Two schedules on the coordinate plane if they are not parallel, are surely crossed in any point. And quite often it is required to find coordinates of this point in algebraic tasks of this kind. Therefore knowledge of instructions for its stay will bring big benefit both to school students, and students.

## Instruction

1. Any schedule can be set a certain function. To find those points in which schedules are crossed it is necessary to solve the equation which has an appearance: f ₁ (x) =f ₂ (x). The result of the decision will also be that point (or points) which you look for. Review the following example. Let value y ₁=k₁x+b ₁, and value y ₂=k₂x+b ₂. For finding of points of intersection on abscissa axis it is necessary to solve y =y equation ₂, that is k₁x+b ₁=k₂x+b ₂.

2. Transform this inequality, having received k₁x-kx=b -b ₁. Now express x: x= (b -b ₁) / (k -k ₂). Thus you will find a point of intersection of schedules which is on OX axis. Find a point of intersection on ordinate axis. Just substitute in any of functions value x which you found earlier.

3. The previous option is suitable for linear function of schedules. If function square, use the following instructions. As well as with linear function, find the same way value x. For this purpose solve a quadratic equation. 2x² + 2x - 4=0 find a discriminant in the equation (the equation is given for an example). For this purpose use a formula: D = b² – 4ac where b – the value before X, and with is a numerical value.

4. Having substituted numerical values, receive expression of a type of D = 4 + 4*4 = 4+16 = 20. Equation roots depend on value of a discriminant. Now to value of variable b with the sign "-" add or take away (in turn) a root from the received discriminant, and divide into the doubled work of coefficient of a. So you will find equation roots, that is coordinates of points of intersection.

5. Schedules of square function have feature: the axis of OX will be crossed two times, that is you will find two coordinates of abscissa axis. If you receive periodic value of dependence of X on Y, then know that the schedule is crossed in infinite number of points with abscissa axis. Check whether correctly you found points of intersection. For this purpose substitute values X in the equation of f (x) =0.