A 2nd order function, when the independent variable X 0, the function value Y 2 and Y 0 are found an

Because the value of the function is 0 when x= -2 and x=1 2, the analytical formula can be set to y=a(x+2)(2x-1), and substituting x=0 and y= -1 into the -1=a(0+2)(0-1) to obtain a=1 2, so the analytical expression of the quadratic function is y=1 2*(x+2)(2x-1)= x 2+3 2*x-1.

A bidirectional function, when the independent variable x=0, the function value y=-1, when x=-2 and [1 2], y=0

Find the analytic formula of this quadratic function.

Let the function twice be y=a(x+2)(x-1 2) when x=0, the value of the function y=-1.

1=a(0+2)(0-1/2)

The analytic formula of the quadratic function of a=1: y=x 2+3 2*x-1 solution 2] function is symmetrical about (-2+1 2) 2=-3 4.

y=a(x+3 4) squared + c=0

x=0, y=-1, 9a/16+c=-1 (1)x=-2,y=0, 25/16a+c=0 (2)a=1c=-25/16

y = (x+3 5) squared - 25 16

The conditions are insufficient, and the specific function cannot be found.

We can use the two known points (-2, -1) and (2, 11) to determine the analytic formula of a function.

In general, the analytic expression of a primary function can be expressed as y = mx + b, where m is the slope and b is the intercept. We need to solve for the values of m and b.

First, we can determine the slope m:, using the slope formula

m = y2 - y1) x2 - x1) Next, we can select any known point (-2, -1) to substitute the slope m into the analytic equation and solve for the intercept b:

1 = 3 * 2) +b

1 = 6 + b

b = 1 + 6

b = 5 Therefore, based on the results obtained from the known points and slopes, the analytic expression of this primary function is:

y = 3x + 5

Let this 2nd order function be y=ax square + bx+c

Substitute x=0 and y=-1.

Substitute x=-2, y=0.

Substitute x=1 2 and y=0.

Three equations are derived! After the same three equations, we can find the unknown number abc, and then substitute abc into y=ax square + bx+c

You can find the analytic formula.

Let the function twice be y=a(x+2)(x-1 2) when x=0, the value of the function y=-1.

1=a(0+2)(0-1/2)

The analytic formula of the quadratic function a=1: y=x 2+3 2*x-1

Solution: Let y=a(x-x1)(x-x2).

Substitute x=0, y=-1, x1=-2, x2=1 2 into y=a(x-x1)(x-x2).

a(0+2)(0-1/2)=-1

a=1 substitute a=1, x1=-2, x2=1 2 into y=a(x-x1)(x-x2), get.

y=(x+2)(x-1/2)

The landlord tries this formula, it's simpler.

First, let the quadratic function f(x)=ax 2+bx+c and then f(0)=-1, we get c=-1, and when x=-2, x=1 2, the independent variable is substituted into the binary linear equation. 【4a-2b-1=0;1 4a+1 2b-1=0] to solve this system of equations. A=1, B=3 2, c=-1, and the quadratic function y=x 2+3 2x-1 is obtained.

Well, I don't know how to do math on the computer.

Let the analytic formula of the quadratic function be y=ax 2+bx+c (a is not equal to 0), and c=-1 from x=0, y=-1

by x=-2, y=0; x=1 2, y=0 gives 4a-2b-1=0

1/4 a +1/2 b-1=0

The solution yields a=1, b=3 2

So the analytic formula is y=x 2+3 2 x -1

Answer: 2: If the origin is passed, the constant term in the analytic formula of the quadratic function must be zero, so you can set y=ax 2+bx, and substitute two points to have.

a-b=-1, a+b=9, a=4, b=5, and the analytic formula y=4x 2+5x

3: The landlord said the topic clearly, did not understand, skipped;

4: According to the relationship between the root and the coefficient, x1+x2=-b 2a=-1 2+3 2=1, x1*x2=c a=-1 2*3 2=-3 4, the intersection point with the y-axis (0, -5), bring in to obtain c=-5, and then solve the above two equations to obtain a=20 3, b=-40 3, and the analytic formula is obtained.

5: Same as above, bring the three points in to get 9a-3b+c=2, a-b+c=-1, a+b+c=3, the simultaneous three-formula solution a=3 8, b=1, c=13 8, and the analytic formula comes out!

Question 1:

Let y=ax +bx+c

put x=0,y=-1; x=-2,y=0 ；x=1 2,y=0 is brought into the solution of the 3-element one-dimensional equation to obtain a=1 b=3 2 c=-1

So y=x +3 2 x -1

Question 2: Let y=ax +bc+c

put x=0, y=0; x=-1,y=-1 ；x=1,y=9 bring in a=4 b=5 c=0

So y=4x +5x

Question 3: y=ax +bx+c

put x=-1, y=-22; x=0,y=-8 ；x=2, y=8 gives a=-2 b=4 c=-8

So y=-2x +4x-8 opening downwards.

b 2a=1 so the axis of symmetry is x=1

b (4a-4ac)=-2 9 so the vertex is (1,-2 9).

Question 4: Let y=ax +bc+c

put x=-1 2,y=0 ; x=3/2,y=0 ；x=0,y=-5 bring in a=20 3 b=-20 3 c=-5

So y=20 3x -40 3x-5

Question 5: Let y=ax +bc+c

put x=-3, y=2; x=-1,y=-1 ；x=1,y=3 bring in a=7 8 b=2 c=1 8

So y=7 8x +2x+1 8

By the way, the test points of the 5 questions are the same, that is, the quadratic function is solved by 3 points, which is actually to solve the 3-element one-time equation. Quadratic functions are the focus of the junior high school entrance examination, so you have to study hard

2011-12-15 20:46 Wolong, China |Level 6 (-2,0) (1 2,0).

y=a(x+2)(x-1/2) (0.-1)-1=a*2*(-1 2) a=1 So y=(x+2)(x-1 2)=x 2+3 2x-1

This correct solution is also the easiest.

Let y=ax 2+b

Bring the numerical value in.

You get A and B

Your question is wrong, x=+-1 2.

In this case, y=-1 4 times x squared minus 1

Let its analytic formula be y=a(x-1 2)(x+2), and then bring x=0, y=-1 into the solution, and get a=1, so the analytic formula is y=(x-1 2)(x+2).