4 5 c=4 15y= 8 15x +4 5x+4 15

Because its vertex m is in the second quadrant, and the function image passes through points a and b, then a<0

Substituting points a and b into the function.

a+b+c=0,c=1

then b = -1-a

Because the vertex is in the second quadrant, the axis of symmetry is to the left of the y-axis.

i.e. -b 2a<0

1+a)/2a<0

1+a>0

a>-1

So -1

Translate the image of y=x2-1 upwards by 2 units to get y=x2+1

Then translate the image of y=x2+1 to the left one unit to get y=(x+1)2+1=x2+2x+2

Reverse thinking, add 2 first, then move one unit to the left to become y=(x+1) 2+1, and then tidy up.

I'll give you a rough idea, and you'll figure out the rest yourself.

First of all, you can find the values of a, b, and c based on the image.

Then you set f(x)=ax2+bx+c-k (a, b, c are all known numbers), and then you make (i.e. discriminant) > 0 (because the problem requires f(x)=0 to have two unequal real roots).

Finally, you can find the range of values for k.

Fourth, look at the image, that is, when the function value is k, there are two x's. The value of the function y can be ranged from =2, and when y=2 there is only x=2, and y 2 has two solutions. So k 2.

Quadratic function iDefinitions and Definition Expressions.

In general, there is a relationship between the independent variable x and the dependent variable y:

y=ax 2+bx+c (a, b, c are constants, a≠0) is called y a quadratic function of x.

To the right of a quadratic function expression is usually a quadratic trinomial.

ii.Three expressions for quadratic functions.

General formula: y=ax 2; +bx+c(a,b,c is constant, a≠0) vertex formula: y=a(x-h) 2; +k [the vertex of the parabola p(h,k)] intersection formula:

y=a(x-x1)(x-x2) [limited to parabolas with intersections a(x1,0) and b(x2,0) with the x-axis].

Note: In the three forms of mutual transformation, there are the following relationships:

h=-b/2a k=(4ac-b^2;)/4a x1,x2=(-b±√b^2;-4ac)/2a

iii.An image of a quadratic function.

If we make the image of the quadratic function y=x 0 5 in the plane Cartesian coordinate system, it can be seen that the image of the quadratic function is a parabola.

iv.The nature of the parabola.

1.A parabola is an axisymmetric figure. The axis of symmetry is a straight line.

x = b/2a。

The only intersection point between the axis of symmetry and the parabola is the vertex p of the parabola.

In particular, when b = 0, the axis of symmetry of the parabola is the y axis (i.e., the straight line x = 0)2The parabola has a vertex p with coordinates.

p [ b/2a ，(4ac-b^2;)/4a ]。

When -b 2a=0, p is on the y-axis; When δ = b 2-4ac = 0, p is on the x-axis.

3.The quadratic term coefficient a determines the direction and magnitude of the opening of the parabola.

When a 0, the parabola opens upwards; When a 0, the parabola opens downwards.

a|The larger it is, the smaller the opening of the parabola.

4.The primary coefficient b and the quadratic coefficient a together determine the position of the axis of symmetry.

When a and b have the same sign (i.e., ab 0), the axis of symmetry is left on the y-axis;

When A and B are different (i.e., AB 0), the axis of symmetry is to the right of the Y axis.

5.The constant term c determines the intersection of the parabola and the y-axis.

The parabola intersects with the y-axis at (0,c).

6.The number of points where the parabola intersects with the x-axis.

b 2-4ac 0, the parabola has 2 intersections with the x-axis.

b 2-4ac=0, the parabola has 1 intersection point with the x-axis.

b 2-4ac 0, the parabola has no intersection with the x-axis.

v.Quadratic functions and unary quadratic equations.

In particular, the quadratic function (hereafter referred to as the function) y=ax 2; +bx+c, when y=0, the quadratic function is a one-dimensional quadratic equation about x (hereinafter referred to as the equation), i.e., ax 2; +bx+c=0

In this case, whether the function image intersects with the x-axis or not, that is, whether the equation has a real root or not.

The abscissa of the intersection of the function and the x-axis is the root of the equation.

I'll answer this question for you! By the way, please give me extra points, thank you, I can help you list the formula of the second connotation, and the rest is too simple. I hope you don't rely too much on others, it's not good for you, hehe....(1) s=x(12 2x)Please simplify by yourself.

Because the adjacent edge is x broken and the opposing side is (12-2x), s=12x-2x 2, x is greater than zero and less than six, (2) because -b 2a=3, substituting 3 into the equation gives s=18, so the maximum value is 18