You will need

- paper;
- - handle.

Instruction

1

From school course of mathematics the students become aware that the number of possible points

**of intersection of**two**graphs**depends on the function. For example, a linear function will have only one*point***of intersection**, linear and square – two, square – two or four, etc.2

Consider the General case of two linear functions (see Fig.1). Let y1=k1x+b1 and y2=k2x+b2. To find the

*point*of their**intersection**we need to solve the equation y1=y2 or k1x+b1=k2x+b2.Transforming the equation you will get: k1x-k2x=b2-b1.Express x as follows:x=(b2-b1)/(k1-k2).3

After finding the values of x – coordinates of the point

**of intersection of**two**graphs**on the x-axis (axis 0X), it remains to calculate the coordinate of the ordinate (axis 0У). It is necessary to substitute in any of the functions, the obtained value x. Thus, the point**of intersection of**U1 and U2 will have the following coordinates: ((b2-b1)/(k1-k2);k1(b2-b1)/(k1-k2)+b2).4

Analyze calculation example find the point

**of intersection of**two**graphs**(see Fig.2).You need to find*the point***of intersection****of the graphs**of the functions f1 (x)=0.5 x^2 and f2 (x)=0.6 x+1,2.Equating f1 (x) and f2 (x), we obtain the following equation:0.5 x y =0.6 x+1,2. Moving all terms to the left side, you will get a quadratic equation:0.5 x^2 x -0,6-1,2=0.The solution of this equation will be two values of x: x1≈2,26,x2≈-1,06.5

Substitute the values of x1 and x2 in any of the expression functions. For example, and f_2 (x1)=0,6•2,26+1,2=2,55, f_2 (x2)=0,6•(-1,06)+1,2=0,56.So, the required points are: t A (2,26;2,55) (-1,06;0,56).