Looking at the schedule of a straight line, it is possible to work out its equation without special difficulties. At the same time two points can be known to you, or is not present – in that case it is necessary to begin the decision with search of two points belonging to a straight line.
Instruction
1. To find coordinates of the point belonging to a straight line choose it on the line and lower perpendicular lines on an axis of coordinates. Define to what number there corresponds the point of intersection, crossing with an axis oh is a value of an abscissa, that is h1, crossing with an axis ou is an ordinate, u1.
2. Try to choose a point which coordinates can be determined without fractional values, for convenience and the accuracy of calculations. For creation of the equation you need at least two points. Find coordinates of one more point belonging to this straight line (h2, u2).
3. Substitute values of coordinates in the equation of the straight line having a general view at =kx+b. At you the system from two equations of y1=kx1+b and y2=kx2+b will turn out. Solve this system, for example, in the following way.
4. Express b from the first equation and substitute in the second, find k, substitute in any equation and find b. For example, the solution of system 1=2k+b and 3=5k+b will look so: b=1-2k, 3=5k+ (1-2nd); 3k=2, k=1.5, b=1-2*1.5=-2. Thus, the equation of a straight line has y=1.5kh-2 appearance.
5. Knowing two points belonging to a straight line try to use the initial equation of a straight line, it looks thus: (x - h1) / (h2 - h1)= (at - u1) / (u2 - u1). Substitute values (h1; u1) and (h2; u2), simplify. For example, points (2;3) and (-1;5) belong to a straight line (x-2) / (-1-2)= (at-3) / (5-3);-3 (x-2) =2 (at-3); - the 3rd +6=2u-6; 2u =12-3kh or at =6-1.5kh.
6. To find the equation of the function having the nonlinear schedule act so. Check all standard schedules of y=x^2, y=x^3, y= √ x, y=sinx, y=cosx, y=tgx, etc. If one of them reminds you your schedule, take it as a basis.
7. Draw the standard schedule of function basis on the same axis of coordinates and find its differences from the schedule. If the schedule is postponed for several units up or down – means this number is added to function (for example, at =sinx+4). If the schedule is transferred to the right or to the left, so the number is added to an argument (for example, at =sin (x + P/2).
8. The extended schedule in height says the schedule that function of an argument is increased by some number (for example, at =2sinx). If the schedule, on the contrary, is reduced in height, so number before function less than 1.
9. Compare the schedule of function basis and your function on width. If it narrower, means before x there is a number more than 1, wide – number less than 1 (for example, at =sin0.5x).
10. Substituting different values in the turned-out function equation x, check whether correctly there is a value of function. If everything is right - you podbrat the function equation according to the schedule.