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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.
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).
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).
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.
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).