Each concrete schedule is set by the corresponding function. Process finding of a point (several points) of crossing of two schedules comes down to the solution of the equation of a type of f1 (x) =f2 (x) which decision will be a required point.
It is required to you
- - paper;
- - handle.
Instruction
1. From a school course of mathematics the pupils will knows that the quantity of possible points of intersection of two schedules directly depends on a type of functions. So, for example, linear functions will have only one point of intersection, linear and square – two, square – two or four, etc.
2. Let's consider the general case with two linear functions (see fig. 1). Let y1=k1x+b1, and y2=k2x+b2. To find a point of their crossing it is necessary to solve the equation of y1=y2 or k1x+b1=k2x+b2. Having transformed equality, you receive: k1x-k2x=b2-b1. Express x as follows: x=(b2-b1)/(k1-k2).
3. After finding of value x – coordinates of a point of intersection of two schedules on abscissa axes (axis 0X), it is necessary to calculate coordinate on ordinate axes (axis 0U). For this purpose it is necessary to substitute in any of functions, the received value x. Thus, the point of intersection u1 and u2 will have the following coordinates: ((b2-b1)/(k1-k2); k1(b2-b1)/(k1-k2) +b2).
4. Analyze an example of calculation of finding of a point of intersection of two schedules (see fig. 2). It is necessary to find a point of intersection of function graphs of f1 (x) =0.5x^2 and f2 (x) =0.6x+1.2. Having equated f1 (x) and f2 (x), receive the following equality: 0.5x^ =0.6x+1.2. Having transferred everything composed in the left part, receive a quadratic equation of a look: 0.5x^2 - 0.6x-1.2=0. Two values will be the solution of this equation x: x1≈2.26, x2 -1.06.
5. Substitute values h1 and h2 in any of expressions of 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, required points are: t. And (2,26;2,55) and t. In (-1,06;0,56).