Instruction

1

To find the coordinates of a point belonging to a straight line, click the line and drop a perpendicular line on a coordinate axis. Determine which number corresponds to the intersection point, the intersection with the axis ox is the value of the abscissa, that is x1, the intersection with the axis OU is the ordinate Y1.

2

Try to choose a point whose coordinates can be determined without fractional values for convenience and accuracy of calculations. To build the equation you need at least two points. Find the coordinates of another point belonging to a given line (x2, Y2).

3

Substitute the coordinate values in the equation of the line having the General form y=kx+b. You'll have a system of two equations Y1=kx1+b and y2=kx2+b. Solve this system, for example, the following method.

4

Express b from the first equation and substitute into the second, find k, substitute in any equation and find b. For example, the solution of the system 1=2k+b and 3=5k+b would be: b=1-2k, 3=5k+(1-2k); 3k=2, k=1.5, b=1-2*1,5=-2. Thus, the equation of the line has the form y=1.5 x-2.

5

Knowing two points belonging to direct, try, use the canonical equation of a line, it looks like this: (x - x1)/(x2 - x1)=(u - U1)/(U2 - U1). Substitute values (x1, Y1) and (x2;Y2), simplify. For example, the point (2;3) and (-1;5) owned direct (x-2)/(-1-2)=(have-3)/(5-3); -3(x-2)=2(y-3); 3x+6=2Y-6; 2Y=12-3x or y=6-1,5 H.

6

To find the equation of a function with a non-linear graph, follow these steps. See all standard graphs of y=x^2, y=x^3, y=√x, y=sinx, y=cosx, y=tgx, etc. If one of them reminds you of your schedule, take it as a basis.

7

Draw on the same axes graph the standard functions of the base and find its difference from the timetable. If chart moved a few units up or down – so features added to this number (e.g. y=sinx+4). If the graph moved right or left, so the number added to the argument (e.g., y=sin (x+P/2).

8

An elongated chart height chart says that a function argument is multiplied by some number (for example, y=2sinx). If the schedule, on the contrary, reduced in height, so the number before the function less than 1.

9

Compare the graph of the function-the basics and your options on width. If it is more narrow, then x is a number greater than 1, a wide – number less than 1 (e.g., y=sin0.5x).

10

Substituting into the function equation for different values of x, check whether there is a value of the function. If everything is correct - you podpali the function equation on the schedule.

Note

Perhaps the graph corresponds to the equation found only at a certain period. In this case, specify for which values of x is performed, the resulting equality.

# Advice 2 : How to write the equation of the line

Straight - line algebraic of the first order. In Cartesian coordinates on the plane the equation of the line defined by the equation of the first degree.

You will need

- Knowledge of analytic geometry. Basic knowledge in algebra.

Instruction

1

The equation of the line is given by the coordinates of two points on the plane through which this line must pass. Make a ratio of the coordinates of these points. Let the first point has coordinates (x1,y1) and the second (x2,y2), then the equation of the line is given by: (x-x1)/(x2-x1) = (y-y1)(y2-y1).

2

Convert the resulting equation of the line and explicitly Express y through x. After this operation, the equation of the line will take its final form: y=(x-x1)/((x2-x1)*(y2-y1))+y1.

Note

If one of the numbers in the denominator equal to zero means that there is a direct parallel to one of the coordinate axes.

Useful advice

After you have compiled the equation of a line, check that it is correct. To do this, substitute the coordinates of the points instead of the corresponding coordinate and verify the validity of the equation.

# Advice 3 : How to solve a function f x

The term of the decision function itself is not used in mathematics. Under this wording should be understood to perform some actions on the given function to find any specific characteristics, as well as clarification of the data required for plotting function.

Instruction

1

It is possible to consider an approximate scheme in which it is appropriate to explore the behavior of functions and to build its graph.

Find the domain of the function. Determine whether the function is even and odd. In the case of finding the right answer, continue testing, only the desired axis. Determine whether the function is periodic. In the case of a positive answer will continue to study just for one period. Find the break points of the function and determine its behavior in the vicinity of these points.

Find the domain of the function. Determine whether the function is even and odd. In the case of finding the right answer, continue testing, only the desired axis. Determine whether the function is periodic. In the case of a positive answer will continue to study just for one period. Find the break points of the function and determine its behavior in the vicinity of these points.

2

Find the point of intersection of the function with the coordinate axes. Find the asymptotes, if any. Investigate using a first derivative function for extrema and intervals of monotonicity. Also do your research by using the second derivative for convexity, concavity and points of inflection. Select points to specify the behaviour of the function and calculate the values of the function. Plot the function, considering the results obtained for all conducted researches.

3

On the axis 0X should highlight the salient point: the break point x=0 , the zeros of the function, extreme points, inflection points. In these pointx and calculate the function values (if they exist) and on the plane 0xy, select the corresponding points of the graph and the points selected for updating. A line drawn through all points is constructed, taking account of the intervals of monotony, convexity and asymptotes, and give a sketch of the graph of a function.

4

So, a specific example of the function y=((x^2)+1)/(x-1) do your research using the first derivative. Rewrite the function in the form y=x+1+2/(x-1). The first derivative is equal to y’=1-2/((x-1)^2).

Find the critical points of the rst kind: y’=0, (x-1)^2=2, the result is two points: x1=1-sqrt2, x2=1+sqrt2. Note the measured value on the scope of the function (Fig. 1).

Determine the sign of the derivative on each of the intervals. Based on the rule of alternating signs from "+" to "-" and "-" to "+", we get that the maximum point of the function x1=1-sqrt2, and the minimum point x2=1+sqrt2. The same conclusion can be drawn and the sign of the second derivative.

Find the critical points of the rst kind: y’=0, (x-1)^2=2, the result is two points: x1=1-sqrt2, x2=1+sqrt2. Note the measured value on the scope of the function (Fig. 1).

Determine the sign of the derivative on each of the intervals. Based on the rule of alternating signs from "+" to "-" and "-" to "+", we get that the maximum point of the function x1=1-sqrt2, and the minimum point x2=1+sqrt2. The same conclusion can be drawn and the sign of the second derivative.

# Advice 4 : As to the points to find a function

In many cases, statistics or measurements of any process are presented in the form of a set of discrete values. But in order to plot a continuous graph, it is necessary at this point to find a function. This can be done by interpolation. For this is good the Lagrange polynomial.

You will need

- paper;
- pencil.

Instruction

1

Determine the degree of the polynomial to use for interpolation. It is: KP*X^n + K(n-1)*X^(n-1) +... + K0*S^0. The number n here is 1 less than the number of known points with different X, through which must pass the resulting function. So just count the points and subtract from the obtained values for the unit.

2

Identify the common view of the desired functions. Since X^0 = 1, it takes the form: f(CP) = KP*X^n + K(n-1)*X^ (n-1) +... + K1*X + K0, where n is found in the first step the value of the degree of the polynomial.

3

Start drafting a system of linear algebraic equations to find the coefficients of the interpolating polynomial. The original set of points specifies the number of matches of values of the coordinates Xn of the unknown function on the x-axis and y-axis f(Xn). Therefore, successive substitution of values of x in the polynomial, the value of which is equal to f(Xn), allows to obtain the desired equation:

KP*CP^n + K(n-1)*Xn^ (n-1) +... + K1*Xn + K0 = f(Xn)

KP*X(n-1)^n + K(n-1)*X(n-1)^ (n-1) +... + K1*X(n-1) + K0 = f(X(n-1))

...

KP*Х1п + K(n-1)*X1^ (n-1) + ... + K1*X1 + K0 = f(X1).

KP*CP^n + K(n-1)*Xn^ (n-1) +... + K1*Xn + K0 = f(Xn)

KP*X(n-1)^n + K(n-1)*X(n-1)^ (n-1) +... + K1*X(n-1) + K0 = f(X(n-1))

...

KP*Х1п + K(n-1)*X1^ (n-1) + ... + K1*X1 + K0 = f(X1).

4

Imagine a system of linear algebraic equations in a convenient form for solution. Calculate the values of CP^n... X1^2 and X1...Xn and then substitute them into the equation. The values (also known) move to the left side of the equations. Get a system of the form:

Spp*KP + SP(n-1)*(n-1) +... + JS1*K1 + K0 - SP = 0

With(n-1)n*KP + (n-q)(n-1)*(n-1) + ... + (n-1)1*K1 + K0 - (n-1) = 0

...

С1п*CP + C1(n-1)*(n-1) +... + C11*K1 + K0 - C1 = 0

Here Spp = CP^n and SP = f(Xn).

Spp*KP + SP(n-1)*(n-1) +... + JS1*K1 + K0 - SP = 0

With(n-1)n*KP + (n-q)(n-1)*(n-1) + ... + (n-1)1*K1 + K0 - (n-1) = 0

...

С1п*CP + C1(n-1)*(n-1) +... + C11*K1 + K0 - C1 = 0

Here Spp = CP^n and SP = f(Xn).

5

Solve the system of linear algebraic equations. Use of any known method. For example, Gaussian elimination or Cramer's rule. The solutions are obtained values of coefficients for the polynomial KP...K0.

6

Find the feature points. Substitute the coefficients KP...K0, found in the previous step, the polynomial KP*X^n + K(n-1)*X^ (n-1) +... + K0*S^0. This expression will be a function equation. I.e., the desired f(X) = KP*X^n + K(n-1)*X^ (n-1) +... + K0*S^0.