Quadratic functions in junior high school mathematics

Let x1 x2, x1-x2=2......（1）

The parabola y=one-half x +x+c has two different intersection points with the x-axis, and the distance between the two intersection points is 2, then 1 2 x1 2+x1+c=0......（2）1/2 x122+x2+c=0……（3）

x1 + x2 = -2 ...... of 2)-(3).(4) Combined with (1) to obtain x1=0, x2=-2, substituting (2) to obtain c=0

Two intersections with the x-axis means that there are two roots in half x +x+c=0, and the expression of the two roots is written using the formula for finding the roots, which can be calculated according to the absolute value of the difference between the two (because I don't know who is bigger) equal to 2.

y=(1/2)x²+x+c

Discriminant =1-2c>0

Get c<1 2

The sum of the two is x1+x2=-2

The product of two is x1*x2=c 2

So. The distance between the two intersections is.

x1-x2|

(x1+x2)²-4x1*x2]

4-4*(c/2)]

4-2c) gives 4-2c=4

So c=0

x=-1，b=1

Then 1=a-b

b=a-1ab=a(a-1)=a -a=(a-1 2) -1 4, so there is a minimum value of -1 4

Only option D

Since there is a set of point errors, then according to the symmetry, it must be the middle set of data that is wrong, then the rest of the data are correct.

Then y=(x-1)(x-3).

Sort it out to get a=1, b=-4, c=3

Quadratic function analytic expressions can also be written.

I wish you the best of your choice in high school, upward, boy!

x=2 is wrong.

The axis of symmetry is x=2

So y=a(x-2) +k

So 3=4a+k

0=a+k, so a=-1, k=1

So y=-x +4x+3

Since there is exactly a set of data errors, then according to symmetry, 2 x,y sides of the axis of symmetry are equal to each other at the distance from the axis of symmetry.

There must be something wrong with the middle set of data.

Then the rest of the data is correct.

Then y=(x-1)(x-3).

Sort it out to get a=1, b=-4, c=3

Quadratic function analytic expressions can also be written.

Solution: Obviously, (0,,3) and (4,3) are symmetrical (1,0) and (3,0) are symmetrical about x=2, because there happens to be a set of data in ** that is miscalculated, so (2,-2) is wrong data;

Find the analytic formula: (1,0) and (3,0) are the solutions of the quadratic function y=ax +bx+c=0.

The quadratic function y=ax +bx+c can be expressed as y=a(x-1)(x-3).

0,,3) to get a(0-1)(0-3)=3 to get a=1

Substituting a=1 into y=x -4x+3 gives the analytic formula y=x -4x+3 of the quadratic function

Any three sets of data determine a quadratic function, so any three sets of data are selected to calculate to obtain a quadratic function, if one of the remaining two groups meets the other set of non-conformity, the secondary quadratic function is obtained.

However, based on the symmetry, it can be seen that x=2 is wrong, i.e. y=(x-1)(x-3).

Sort it out to get a=1, b=-4, c=3

Quadratic function analytic expressions can also be written.

Because the quadratic function is symmetrical with respect to the axis of symmetry, the only mistake can be on the axis of symmetry (x=2)3=c0=a+b+c

0=9a+3b+c

The solution yields a=1, b=-4, c=3

i.e. y=x 2-4x+3

You draw these 5 points on the coordinate axis, and once you draw them, you can see at a glance which one is the wrong point.

When x=0, y=c=3, when y=0, x=3 or 1, because the point at y=0 is equal to x=0, y=4 and x=4, y=3, so these four points are correct, and the function is brought into the function to find the analytic formula. In turn, verifying that x=2, y=-2 is wrong.

Put the generations in, find different ones and solve them.

1.Let a(0,a),b(b,0).

a=4,b^2+(k-1)b+4=0

1/2)*4*(-b)=6,b=-3

So a(0,4),b(-3,0);

2.-b^2+(k-1)b+4=0

9-3(k-1)+4=0

k=-2/3

y=-x^2-(5/3)x+4;

3.Let p(x,0).

pa=√(4^2+x^2)

ab=√(4^2+3^2)=5

pb=3+|x|

1)pa=ab

4^2+x^2)=5

x= 3 (negative value rounded).

p(3，0）;

2)pa=pb

4^2+x^2)=3+|x|

16+x^2=x^2+6|x|+9

x|=7/6

x=±7/6

p(7/6，0）,p(-7/6，0）;

3)pb=ab

3+|x|=5

x=±2p(2，0）,p(-2，0）;

So p(3,0),p(7 6,0),p(-7 6,0),p(2,0),p(-2,0) are sought.