Parabola – one of curves of the second order, its point are constructed according to a quadratic equation. The main thing is to find parabola top in creation of this curve. It can be done in several ways.

## Instruction

1. To find parabola top coordinates, use the following formula: x =-b/2a, where and – coefficient before x in a square, and b – coefficient before x. Substitute your values and calculate its value. Then substitute the received value instead of x in the equation and count top ordinate. For example, if you were given the equation at =2х^2-4Õ +5, then find an abscissa as follows: x =-(-4)/2*2=1. Having substituted x =1 in the equation, calculate value at for parabola top: at =2*1^2-4*1+5=3. Thus, the top of a parabola has coordinates (1;3).

2. The value of ordinate of a parabola can be found also without predesign of an abscissa. For this purpose use a formula at =-b^2/4ас + page.

3. If you are familiar with a concept of a derivative, find parabola top by means of derivatives, having used the following property of any function: the first derivative of function equated to zero indicates extremum points. As the parabola top irrespective of, its branches up or down are directed, is an extremum point, calculate a derivative for your function. In a general view it will have an appearance of f (x) =2akh +b. Equate it to zero and receive coordinates of top of the parabola corresponding to your function.

4. Try to find parabola top, having used its such property as symmetry. For this purpose find parabola points of intersection with an axis oh, having equated function to zero (having substituted at =0). Having solved a quadratic equation, you will find h1 and h2. As the parabola is symmetric concerning the head mistress passing through top, these points will be equidistanted from a top abscissa. That to find it, we will halve distance between points: x = (Ikh1-h2i)/2.

5. If any of coefficients it is equal to zero (except a), calculate parabola top coordinates by the facilitated formulas. For example, if b=0, that is the equation has an appearance at = ах^2+ with, then the top will lie on an axis ou and its coordinates will be equal (0; с). If not only b=0 coefficient, but also with =0, then the top of a parabola is in the beginning of coordinates, a point (0;0).