# Advice 1: How to determine the vertex of a parabola

Parabola – one of the curves of the second order, it is constructed in accordance with a quadratic equation. Most importantly in the construction of this curve is to find the vertex of a parabola. This can be done in several ways. Instruction
1
To find the coordinates of the vertex of the parabola, use the following formula: x=-b/2A, where a is the coefficient in front of x squared, and b is the coefficient in front of X. Substitute your values and calculate its value. Then substitute the resulting value is x in the equation and calculate the y coordinate of the vertex. For example, if you are given the equation y=2x^2-4x+5, then find the abscissa in the following way: x=-(-4)/2*2=1. Substituting x=1 into the equation, calculate the value of y for the vertex of a parabola: =2*1^2-4*1+5=3. Thus, the vertex of the parabola has coordinates (1;3).
2
The value of the ordinates of the parabola can be found without preliminary calculation of the abscissa. Use the formula y=-b^2/4 ° C+S.
3
If you are familiar with the concept of the derivative, find the vertex of a parabola using derivatives, using the following property of any functions: the first derivative equal to zero indicates extreme points. Since the vertex of the parabola, regardless of the sent its branches upwards or downwards, is a point of extremum, calculate derivative for your function. In General it will be of the form f(x)=2ах+b. Paranaita it to zero and get the coordinates of the vertex of the parabolacorresponding to your function.
4
Try to find the vertex of a parabolausing such property as symmetry. To do this, find the point of intersection of the parabola with the axis ox, equating the function to zero (substituting y=0). Solving the quadratic equation, you will find x1 and x2. Since the parabola is symmetric with respect to the directrix passing through the vertex, those points will be equidistant from the abscissa of the vertex. To find it, divide the distance between points in half: x=(Іх1-х2І)/2.
5
If any of the coefficients is zero (except a), calculate the coordinates of the vertex of a parabola on a lightweight formula. For example, if b=0, the equation has the form y=Ah^2+C, the vertex will lie on the axis Oy and its coordinates will be zero (0;off). If not only the coefficient b=0 but C=0, the vertex of the parabola is at the origin, the point (0;0).

# Advice 2 : How to graph the parabola

A parabola is the graph of a quadratic function of the form y=A·x2+B·x+C. Before plotting it is necessary to undertake an analytical study of the function. Usually I draw a parabola in the Cartesian rectangular coordinate system which is represented by two perpendicular axes Ox and Oy. Instruction
1
The first item, write down the domain of the function D(y). The parabola is defined on the entire number line, if you do not set any additional conditions. This is usually indicated by entry D(y)=R, where R is the set of all real numbers.
2
Find the vertex of a parabola. Coordinate on the x-axis x0=-B/2A. Substitute x0 into the equation of a parabola and count the coordinate of the vertex on the axis of ordinates Oy. So, the second item should be recorded: (x0;y0) is the coordinates of the vertex of the parabola. Of course, instead of x0 and y0, you must have a specific number. Mark this point on the drawing.
3
Comparing senior with A coefficient of x2 zero, draw a conclusion about the direction of the branches of the parabola. If A>0, the branches of the parabola are directed upwards. A negative value indicates A branch of a parabola facing down.
4
Now you can find the set of values of the function E(y). If the branches are directed upwards, the function y takes all values above y0. When the branches down function takes values below y0. For the first case write: E(y)=[y0,+∞) and the second as E(y)=(-∞;y0]. Square bracket says that the last number is included in the interval.
5
Write the equation for the axis of symmetry of the parabola. It will be of the form: x=x0 and passing through the vertex. Draw this axis perpendicular to the Ox axis.
6
Find the "zeros" of the function. These points will intersect the coordinate axis. Paranaita x is zero and count y for this case. Then find for what values of the argument function y will go to zero. To do this, solve the quadratic equation A·x2+B·x+C=0. Mark the points on the graph.
7
Find additional points to construct the parabola. Will be issued in the form of a table. The first line write the argument x, the second a function of y. It is better to choose such numbers for which x and y are integers because fractional numbers to portray uncomfortable. Note plotting points on the chart.