How do I find the analytic expression of a function using the special value method?

The general form of a quadratic function is: y=ax2+bx+c

Any three points are known).

Vertex formula: y=a(x+d)2+h

Known vertices and any vertices other than vertices).

Some versions of the textbook are also noted.

The principle is the same. For example, if the vertices of a quadratic function image (-2,1) are known and (1,0) are passed, the quadratic function is found analytically.

Solution: Let y=a(x+2)2+1

Note: y=a(x-d)2+h where d is the abscissa of the vertex and h is the ordinate of the vertex.

Because. Quadratic function image over the point (1,0).

Therefore. a*3 squared + 1 = 0

The solution is a=-1 9

Therefore, the analytical formula for the quadratic function is .

y=-1/9(x+2)2+1

This problem is a sample problem, so it will not be further simplified into a general form) Two roots: the two intersections of the known function image and the x-axis and another point.

First there must be an intersection point (b2-4ac>0).

y=a(x-x1)(x-x2)

where x1 and x2 are the intersection points between the image and the x-axis.

And it's two ax2+bx+c=0.

If the general form of the quadratic function and one intersection point with the x-axis are known, another intersection point can be found.

Take advantage of the relationship between roots and coefficients.

For example, if y=x2+4x+3 intersects with the x-axis at an intersection point is (-1,0), find the coordinates of its intersection with the x-axis.

Solution: From the relationship between the root and the coefficient:

x1+x2=-b/a=-4

then x2=-4-x1=-4-(-1)=-3

So the coordinates of the other intersection point with the x-axis are (-3,0).

In addition, Y=AX2+BX+C can be translated to the right by 2 units.

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

Translate 2 units downwards to get: y=a(x-2)2+b(x-2)+c-2 Remember: "Add left and subtract right."

plus up and down minus".

The analytical formula of the function with special values is generally used to find abstract functions, which depends on the specific topic, but there are also general ways to take the following points:

1) The special value is generally taken as 0, 1, -1, some numbers, generally around 0, because it is easier to calculate, and it is not far from what the question asks.

2) If the question tells you that a function f(x) is an odd function and its definition domain contains an origin, then there is f(0)=0.

3) Another way to find the analytic formula of an abstract function is the system of equations

Example: Knowing that f(x) satisfies f(x)+2f(1 x)=2x+1, find the analytic formula of f(x).

Solution: Since the original function definition domain is x is not equal to 0, it can be obtained by replacing all x in the original equation with 1 x.

f（1/x）+2f（x）=2*1/x+1

Then the two equations can be solved by concentrating the f(x).

I don't have any long remarks about the analytic formula of quadratic functions, but a concise and solid foundation can be conducive to improving Ah.

The general form of a quadratic function: y=ax2+bx+c (any three points are known).

Vertex formula: y=a(x+d)2+h (known vertices and any vertex other than vertices) Some versions of the textbook also note the same principle.

For example, if the vertices of a quadratic function image (-2,1) are known and (1,0) are passed, the quadratic function is found analytically.

Let the letter calendar y=a(x+2)2+1 Note: y=a(x-d)2+h, where d is the abscissa of the vertex and h is the ordinate of the vertex.

Since the quadratic function image passes through the point (1,0).

Therefore, the square of a*3 +1=0 solves a=-1 9

Therefore, the analytical formula of the quadratic function is y=-1 9(x+2)2+1

This question is a sample question, so it will not be further reduced to a general form).

Two-root formula: the intersection of the known function image and the x-axis and the other point must first have an intersection (b2-4ac>0).

y=a(x-x1)(x-x2) where the scroll x1, x2 is the intersection of the image and the x-axis, and is the two ax2+bx+c=0.

If we know the general form of the quadratic function and one intersection point with the x-axis, we can find another intersection using the relationship between the root and the coefficient.

For example, if y=x2+4x+3 intersects with the x-axis at an intersection point is (-1,0), find the coordinates of its intersection with the x-axis.

From the relationship between the root and the coefficient:

x1+x2=-b a=-4 then x2=-4-x1=-4-(-1)=-3

So the coordinates of the other intersection point with the x-axis are (-3,0).

In addition, Y=AX2+BX+C can be translated to the right by 2 units.

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

Translate 2 units downwards to obtain: y=a(x-2)2+b(x-2)+c-2

Remember: "add left and subtract right plus add up and subtract".

Finding a function analytic with a special value is generally used to find an abstract function.

Yes, this depends on the specific topic, but there are also general ways to take the following points:

1) The special value is generally taken 0,1,-1, some numbers of imaginary and wanton, generally taken around 0, slippery because it is easier to calculate, and it is not far from the problem.

2) If the title tells you that a function f(x) is odd.

and its definition field contains the origin, then there is f(0) reputation key = 0

3) Another way to find the analytic formula of an abstract function is the system of equations

Example: Knowing that f(x) satisfies f(x)+2f(1 x)=2x+1, find the analytic formula of f(x).

Solution: Since the original function defines the domain.

Since x is not equal to 0, replace all x in the original equation with 1 x, and you can get it.

f（1/x）+2f（x）=2*1/x+1

Then the two equations can be solved to the satisfaction of f(x).

The general form of a quadratic function is: y=ax2+bx+c

Any three points are known).

Vertex formula: y=a(x+d)2+h

Known vertices and any vertices other than vertices).

Some versions of the textbook are also noted.

The principle is the same. For example, if the vertices of a quadratic function image (-2,1) are known and (1,0) are passed, the quadratic function is found analytically.

Solution: Let y=a(x+2)2+1

Note: y=a(x-d)2+h, where d is the abscissa of the vertex, and h is the ordinate of the vertex due to the quadratic function image passing over the point (1,0).

So the square of a*3 + 1 = 0

The solution is a=-1 9

Therefore, the analytical formula for the quadratic function is .

y=-1/9(x+2)2+1

This problem is a sample problem, so it will not be further simplified into a general form) Two roots: the intersection of the known function image and the x-axis and the other point must first have an intersection (b2-4ac>0).

y=a(x-x1)(x-x2)

where x1 and x2 are the intersection points between the image and the x-axis.

And it's two ax2+bx+c=0.

If you know the general form of the quadratic function and one intersection point with the x-axis, you can find another intersection using the relation of the root to the coefficient.

For example, if y=x2+4x+3 intersects with the x-axis at an intersection point is (-1,0), find the coordinates of its intersection with the x-axis.

Solution: From the relationship between the root and the coefficient:

x1+x2=-b/a=-4

then x2=-4-x1=-4-(-1)=-3

So the coordinates of the other intersection point with the x-axis are (-3,0).

In addition, by translating y=ax2+bx+c to the right by 2 units, we get y=a(x-2)2+b(x-2)+c

Translate 2 units downwards to get: y=a(x-2)2+b(x-2)+c-2 Remember: "Add left and subtract right."

plus up and down minus".

The analytical formula of the function with special values is generally used to find abstract functions, which depends on the specific topic, but there are also general ways to take the following points: (1) Some numbers of 0, 1, -1, and special values are generally taken around 0, because it is easier to calculate, and it is not far from what the problem is found. (2) If the question tells you that a well-debated function f(x) is an odd function, and its definition domain contains the buried origin, then there is f(0)=0.

3) Another way to find the analytic formula of an abstract function is the system of equation method: Example: Knowing that f(x) satisfies f(x)+2f(1 x)=2x+1, find the analytical solution of f(x).

Since the original function definition domain is not equal to x, replacing all x in the original equation with 1 x can obtain f(1 x)+2f(x)=2*1 x+1 and then the two square spikes can be solved to f(x) satisfaction.