Junior 3 Math Problem How to make a general formula of a quadratic function into a vertex formula?

General formula: y=ax 0 5+bx+c(a,b,c is constant, a≠0)vertex formula: y=a(x-h) 0 5+k [the vertex of the parabola p(h,k)]intersection formula:

y=a(x-x1)(x-x2) [only for parabolas with intersections a(x1,0) and b(x2,0) with the x-axis].

Note: In the three forms of mutual transformation, there are the following relationships:

h=-b 2a k=(4ac-b 0 5) 4a x1,x2=[-b (b 0 5-4ac)] 2a fixed-point applications are often seen in multiple-choice questions and fill-in-the-blank questions, and general applications are common in large questions.

Another less common is the conversion of intersection to vertex.

y=a(x-x1)(x-x2) y=a[x-(x1+x2)/2]�0�5-[a(x1-x2)�0�5] /4

First, 5, y=5(x 2+4 5 x+1 5), add 2 (root number 5) in parentheses, and subtract 2 (root number 5), then y=5(x 2+4 5 x+2 (root number 5)-2 (root number 5)+1 5)=5(x+2 (root number 5)) 2+1-2*(root number 5)).

The general expression of quadratic function is as follows: y=a(x+b 2a)+(4ac-b) 4a, and the basic representation of quadratic function is y=ax Zheng Li + bx+c(a≠0). The quadratic function must be quadratic at its highest order, and the image of the quadratic function is a lodging plexus guessing line with the axis of symmetry parallel to or coincident with the y-axis.

The expression of a quadratic function is y=ax +bx+c (and a≠0) is unformed, and it is defined as a quadratic polynomial (or monomial). If the value of y is equal to zero, a quadratic equation is obtained. The solution of this equation is called the root of the equation or the zero point of the function.

There are two ways to generalize the quadratic function into a vertex formula, the matching method or the formula method, 1. The matching method example, <>

2. The vertex formula can be obtained through the formula - the formula is formed:

Quadratic functions are generally formulated to vertex method analysis:

Matching method:

y=ax+bx+c=a(x+bx a)+c=a(x+bx a+b 4a-b 4a)+c=a(x+b 2a)-b 4a+c=a(x+b 2a)+(4ac-b) 4a.

Basic Qi Modulus Definition of Quadratic Functions:

"variable" is different from "independent variable", and it cannot be said that "a quadratic function is a polynomial function in which the highest degree of a variable is quadratic". An "unknown" is just a number (the exact value is unknown, but only one value is taken), and a "variable" can be arbitrarily taken within the range of real numbers.

The concept of "unknowns" is applied to equations (in functional equations and differential equations, it is an unknown function, but whether it is an unknown number or an unknown function, it generally represents a number or a number of worlds - there are also special cases), but the letters in the function represent variables, and the meaning is different. The difference between the two can also be seen from the definition of the function, just as the relationship between the function is not equal to the function.

The property of the quadratic function: the balance of the limbs

The image of a quadratic function is a parabola, but a parabola is not necessarily a quadratic function. The opening of the parabola up or down is a quadratic function. The parabola of the balance limb is an axisymmetric figure. The axis of symmetry is a straight line. The only intersection point between the axis of symmetry and the parabola is the vertex p of the parabola.

In particular, when b = 0, the axis of symmetry of the parabola is the y axis (i.e., the straight line x = 0). The quadratic term coefficient a determines the direction and magnitude of the opening of the parabola. When a>0, the parabola opening is upward; When a<0, the parabolic opening is downward.

a|The larger it is, the smaller the opening of the parabola;a|The smaller it is, the larger the opening of the parabola.

Summary:

It is one of the basic contents of learning quadratic functions to modulate the general formula of quadratic functions into vertex formulas. As long as you master the mathematical methods such as the matching method, solving the quadratic equation, and finding the square, you can easily deform the general formula to obtain the vertex formula. At the same time, through more practice and example exercises, you can also master this knowledge point more proficiently.

y=ax +bx+c, which is converted into a vertex formula: y=a(x+b 2a) +4ac-b ) 4a The recipe process is as follows: y=ax +bx+c=a(x +bx a)+c=a(x +bx a+b 4a -b 4a)+c=a(x+b 2a) -b 4a+c=a(x+b 2a) +4ac-b ) 4a

On the image of a quadratic function:

Vertex formula: y=a(x-h) +k, the vertex coordinates of the parabola p(h,k): for the general quadratic function y=ax 2+bx+c its vertex coordinates are (-b 2a, (4ac-b) 4a).

If 2 of the 3 intersection points are the intersection points of the quadratic function and the x-axis, then the analytical formula of the quadratic function can be set as: y=a(x-x1)(x-x2) (x1,x2 is the coordinates of the 2 intersection points of the quadratic function and the x-axis), and the analytical formula of the quadratic function can be found according to the other point, if you know that the vertex coordinates are (h,k), then you can set the analytical formula of y=a(x-h)2+k, and the quadratic function analytic formula can be found according to another point.

The purpose of vertex is to more intuitively obtain the symmetry axis and vertex coordinates of the parabola.

The axis of symmetry of y=a(x-h) 2+k is x-h=0 and the vertices are (h, k).

How to convert y=ax 2+bx+c to vertex y=a(x-h) 2+k.

y=ax^2+bx+c

a（x^2+b/ax+c/a)

a〔〔x+b/(2a)〕〕2+(4ac-b^2)/4a

That is, the axis of symmetry of y=ax 2+bx+c is x=-b (2a), and the vertex coordinates are -b (2a ac-b 2) 4a

The general formula of the quadratic function is y=ax +bx+c, and the vertex formula is y=a(x+b 2a) +4ac-b ) 4a.

The basic representation of a quadratic function is y=ax +bx+c(a≠0). The quadratic function must be quadratic at its highest order, and the image of the quadratic function is a parabola with the axis of symmetry parallel to or coincident with the y-axis.

The quadratic function expression is y=ax +bx+c (and a≠0) and is defined as a quadratic polynomial (or monomial).

If the value of y is equal to zero, a quadratic equation is obtained. The solution of this equation is called the root of the equation or the zero point of the function.

y=ax +bx+c, which is converted into a vertex formula: y=a(x+b 2a) +4ac-b ) 4a The recipe process is as follows: y=ax +bx+c=a(x +bx a)+c=a(x +bx a+b 4a -b 4a)+c=a(x+b 2a) -b 4a+c=a(x+b 2a) +4ac-b ) 4a

On the image of a quadratic function:

Vertex formula: y=a(x-h) +k, the vertex coordinates of the parabola p(h,k): for the general quadratic function y=ax 2+bx+c its vertex coordinates are (-b 2a, (4ac-b) 4a).

If 2 of the 3 intersection points are the intersection points of the quadratic function and the x-axis, then the analytical formula of the quadratic function can be set as: y=a(x-x1)(x-x2) (x1,x2 is the coordinates of the 2 intersection points of the quadratic function and the x-axis), and the analytical formula of the quadratic function can be found according to the other point, if you know that the vertex coordinates are (h,k), then you can set the analytical formula of y=a(x-h)2+k, and the quadratic function analytic formula can be found according to another point.

Quadratic functions are generally styled into vertex methods for teaching.

Recipe and factorization are sufficient.

Draw a picture, or make a draft, and see if it's possible to become a vertex, thank you.

Vertex formula: y=a(x-h)2+k (a≠0, k is constant)Vertex coordinates: [-b 2a,(4ac-b2) 4a] Vertex coordinates are used to represent the position of parabolic vertices of quadratic functions.

The expression is y=ax 2+bx+c (a is not equal to 0) Example: y=2x 2+4x+6

Extract the common factor 2 first: y=2(x 2+2x+3) and then turn it into a perfect square formula (the excess in parentheses is kicked out): y=2(x+1) 2+4 (don't forget to multiply it by 2 when kicking it out).

This is a vertex formula, so its coordinates are (-1,4).

The intersection formula of the quadratic function is.

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

We know that the vertex formula of a quadratic function is y=y=a(x+b 2a) +4ac-b ) 4a, so in order to convert the intersection formula to the vertex formula, we must first find the general answer supplement.

Finally, it was simplified.

4ac-b ) 4a=-[a(x1-x2) ] 4 so the vertex formula is.

y=a[x-(x1+x2)/2]²-a(x1-x2)²]/4(a≠0)

For example: (x+5)(x-9)=0

Simplify: x 2-4x-45 = 0

Bring in the formula.

y=1×[x-(-5+9)/2]²-1×(-5-9)²]/4=(x-2)²-49