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

# 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 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 4: How to find the slope of the tangent

Y=f(x) will be tangent to the shown in figure the schedule at the point x0 if it passes through the point with coordinates (x0; f(x0)) and has the slope f'(x0). To find such a factor, knowing the characteristics of a tangent, it's easy.

You will need

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

Instruction

1

Please note that the schedule is differentiable at the point x0 of the function f(x) does not differ from the segment tangent. In view of this, it is rather close to the segment l, which passes through the point (x0, f(x0)) and (x0+Δx f(x0 + Δx)). To set a line that passes through a certain point And the coefficients (x0, f(x0)), specify its angular coefficient. The angular coefficient is equal to Δy/Δx secant tangent (DF→0) and tends to the number f‘(x0).

2

If the values of f‘(x0) exists, then a tangent or not, or it is held vertically. Because of this, the presence of the derivative of the function at the point x0 due to the existence applied on other than vertical tangent touching the graph of the function at the point (x0, f(x0)). In this case, the slope of the tangent is equal to f'(x0). Thus, it becomes clear the geometric meaning of derivative – calculation of the angular coefficient of the tangent.

3

Draw the picture for more tangents that would be in contact with the graph of the function at the points x1, x2 and X3, and also note the angles formed by these tangents with the abscissa axis (this angle is counted positive in the direction from the axis to the tangent line). For example, the first angle, that is, α1, will be sharp, the second (α2) is stupid, and third (α3) is equal to zero, as conducted straight line parallel to OX axis. In this case, the tangent of the obtuse angle is a negative value, the tangent of an acute angle is positive, and tg0 if the result is zero.

Note

Correctly determine the angle formed by a tangent. To do this, use the protractor.

Useful advice

Two straight inclined will be parallel in that case, if their angular coefficients are equal; perpendicular if the product of the angular coefficients of these tangents equal -1.

# Advice 5: 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 6: How to find the tangent of an angle tangent

The geometric meaning of the first order derivative of the function F(x) is a tangent line to its graphics, which passes through a given point of the curve and matching it at this point. Moreover, the value of the derivative at a given point x0 is the slope or tangent of the angle of inclination of the tangent line k = tg a = F`(x0). The calculation of this ratio is one of the most common tasks of the theory of functions.

Instruction

1

Write the given function F(x), e.g. F(x) = (x3 + 15x +26). If the task explicitly specify the point through which the tangent is held, for example, its coordinate x0 = -2, we can dispense with plotting functions and additional straight lines on the Cartesian system OXY. Find the derivative of the first order from the given function F`(x). In this example, F`(x) = (3x2 + 15). Substitute the given value of the argument x0 in the derivative function and calculate its value: F`(-2) = (3(-2)2 + 15) = 27. So you found tg a = 27.

2

When considering tasks where it is required to determine the tangent of the angle of inclination of the tangent to the graph of the function at the point of intersection of this graph with the abscissa axis, you need first to find the numerical value of the coordinates of the point of intersection of the function with OH. For clarity, it is best to build a graph of a function on the two-dimensional plane OXY.

3

Specify the coordinate range for x, for example, from -5 to 5 with step 1. Substituting in the function values x, calculate the corresponding ordinates y and put on the coordinate plane plotting points (x, y). Connect the dots smooth line. You will see the completed chart, the intersection of function and the x-axis. The ordinate of the function at a given point is equal to zero. Find the numerical value of the corresponding argument. For this specified function, for example F(x) = (4x2 - 16), Paranaita to zero. Solve the resulting equation with one variable and calculate x: 4x2 - 16 = 0, x2 = 4, x = 2. Thus, according to the condition of the problem, the tangent of the angle of inclination of the tangent to the graph of the function is to find the point with coordinates x0 = 2.

4

Similarly to the previously described method determine the derivative function: F`(x) = 8*x. Then calculate its value at the point with x0 = 2, which corresponds to the intersection point of the original function with OH. Substitute the obtained value into the derivative function and calculate the tangent of an angle tangent: tg a = F`(2) = 16.

5

When finding the slope at the point of intersection of the function with the ordinate axis (Oh) do the same. Only the coordinate of the initial point x0 should be set to zero.