Instruction

1

This problem can be solved analytically, therefore, we can not draw graphs of

**straight**and*parabola*. This often gives an advantage in sample solution, as the task can be given such functions, that it is easier and faster to draw.2

According to the textbooks on algebra parabola set function of the form f(x)=ax^2+bx+c, where a,b,c are real numbers, while the coefficient a is different it is zero. The function g(x)=kx+h, where k h is a real number, defines a line in the plane.

3

The point

**of intersection of****straight**and*parabola*is a common point of both curves, so it functions take the same value, that is, f(x)=g(x). This statement allows to write the equation: ax^2+bx+c=kx+h, which will give the opportunity to find many points**of intersection**.4

In the equation ax^2+bx+c=kx+h, you must move all terms to the left side and lead like this: ax^2+(b-k)x+c-h=0. It now remains to solve the obtained quadratic equation.

5

All found the "x" is not the answer to the problem, as a point in the plane is characterized by two real numbers (x,y). To complete the solution, it is necessary to calculate the corresponding "allocated prizes as stated". To do this, substitute the "x" in either the function f(x) or the function g(x), as for the point

**of intersection**is true: y=f(x)=g(x). Then you will find all common points*of the parabola*and**straight**.6

To consolidate the material it is very important to consider the solution to the example. Let the parabola is given by f(x)=x^2-3x+3 and the line – g(x)=2x-3. Write down the equation f(x)=g(x), that is x^2-3x+3=2x-3. Transferring all terms to the left side, and bringing a similar, we get: x^2-5x+6=0. The roots of this quadratic equation: x1=2, x2=3. Now find the corresponding "allocated prizes as stated": y1=g(x1)=1, y2=g(x2)=3. Thus, all points

**of intersection**: (2,1) and (3,3).# Advice 2: How to write the equation of a parabola

**The equation**

**of a parabola**is a quadratic function. There are several options for setting this equation. It all depends on what options are provided in the problem statement.

Instruction

1

Parabola is a curve that resembles the arc is the graph of the exponential function. Regardless of what features of the parabola, this function is even. Even called such a function which for all values of the argument out of scope when you change the sign of the argument value is not changed:f(-x)=f(x)Starting with the most simple functions: y=x^2. Of its kind it can be concluded that it increases both positive and negative values of the argument x. The point at which x=0, and thus y =0 is a minimum point of the function.

2

Below are all the basic options for constructing this function and its equation. As the first example below considers the function of the form:f(x)=x^2+a, where a is an integer of cyclods in order to graph this function, you must move the graph of the function f(x) on a units. An example is the function y=x^2+3, where along the y axis, move the function up by two units. If given a function with the opposite sign, for example y=x^2-3, then its graph move down on the y-axis.

3

Another type of function which can be set parabola - f(x)=(x +a)^2. In such cases, the timing, on the contrary, moves along the x-axis (x-axis) on a units. For example, consider the function: y=(x +4)^2 and y=(x-4)^2. In the first case, where there is a function with a plus sign, shift the graph along the x-axis to the left, and in the second case to the right. All these cases shown in the figure.

4

There are also parabolic dependence of the form y=x^4. In such cases, x=const, and y increases dramatically. However, this applies only to even functions.Graphs

**of the parabola**and are often present in physical problems, for example, the flight of the body describes a line that is similar to a parabola. It is also a kind**of a parabola**is a longitudinal section of a reflector of a headlamp, a flashlight. Unlike sine waves, this schedule is non-periodic and increasing.