In the equation y ax bx c

1) is a system of ternary equations:

When x=-1, y=0;Substituting y=ax +bx+c, we get 0=a(-1) +b(-1)+c, that is, a-b+c=0.

When x=1, y=4;Substituting y=ax +bx+c, we get 4=a(1) +b(1)+c, that is, a+b+c=4.

When x=2, y=3;Substituting y=ax +bx+c, we get 3=a(2) +b(2)+c, that is, 4a+2b+c=3.

The solution yields a=-1, b=2, c=3.

2) The substitution method of the first year of junior high school can solve the equation. Two unknowns are used in the same generation. I'm going to get off work, so I'll come back next time I have time.

This is a system of ternary equations, and the two phases are solved to get the binary.

1）{a-b+c=0 1

a+b+c=4 2

a+b+c=3 3

2-1 gets 2b=4 gets b=2;

3-1 to get 3a+3b=3 to get a=-1;

Substituting the value of a b into 1 gives c equals 3

2）｛x-y=-1 1

2x+y-z=3 2

3x-2y+z=4 3

3-2 gives 5x-y=7 4

1 and 4 are solved to x=2 y=3, and the solution z, which is substituted for 2, is equal to 4.

Bringing in xy has a-b+c=0 a+b+c=4 4a+2b+c=3 ternary three formulas The first + second has a+c=2

2*1st + 3rd has 6a+3c=3

Then there is a=-1 c=3 and b=2 can be obtained

The first thing that comes to mind when doing this kind of problem is to eliminate one of the unknowns, which can be added up by equations. You can simulate the above method and calculate the following... Come on, what you learn is your own. Give you the answer to the second question x=2

y=3 z=4

x=-1, y=0 substituting the equation to get a-b+c=0(1); When substituting x=1, y=4, we get a+b+c=4(2); When substituting x=2, y=3 gives 4a+2b+c=3(3). The equations established by (1), (2), and (3) are solvable.

1) Column equations: a-b+c=0

a+b+c=4

4a+2b+c=3

This gives us the following results: a=-1 b=2 c=3

2)x=2 y=3 z=4

1>.When x=1, a+b+c=4; <2>x=-1,a-b+c=0;<3>x=2,4a+2b+c=3

1>, <2>, <3> simultaneous equations yield: b=2 a=-1 c=3

bring in x=-1 y=0 0=a-b+c;

bring in x=1 y=4 4=a+b+c;

bring in x=2 y=3 3=4a+2b+c;

The solution yields a=-1;

b=2；c=3；

One... Substituting three sets of numbers yields abc as -1, 2, 3

Two... xyz is 2,3,4

Substituting the known into the equation to get.

a-b+c=0（1）

a+b+c=4（2）

4a+2b+c=3（3）

Synthesis (1), (2), and (3) solution.

a=-1，b=2，c=3

The content of elementary and junior high school is too simple.

The halo first substituting a numerical value and building a system of equations similar to 2 can be solved.

The above urban 3 yuan equation system is very simple, should have been studied in high school, and should be taught in person.

Firstly, the values of the above three sets of x and y are brought into the solution to obtain a 28, and then the analytical formula is brought into x for evaluation.

From the known system of equations:

2=a-b+c

8=4a+2b+c

158=25a+5b+c

Solution: a=8, b=-6, c=-12, y=8x 2-6x-12

When x=-2, y=8 4-6 (-2)-12

(1) Substituting x,y into the equation (2) y=-16x -6x+12

Then 2=a-b+c when x=-2.

8=4a+2b+c y=-16*（-2）²—6*（-2）+12

58=25a+5b+c =-16*4+12+12

Solution = -64+24

a=-16 =-40

b=-6c=12

a-b+c=0②

4a+2b+c=3②

25a+5b+c=-60③

3a+3b=3, i.e., a+b=3

21a+3b=-63, i.e., 7a+b=-21, and finally a=-4, b=7, c=11

3a+3b=3, then 3(a+b)=3, a+b=1

Upstairs is wrong!

I figured out b = 7, but nothing else is correct.

Substituting the value of x and the value of y gives us three equations about a, b, and c.

Couple the three more equations and solve the equations.

①0=a-b+c

3=4a+2b+c

60=25a+5b+c

minus to get 1=a+b, i.e. a=1-b

Multiply by 2 and add to give 1=2a+c, c=1-2a=1-2(1-b)=2b-1

Substituting , we get 60 = 25 (1-b) + 5b + (2b -1) 60 = 24-18b

b=-2, a=1-(-2)=3, c=2 (-2)-1=-5, so a=3, b=-2, c=-5

In the equation y=ax -bx+c, when x=-1, y=4;

a*(-1)²-b*(-1)+c=4

a+b+c=4 (1)

When x=2, y=4;

a*2²-b*2+c=4

4a-2b+c=4 (2)

When x=1, y=2

a*1²-b*1+c=2

a-b+c=2 (3)

Obtained from (1)-(3).

2b=2b=1a=1

c=2y=ax²-bx+c

y=x²-x+2

When x=-2, y=(-2)-2)+2=8

Solution 1 consists of y=4 when x=-1When x=2, y=4;When x=1, y=2 gives a+b+c=4....4a-2b+c=4...a-b+c=2...

-2b = -2 from -

i.e. b=1 is obtained by and .

a+c=34a+c=6

The solution yields a=1 and c=2

i.e. a=1, b=1, c=2

2 is known by 1. y=ax²-bx+c

x²-x+2

When x=-2.

y=（-2）²-2）+2=8

In the equation y=ax -bx+c, when x=-1, y=4;When x=2, y=4;When x=1, y=2(1) Find the values of a, b, c, and more. (2) When x=-2, find the value of y.

Solution: (1) Substituting the coordinates of the three points into the equation obtains:

a+b+c=4...1)；4a-2b+c=4...2)；a-b+c=2...3)

1)-(3) 2b=2, so b=1;Substituting formula (1) yields a+c=3, i.e., c=3-a; Substituting equation (2) gives 4a-2+3-a=3a+1=4, so a=1;

c=4-a-b=4-1-1=2；i.e. a=1, b=1, c=2y=x²-x+2；

2)。 y(-2)=(-2)²-2)+2=4+2+2=8.That is, when x = -2, y = 8

x=-1;y=a-b+c=5(1)

x=1;y=a+b+c=-1(2)

x=2;y=4a+2b+c=-1(3)

2)-(1) De:

2b=-6;

b=-3;So a+c=2; (4)

4a+c=5(5)

5)-(4) De:

3a=3;a=1;

Bring in (4);

c=2-1=1;

a=1;b=-3;c=1;

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Substituting x = 1 and y = 5 yields:

a－b＋c=5 --

Substituting x = 1 and y = 1 yields:

a b c = 1 -- substituting x = 2 and y = 1 to get :

4a 2b c= 1 --solution , get:

a=1、b=－3、c=1