How is the vertex coordinate formula for a parabola found?????????

1 understanding: There is a parabola y=ax 2+bx+c, if it intersects the x-axis, then the x-coordinate of the intersection is the solution of the equation ax 2+bx+c=0 when y=0, x1=[-b+(b 2-4ac) 1 2] 2a, x2=[-b-(b 2-4ac) 1 2] 2a (if (b 2-4ac) 1 2 is zero, then x1=x2=-b 2a), then (x1+x2) 2=-b 2a

It's the axis of symmetry. There are two understandings, in fact, this understanding covers the above situation, that is, the x value of the parabolic axis of symmetry will make the parabola y=ax 2+bx+c have an extreme value (maximum or minimum), but y=ax 2+bx+c is variable to form y=a(x+b 2a) 2-(4ac-b 2) 4a

However, since (x+b 2a) 2 can only be greater than or equal to 0, the parabola with the opening of a>0 facing upwards has a minimum value of (4ac-b 2) 4a only when (x+b 2a) 2=0, and conversely, in the parabola with the opening of a<0 pointing downward, y has a maximum value of (4ac-b 2) 4a only when (x+b 2a) 2=0. In any case, y will only take the extreme value when (x+b 2a) 2=0, so that the x value of (x+b 2a) 2=0 is the position of the axis of symmetry, then x=-b 2a

Let it intersect with the x-axis.

The coordinates are. x1,x2

Then the abscissa of the intersection point is half of the sum of the two coordinates.

Then use the relation of the root to the coefficient.

Represent xi+x2.

The 2 at the end is the origin of the abscissa of the vertex formula.

And then. Bring in the analytical formula to find the ordinate.

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

For vertex poses.

y=a(x-h)^2+k

H is. b/2a

k is (4ac-b 2) 4a

This is the most basic knowledge!! Eldest brother!!!

Vertex formula: y=a(x-h) +k Vertex coordinates of the parabola p(h,k): For the quadratic function y=ax +bx+c(a≠0) its vertex coordinates are [-b 2a,(4ac-b) 4a].

Knowing the vertices of the parabola, you can find the analytic equation by simply giving the coordinates of another point.

For example, the vertices of a parabola are known to be (-3,2) and (.

The analytic formula can be set to y=a(x+3) +2. Then substitute x=2 and y=1.

Find a=-1 25, i.e., y=-1, 25(x+3) +2.

y=ax^2+bx+c:（-b/2a,4ac-b^2/4a)

Or the quadratic function is vertice: y=a(x-h) 2+k

The vertex coordinates are (h,k).

The vertex coordinates formula for y=ax +bx+c(a≠0) is (-b 2a, (4ac-b) 4a).

The vertex coordinates of y=ax +bx are.

b/2a，-b²/4a)

Parabolic formula:

General formula: y=ax2+bx+c (a, b, c are constants, a≠0) vertex formula: y=a(x-h)2+k (a, h, k are constants, a≠0) intersection formula (two-root formula):

y=a(x-x1)(x-x2) (a≠0) where is the coordinates of the intersection of the parabola y=ax2+bx+c (a, b, c are constants, a≠0) and the x-axis, i.e., the two real roots of the equation ax2+bx+c=0.

General formula: y=ax 2+bx+c (a,b,c is constant, a≠0) vertex formula: y=a(x-h) 2+k

The vertices of the parabola p(h,k)].

For the quadratic function y=ax 2+bx+c

Its vertex coordinates are (-b 2a,(4ac-b 2) 4a) intersection formula: y=a(x-x) (x-x) is limited to parabolas with intersection points a(x,0) and b(x,0) with the x-axis].

where x1,2 = -b b 2 4ac

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

H=-b 2A= (x +x) 2 k=(4ac-b 2) 4a intersection with x-axis: x, x =(-b b 2-4ac) 2a

The general formula for vertex coordinates of the unary quadratic function y=ax +bx+c(a≠0) is (-b 2a, (4ac-b) 4a).

So take c=0, and the vertex coordinates of y=ax +bx(a≠0) are (-b 2a, -b 4a).

To require the vertex coordinates of a parabola, the following formula can be used: for the general form of the parabolic equation y = ax 2 + bx + c, where a, b, c are constants, and the x coordinates of the vertices can be obtained by the formula x = b 2a).

There are also several ways to solve for the vertex coordinates of a parabola

Method 1: Use a perfectly squared formula.

For the parabolic equation of the general form y = ax 2 + bx + c, where a, b, c are constants, the x-coordinates of the vertices can be found quietly by the formula x = b 2a). Then, the obtained x-coordinate is substituted into the parabolic equation to calculate the corresponding y-coordinate.

For example, for the parabolic equation y = 2x 2 + 4x + 1, first calculate the x coordinates: x = b 2a) =4 2*2) =1 and then substitute x = 1 into the parabolic equation to calculate the y coordinates: y = 2*(-1) 2 + 4*(-1) +1 = 2 + 4) +1 = 1 So, the vertex coordinates of the parabola are (-1, -1).

Method 2: Complete the square.

For the parabolic equation of the general form y = ax 2 + bx + c, it can be written as the standard form y = a(x - h) 2 + k, where (h, k) are the vertex coordinates. First, the parabolic equation is squared, i.e., the coefficients of the x 2 and x terms are moved to one side of the equation respectively, giving y - c = a (x 2 + bx a). Then, divide the coefficient of the x 2 term by a and square half of the coefficient of the x term to get y - c = a (x 2 + bx a + b 2a) 2).

Then square the contents in the right parentheses to get y - c = a(x + b 2a) 2 + b 2 - 4ac) 4a. Finally, move the constant term on the right to the side of the equation and get y = a(x + b 2a) 2 + b 2 - 4ac) 4a + c. From this standard form, you can read the vertex coordinates directly as (-b 2a, (b 2 - 4ac) 4a + c).

For example, for the parabolic equation y = 2x 2 + 4x + 1, the coordinates of the vertices can be obtained as (-4 (2*2), 4 2 - 4*2*1) (4*2) +1) =1, -1 according to the standard formula. So, the vertex coordinates of the parabola are (-1, -1).

These are common ways to solve parabolic vertex coordinates, and depending on the situation, you can choose the most suitable method for the calculation.

Let y=ax 2+bx+c

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

Therefore, the vertex coordinate number is imaginary x=-b 2a

When a>0, a(x+b 2a) 2 brigade digging 0 , y min: (c-b 2 4a) when a

Question 1: How to find the vertex of a general parabola? Vertex Formula:

y=a(x-h)2+k The vertices of the parabola p(h,k) [At the same time, the line x=h is the axis of symmetry of this quadratic function] Verex coordinates: For the quadratic function y=ax2+bx+c(a≠0), its vertex coordinates are [-b 2a,(4ac-b2) 4a].

Problem 2: How to find y=ax2+bx+c(a≠0) for mathematical parabola vertex coordinates formula

a[x2+(b/a)x+(b/2a)2]+c-(b2/4a)

a(x+b/2a)2+(4ac-b2)/4a

The vertex coordinates are (-b 2a, (4ac-b2) 4a).

Problem 3: Parabolic vertex coordinate formula Parabolic vertex coordinate formula:

When h>0, the image of y=a(x-h)2 can be changed from the parabola y=ax2; When h0, k>0, move the parabola y=ax2 parallel to the right by moving h units, and then move k units upwards to get the image of y=a(x-h)2+k;

When h>0,k0, move the parabola parallel to the left |h|units, and then move k units upwards to get the image of y=a(x-h)2+k;

When h0, the opening is pointing upwards when a