At mathematics lessons the school students and students constantly face lines on the coordinate plane — schedules. And not less often it is required to find crossing of these lines in many algebraic tasks that in itself is not a problem at knowledge of certain algorithms.
Instruction
1. The quantity of possible points of intersection of two certain schedules depends on a type of the used function. For example, linear functions always have one point of intersection, and existence at once of several points – two, four and more is characteristic of square. Consider this fact on a concrete example of finding of a point of intersection of two schedules with two linear functions. Let it will be functions of the following look: y ₁=k₁x+b ₁ and y ₂=k₂x+b ₂. To find a point of their crossing, you have to solve the equation of the k₁x+b type ₁=k₂x+b ₂ or y =y ₂.
2. Transform equality as a result of which the following will turn out: k₁x-kx=b -b ₁. Then express a variable x thus: x= (b -b ₁) / (k -k ₂). Now find value x, that is the coordinate of a point of intersection of two available schedules on abscissa axis. Then calculate the corresponding coordinate on ordinate axis. For this purpose substitute the received value in any of the functions presented earlier x. As a result you receive coordinates of a point of intersection at ₁ and at ₂ which will look as follows: ((b -b ₁) / (k -k ₂); k ₁ (b -b ₁) / (k -k ₂) +b ₂).
3. This example was reviewed in a general view, that is without use of numerical values. For descriptive reasons consider one more option. It is required to find a point of intersection of two function graphs, such as f ₂ (x) =0.6x+1.2 and f ₁ (x) =0.5x². Equate f ₂ (x) and f ₁ (x), as a result at you equality of the following look has to turn out: 0.5x²=0.6x+1.2. Transfer everything available composed in the left part, at the same time at you the quadratic equation of a look 0.5x² - 0.6x-1.2=0 will turn out. Solve this equation. The following values will be the correct answer: x ₁≈ 2.26, x ₂≈-1.06. Substitute result in any of expressions of functions. Finally you calculate required points. In our example – it is t. And (2,26;2,55) and t. In (-1,06;0,56). Leaning on the considered options, you will always be able independently to find a point of intersection of two schedules.