How can coordinates be substituted as quadratic functions

Generally, you need the coordinates of three points to find it, but you only give the coordinates of two points, so it is a bit troublesome to find, but you can also find it Here's how to find it.

Substituting two points into the equation both:

0 = 4a + 2b + c 1 formula.

6 = a-b + c 2 formula.

Multiply 2 by 2 to get 12=2a-2b+2c 3.

3 + 1.

12=2a+3c=(12-3c) 2 4.

Substituting 4 into 1 and 2 gives you a system of binary equations, which is very troublesome to find, so I won't ask for it, but you can try.

Actually, it's easy to find three points, so I'll add the coordinates of a point in the original problem (assuming that the parabola passes through the point and substituting it into the equation to get c=0.)

All that's left is to substitute the other two points into the equation and solve the same associature as Eq. 1 plus 2 by Eq. 2 to get a=2

The equation b=-4 is solved as y=2x 2-4x

Generally, the exam comes up with three coordinate points for you to find the parabolic equation.

Speaking of which, I think you should understand, don't understand how to add me, I'll give you a detailed explanation of learning progress.

Solution: The meaning of the point (2,0) is that when x=2, y=0 puts x=2, y=0 generation y=ax 2+bx+c, and gets 0=4a+2b+c, and the same as 6=a-b+c

Three coefficients to be determined, three conditions are required to find the specific number. Three letters, two conditions, two of which are to be represented by one other letter. The value of the function can be found analytically.

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".

There is one parameter in the result. You can substitute two points to get a system of ternary equations (there are only two equations), and you can eliminate one parameter first. This allows one parameter to represent the other two parameters.

For example: y=ax 2+bx+c(a≠0) past (1,0) and (2,1) two points, find f(x). f(x)=ax^2+(1-3a)x+2a-1(a≠0）.

This is generally given the coordinates of three points, and if it is not a special coordinate, it can only be a system of ternary equations.

After the coordinates are substituted, each formula a, b, c three unknowns, suppose the first formula is 3a+4b+5c=0, and then use b and c to represent a, that is: a=-(4a+5b) 3, and then bring this formula into the second and third equations of the equation, and then the second and third equations become a system of binary equations (you will solve the binary equations, right?). ), find b,c, and then substitute the values of b,c into a=-(4b+5c) 3, and get the three coefficients of a,b,c of the quadratic function.

The two coordinates correspond to the x y

However, two coordinates cannot be used to find a quadratic function, and three are needed.

The methods of the first two will work, just substitute them one by one.

The abscissa of the axis of symmetry is the midpoint of the segment between the two intersections.

x=(-4+2)/2=-1

The axis of symmetry of the parabola is x=-1

Let the analytic formula of the quadratic function be .

y＝ax^2+bx+c

Then, according to the three points passed, the coordinates of the three points are substituted.

You can get a system of equations composed of three equations about a, b, and c, and by solving this system of equations, you can get the values of a, b, and c, and you can find the quadratic function.

Hope it helps

According to the expression, the generation coordinates will do, find a few unknowns, and the expression will do.

First set ax2+bx+c=0 according to the standard formula of the quadratic function, and then bring in its coordinates for you, generally there will be three coordinates, or two coordinates with one can get the conditions about a, b, c relations, and finally the simultaneous equation to get a, b, c is good.

First, through the analytic formula of the origin, y=a x 2

Then substitute the point into the formula, (substituting a coordinate is actually enough) a=

The analytic formula is: y= x 2