The quadratic function y ax2 bx c a 0 is known, according to the conclusions given,

The quadratic function y ax2 bx c(a≠0) is known, and according to the conclusions given, the abscissa of the two intersections with the x-axis, one is larger than 1 and the other is smaller than 1.

What is this question for? Finding the analytic formula of a function?

Analysis: Quadratic function y ax2 bx c(a≠0)y ax 2+bx+c=a[(x+b (2a)] 2+(4ac-b 2) (4a).

Let y ax 2+bx+c=0

x=-b (2a) b 2-4ac) (2a) ling-b (2a)=1==>-b=2a

i.e. a, b symbols are opposite, |b|=2|a|, ensure that the symmetry axis of the function is x=1b 2-4ac>0, and ensure that the function image has two intersections with the x-axis.

Example: f(x)=x 2-2x-1 2, f(x)=-2x, 2+4x+3, when the axis of symmetry is not necessarily at x=1.

1+b (2a)< b 2-4ac) (2a)(2a+b) (2a)] b 2-4ac) (2a) (2a) when a>0, 4a 2+b 2+4aba+b<-cWhen a<0, 4a 2+b 2+4aba+b>-cb 2-4ac>0, ensuring that the function image has two intersections with the x-axis.

Example: f(x)=2x2-3x+1 2, f(x)=-x 2+5x-1

Solution: Obtained from the relationship between the root and the coefficient.

x1+x2=-b/a,x1x2=c/a

From the meaning of the title. x1<1,x2>1

x1-1<0,x2-1>0

x1-1)(x2-1)

x1x2-(x1+x2)+1<0

c/a-b/a+1<0

a+c-b)/a<0

a(a+c-b)<0

Therefore, conditions a(a+c-b)<0 can be added

Solution: Since the image opens upwards, a 0

The parabolic axis of symmetry is to the left of the y-axis, and if a and b have the same sign, then b 0

The parabola intersects with the negative semi-axis of the y-axis, c 0, then abc 0, is right.

As can be seen from the figure, the parabola passes through the point (1,2), and the point (1,2) is substituted into y ax bx c, and a+b+c=2 is correct.

As can be seen from the figure, when x -1, the corresponding point is in the third quadrant, and x -1 is substituted into y ax bx c, and a-b+c 0 is obtained

Subtract a-b+c 0 from a+b+c=2 to get -2b -2

b 1 is wrong.

Solution: A 0 can be pushed downward from the opening of the parabola, because the axis of symmetry is to the right of the y-axis, and the axis of symmetry is x=-b2a

0, and a 0, so b 0, from the intersection of the parabola and the y axis on the positive half axis of the y axis, we can know c 0, so abc 0, false;

As can be seen from the image, the axis of symmetry x=-b2a

0 and axis of symmetry x=-b2a

1, so 2a+b 0, correct;

As can be seen from the image: when x=-1, y 0 a-b+c 0, error;

When x=-1, y 0, a-b+c 0, a+c b, and b 0, so a+c 0, so correct

In summary: Correct

Therefore, it is not difficult to choose C.

Substituting x=1 gives a+b+c=2, so a+c=2-b substituting x=0, and c 0 can be known from the image

By the axis of symmetry formula, the axis of symmetry is x=-(b 2a).

As can be seen from the figure, -1 -(b 2a) 0

So a, b have the same sign, so abc 0

By the image trend, a 0, so b 0

Substituting x=-1, a-b+c 0 is 2-b-b 0, so b 1, conclusion 2 is correct.

(1) y = (t 1) x 2 (t 2) x 3 The function values of 2 at x 0 and x 2 are equal.

Substitute x 0 and x 2.

t＋1)*0²＋2(t＋2)*0＋3/2=(t＋1)*2²＋2(t＋2)*2＋3/2

4(t+1)+4(t+2)=0

t=-3/2＝

The analytic formula of the quadratic function is y=-x 2 x 3 2

2) The image of the primary function y kx 6 and the image of the quadratic function pass through the point a(-3,m), m=k*(-3)+6

m=-(-3)²/2-3+3/2

The solution is m=-6 k=4

3）y=-x²/2＋x＋3/2=0

x-3)(x+1)=0

The solution is x1=3 x2=-1

b(-1,0) c(3,0)

y=-x 2 x 3 2 into vertices.

y=-x²/2+x+3/2

1/2(x+1)(x-3) -1≤x≤3

Left shift n(n>0), y=-1 2(x+1+n)(x-3+n) (n-1<=x<=-n+3).

y kx 6 translates upwards by n units, y kx 6+n = 4x+6+n

If the translated straight line is tangent to the translated quadratic function, then the equation 4x+6+n=-1 2(x-3+n)(x+1+n) has two equal real solutions.

i.e. x + (2n + 6) x + n + 9 = 0 has two equal solutions to real numbers.

Discriminant 0

2n+6)²-4(n²+9)=0

Solution n=0, contradictory.

Therefore, the translated straight line is not tangent to the translated parabola.

Combined with the image, it can be seen that if the translated line and the parabola have common points, the intersection points of the two critical points are (-n-1,0) and (3-n,0).

then 0=4(-n-1)+6+n, 0=4(3-n)+6+n

The solution gives n=2 3, n=6

Therefore, 2 3 < = n< = 6

The first year of high school, it's a piece of cake, but there is no wealth value, if the wealth is worth more, I will help you do it.

B test question analysis: the image opening is downward, a 0, when x = 0, y 0, then c 0

Axis of symmetry <>

then abc 0Mistake.

Because b=-2a, 4a+2b+c=4a-4a+c=c 0 is correct.

Because a 0, b 0, c 0, so: <>

c-a)(3a+c)

It is easy to know c-a 0, and when 3a+c is substituted into the original formula, y=3a+c+bx, it is easy to know x=<>

At this point y 0, so 3a+c+<>

b 0, so.

That's right; Because a 0, b 0, c 0, <>

When x=3, y=9a+3b+c, 2y=18a+6b+2c

So 3b = y-9a-c, 2c = 2y-18a-6b, then 2c-3b = 2y-18a-6b-(y-9a-c) = y-9a-5b, because b = -2a. Original formula = y-9a+10a=y+a

According to the image, when x=3 is, y 0So y+a 0then 2c-3b 0, 2c 3b. That's right;

Because a 0, b 0, then a+b 0, and when m 0, a m2

0, bm 0, then <>

Correct, when m is less than 0, then bm 0, then <>

Mistake. Comments: This question is more difficult, mainly to test students' mastery of the properties of one-dimensional quadratic equations, and to pay attention to the cultivation of the idea of combining numbers and shapes to be applied to the examination.

Summary. These two questions need to be written about the process.

Find the quadratic function analytically according to the following conditions:

1. I know that the image of a quadratic function passes through the points a(0,-1),b(1,0),c(-1,4);

2. The parabolic vertex p(-1,-3) is known, and the point a(o,-6) is crossed;

These two questions need to be written about the process.

2. The parabolic vertex p(-1,-3) is known, and the point a(o,-6) is crossed;

Thank you. Find the quadratic function analytically according to the following conditions:

These two questions need to be written about the process.

2. The parabolic vertex p(-1,-3) is known, and the point a(o,-6) is crossed;

1. I know that the image of a quadratic function passes through the points a(0,-1),b(1,0),c(-1,4);

Find the quadratic function analytically according to the following conditions:

These two questions need to be written about the process.

2. The parabolic vertex p(-1,-3) is known, and the point a(o,-6) is crossed;

1. I know that the image of a quadratic function passes through the points a(0,-1),b(1,0),c(-1,4);

Find the quadratic function analytically according to the following conditions:

It's not difficult. Solution: (1) a, b are the two roots of the equation x 2-(m-1)x+m+4=0 a+b=m-1 a*b=m+4

ab2=52=a2+b2=（a+b）2-2ab=(m-1)2-2（m+4）

The solution is m1=6 m2=-2 ( a+b 0 rounded) and substituted a+b=6-1 a*b=6+4

Solve a= b=

2) By the question of similarity according to the overlapping part rt bc'm∽rt△abcy=s△bc'm=s△abc*(a-x)/a=ab/2*(a-x)/a=b(a-x)/2

0 x aa b has been obtained, directly substituted, I am a little lazy, ha) and then substituted when y=3 8 b(a-x) 2=3 8 can be directly found x=

Question Analysis: When x=1, y=a+b+c=0, this option is incorrect;

When x= 1, the negative half axis of the intersection point between the image and the x-axis is obviously greater than 1, y=a b+c 0, so this option is correct.

From the opening of the parabola downward, we know a 0, and the axis of symmetry is 1 x=<>0, 2a+b 0, so this option is correct;

The axis of symmetry is x=<>

0, a, b heterogeneous, i.e. b 0, abc 0, so this option is wrong so the answer is

Test point: The relationship between the image of a quadratic function and the coefficients