You will need

- mathematical Handbook;
- - notebook;
- - pencil;
- - handle;
- - protractor;
- a pair of compasses.

Instruction

1

Please note that the graph of a differentiable function f(x) at point x0 has no differences to the segment of the tangent. So it is rather close to the segment l passing through the points (x0, f(x0)) and (x0+Δx f(x0 + Δx)). To specify the straight line passing through the point A with the coefficients (x0, f(x0)), specify its angular coefficient. In this case it is equal to Δy/Δx secant tangent (DF→0) and tends to f‘(x0).

2

If the values of f‘(x0) does not exist, it is possible that a tangent there, or it is held vertically. On this basis, the presence of the derivative of the function at the point x0 due to the existence applied on other than vertical tangent, which touches the graph of the function at the point (x0, f(x0)). In this case, the slope of the tangent is f'(x0). Becomes clear the geometric meaning of derivative, that is, the calculation of the angular coefficient of the tangent.

3

That is, in order to find the slope of the tangent, you need to find the value of the derivative of the function at the point of tangency. Example: to find the slope of the tangent to the graph of the function y = X3 at the point with abscissa X0 = 1. Solution: Find the derivative of this function y(x) = 3x2; find the value of the derivative at the point X0 = 1. have(1) = 3 × 12 = 3. The slope of the tangent at the point X0 = 1 is equal to 3.

4

Draw the picture for more tangents so that they adjoined to the graph of the function at the following points: x1, x2 and X3. Mark the angles that are formed by tangent data with the abscissa axis (angle measured in the positive direction from the axis to the tangent line). For example, the first angle α1 is an acute, the second (α2) is stupid, but the third (α3) will be zero, as performed the straight line is parallel to the OX axis. In this case, the tangent of the obtuse angle has a negative value, and tangent of an acute angle is positive, when tg0 and the result is zero.

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

The tangent to the curve is a straight line, which is adjacent to this curve at a given point, that is, passes through it so that in a small area around this point is possible without much loss of accuracy, replace curve segment tangent. If this curve is the graph of a function, tangent to it is possible to build a special equation.

Instruction

1

Suppose you have a graph of some function. Through two points lying on the graph, you can draw a straight line. Such a line crossing the graph of the given function at two points is called a secant.

If, leaving the first location on the ground, gradually moving in the direction of the second point, then the secant line will gradually turn around, trying to any particular position. In the end, when the two points will merge into one, clipping will fit tightly to your sense of in this one point. In other words, the secant will become a tangent.

If, leaving the first location on the ground, gradually moving in the direction of the second point, then the secant line will gradually turn around, trying to any particular position. In the end, when the two points will merge into one, clipping will fit tightly to your sense of in this one point. In other words, the secant will become a tangent.

2

Any sloping (i.e. not vertical) straight line on the coordinate plane is the graph of the equation y = kx + b. The secant passing through the points (x1, y1) and (x2, y2), must, therefore, satisfy the conditions:

kx1 + b = y1, kx2 + b = y2.

Solving this system of two linear equations, we get: kx2 - kx1 = y2 - y1. Thus, k = (y2 - y1)/(x2 - x1).

kx1 + b = y1, kx2 + b = y2.

Solving this system of two linear equations, we get: kx2 - kx1 = y2 - y1. Thus, k = (y2 - y1)/(x2 - x1).

3

When the distance between x1 and x2 tends to zero, the differences become differentials. Thus, the equation of the tangent passing through the point (x0, y0) the factor k is equal to ∂y0/∂x0 = f'(x0), that is, the value of the derivative of f(x) at the point x0.

4

To find the coefficient b, substitute the computed value of k in the equation f'(x0)*x0 + b = f(x0). Solving this equation for b, we get b = f(x0) - f'(x0)*x0.

5

The final version of the equation of the tangent to the graph of the given function at the point x0 looks like this:

y = f'(x0)*(x - x0) + f(x0).

y = f'(x0)*(x - x0) + f(x0).

6

As an example, consider the equation of the tangent to the function f(x) = x^2 at the point x0 = 3. Derivative of x^2 equal to 2x. Therefore, the equation of the tangent becomes:

y = 6*(x - 3) + 9 = 6x - 9.

The correctness of this equation easy to check. Graph the line y = 6x - 9 passes through the same point (3;9) as the original parabola. By building two graphs, you can see that this video really fits to the parabola at this point.

y = 6*(x - 3) + 9 = 6x - 9.

The correctness of this equation easy to check. Graph the line y = 6x - 9 passes through the same point (3;9) as the original parabola. By building two graphs, you can see that this video really fits to the parabola at this point.

7

Thus, the graph of the function has a tangent at the point x0 only when the function has a derivative at this point. If at the point x0, the function has a discontinuity of the second kind, then the tangent becomes vertical asymptote. However, the mere existence of the derivative at the point x0 does not guarantee the indispensable existence of the tangent at that point. For example, the function f(x) = |x| at the point x0 = 0 is continuous and differentiable, but draw a tangent to it at this point is impossible. The standard formula in this case yields the equation y = 0, but this line is not tangent to the graph module.

# Advice 3 : How to solve function graph and the tangent

The task of drawing up the equation of the tangent to the

**graphics****functions**is reduced to the necessity of making a selection of the many direct to those that can satisfy specified requirements. All this straight can be defined either by points or angular factor. In order to solve graph**the function**and the tangent, you need to perform certain actions.Instruction

1

Read carefully the task of writing the equation of a tangent. As a rule, there is a certain equation graph

**functions**expressed in x and y and the coordinates of one point tangent.2

Plot the

**function**in the x and y coordinates. To do this, make a table of the ratio of equity y at a given x value. If the schedule**function**is nonlinear, then to build will need at least five coordinates. Draw coordinate axes and the graph**of the function**. Put also the point, which is specified in the problem statement.3

Find the value of the abscissa of the touch point, which is marked with the letter "a". If it coincides with the given point tangent, then "a" will be equal to its x-coordinate. Determine the value

**of the function**f(a), substituting in the equation**of the function**value of the abscissa.4

Determine the first derivative equation

**of the function**f'(x) and substitute in it the value of the point "a".5

Take the General equation of the tangent, which is defined as y = f(a) = f (a)(x – a), and substitute into it the values of a, f(a), f '(a). The result will be found the solution of functions graph and a tangent.

6

Let's solve the problem another way, if the given tangent point does not coincide with the point of tangency. In this case, the equation of the tangent instead of numbers to substitute the letter "a". After that, instead of letters "x" and "y" substitute the coordinate value of the set point. Solve the resulting equation in which the letter "a" is unknown. Put the obtained value in the equation of the tangent.

7

Write down the equation of the tangent with the letter "a" if the task is set to the equation

**of the function**and the equation of the parallel line for the desired tangent. After that you need to find the derivative**of the function**is parallel to the line to determine the y-coordinate of the point "a". Substitute the appropriate value for the equation of the tangent and the solve function.# Advice 4 : How to find the x-coordinate of the touch point

When setting up the equation of the tangent to the graph of the function, we use the term "abscissa of the touch point". This value can be set initially in terms of the problem or you must define it yourself.

Instruction

1

Draw on a sheet in the box of axes x and y. Examine the given equation for the graph of a function. If it is linear, it is enough to know two values for the parameter d for any x, and then build the found point on the coordinate axis and connect them with a straight line. If the graph is nonlinear, then make a table of the dependence from x and pick up at least five points to plot.

2

Plot the function and put it on a coordinate axis specified point tangent. If it coincides with a function, its x-coordinate is equal to the letter a, which is denoted by the x coordinate of the touch point.

3

Determine the value of the abscissa of the touch point when the specified point is not a tangent coincides with the graph of the function. Specify a third parameter with the letter "a".

4

Write down the equation of the function f(a). To do this in the original equation instead of x and substitute. Find the derivative of the function f(x) and f(a). Substitute required data in the General equation of the tangent, which has the form: y = f(a) + f '(a)(x – a). The result is an equation that consists of three unknown parameters.

5

Substitute in place of x and y coordinates of the given point through which the tangent passes. Then find the solution of this equation for all and. If it is a square, there will be two values of the abscissa of the touch point. This means that the tangent line passes twice near the graph of a function.

6

Draw a graph of the given function and is parallel to the line defined according to the problem. In this case, you must also specify the unknown parameter a and substitute it into the equation f(a). Paranaita the derivative f(a) to the derivative of the equation of a parallel line. This action comes from the condition of parallelism of two functions. Find the roots of the resulting equation, which will be the abscissa of the touch point.

# Advice 5 : How to find the equation of the tangent to the graph of the function

This statement contains the answer to the question of how to find the equation of the tangent to the graph of the function. Provide complete reference information. The use of theoretical calculations analysed with specific examples.

Instruction

1

Reference material.

To begin with we define a tangent. Tangent to the curve at a given point M is called the limiting position of the secant NM, when the point N along the curve is approaching to the point M.

Find the equation of the tangent to the graph of the function y = f(x).

To begin with we define a tangent. Tangent to the curve at a given point M is called the limiting position of the secant NM, when the point N along the curve is approaching to the point M.

Find the equation of the tangent to the graph of the function y = f(x).

2

Determined by the slope of the tangent to the curve at the point M.

The curve representing the graph of the function y = f(x), continuous in some neighborhood of the point M (including the M).

Spend section MN1, forming with the positive direction of the Ox axis the angle α.

The coordinates of the point M (x; y), the coordinates of a point N1(x+∆x, y+∆y).

From the resulting triangle MN1N you can find the slope of this secant:

tg α = Δy/Δx

MN = ∆x

NN1 = ∆y

As the point N1 on the curve the point M to the secant MN1 turns around a point M, and the angle α tends to the angle ϕ between the tangent MT and the positive direction of the Ox axis.

k = tg ϕ =〖 lim〗┬(∆x→0)〖 〗 Δy/Δx = f`(x)

Thus, the slope of the tangent to the graph of the function is equal to the value of the derivative of this function at the tangent point. This is the geometric meaning of the derivative.

The curve representing the graph of the function y = f(x), continuous in some neighborhood of the point M (including the M).

Spend section MN1, forming with the positive direction of the Ox axis the angle α.

The coordinates of the point M (x; y), the coordinates of a point N1(x+∆x, y+∆y).

From the resulting triangle MN1N you can find the slope of this secant:

tg α = Δy/Δx

MN = ∆x

NN1 = ∆y

As the point N1 on the curve the point M to the secant MN1 turns around a point M, and the angle α tends to the angle ϕ between the tangent MT and the positive direction of the Ox axis.

k = tg ϕ =〖 lim〗┬(∆x→0)〖 〗 Δy/Δx = f`(x)

Thus, the slope of the tangent to the graph of the function is equal to the value of the derivative of this function at the tangent point. This is the geometric meaning of the derivative.

3

The equation of the tangent to a given curve at a given point M has the form:

y - y0 = f`(x0) (x - x0),

where (x0; y0) is the coordinate of the touch point

(x; y) the current coordinates, i.e. the coordinates of any point belonging to the tangent,

f`(x0) = k = tg α is the slope of the tangent.

y - y0 = f`(x0) (x - x0),

where (x0; y0) is the coordinate of the touch point

(x; y) the current coordinates, i.e. the coordinates of any point belonging to the tangent,

f`(x0) = k = tg α is the slope of the tangent.

4

Find the equation of the tangent, for example.

Given the graph of the function y=x2 – 2x. Need to find the equation of the tangent at the point with abscissa x0 = 3.

From the equation of the given curve find the ordinate of the touch point y0 = 32 - 2∙3 = 3.

Find the derivative and then calculate its value at the point x0 = 3.

Have:

y`=2x – 2

f`(3) = 2∙3 – 2 = 4.

Now that we know the point (3; 3) on the curve and the angular factor f`(3) = 4 the tangent at that point, we get the desired equation:

y – 3 = 4 (x – 3)

or

y – 4x + 9 = 0

Given the graph of the function y=x2 – 2x. Need to find the equation of the tangent at the point with abscissa x0 = 3.

From the equation of the given curve find the ordinate of the touch point y0 = 32 - 2∙3 = 3.

Find the derivative and then calculate its value at the point x0 = 3.

Have:

y`=2x – 2

f`(3) = 2∙3 – 2 = 4.

Now that we know the point (3; 3) on the curve and the angular factor f`(3) = 4 the tangent at that point, we get the desired equation:

y – 3 = 4 (x – 3)

or

y – 4x + 9 = 0