Tangent is a straight line passing through a point on the curve and coinciding with it at this point up to first order (Fig. 1).

Another definition: this is the limiting position of the secant at Δ x→0.

Explanation: Take a straight line intersecting the curve at two points: A And b(see picture). This is a secant. We will rotate it clockwise until it finds only one common point with the curve. This will give us a tangent.

Strict definition of tangent:

Tangent to the graph of a function f, differentiable at the point xO, is a straight line passing through the point ( xO; f(xO)) and having a slope f′( xO).

The slope has a straight line of the form y =kx +b. Coefficient k and is slope this straight line.

Slope factor equal to tangent acute angle, formed by this straight line with the abscissa axis:

k = tan α

Here angle α is the angle between the straight line y =kx +b and positive (that is, counterclockwise) direction of the x-axis. It is called angle of inclination of a straight line(Fig. 1 and 2).

If the angle of inclination is straight y =kx +b acute, then the slope is positive number. The graph is increasing (Fig. 1).

If the angle of inclination is straight y =kx +b is obtuse, then the slope is a negative number. The graph is decreasing (Fig. 2).

If the straight line is parallel to the x-axis, then the angle of inclination of the straight line is zero. In this case, the slope of the line is also zero (since the tangent of zero is zero). The equation of the straight line will look like y = b (Fig. 3).

If the angle of inclination of a straight line is 90º (π/2), that is, it is perpendicular to the abscissa axis, then the straight line is given by the equality x =c, Where c– some real number (Fig. 4).

Equation of the tangent to the graph of a functiony = f(x) at point xO:

Example: Find the equation of the tangent to the graph of the function f(x) = x 3 – 2x 2 + 1 at the point with abscissa 2.

Solution .

We follow the algorithm.

1) Touch point xO is equal to 2. Calculate f(xO):

f(xO) = f(2) = 2 3 – 2 ∙ 2 2 + 1 = 8 – 8 + 1 = 1

2) Find f′( x). To do this, we apply the differentiation formulas outlined in the previous section. According to these formulas, X 2 = 2X, A X 3 = 3X 2. Means:

f′( x) = 3X 2 – 2 ∙ 2X = 3X 2 – 4X.

Now, using the resulting value f′( x), calculate f′( xO):

f′( xO) = f′(2) = 3 ∙ 2 2 – 4 ∙ 2 = 12 – 8 = 4.

3) So, we have all the necessary data: xO = 2, f(xO) = 1, f ′( xO) = 4. Substitute these numbers into the tangent equation and find the final solution:

y = f(xO) + f′( xO) (x – x o) = 1 + 4 ∙ (x – 2) = 1 + 4x – 8 = –7 + 4x = 4x – 7.

Answer: y = 4x – 7.

Instructions

We determine the angular coefficient of the tangent to the curve at point M.
The curve representing the graph of the function y = f(x) is continuous in a certain neighborhood of the point M (including the point M itself).

If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, it becomes clear geometric meaning derivative – calculation of the slope of the tangent.

Find the abscissa value of the tangent point, which is denoted by the letter “a”. If it coincides with a given tangent point, then "a" will be its x-coordinate. Determine the value functions f(a) by substituting into the equation functions abscissa value.

Determine the first derivative of the equation functions f’(x) and substitute the value of point “a” into it.

Take general equation tangent, which is defined as y = f(a) = f (a)(x – a), and substitute the found values ​​of a, f(a), f "(a) into it. As a result, the solution to the graph and the tangent will be found.

Solve the problem in a different way if the given tangent point does not coincide with the tangent point. In this case, it is necessary to substitute “a” instead of numbers in the tangent equation. After this, instead of the letters “x” and “y”, substitute the value of the coordinates of the given point. Solve the resulting equation in which “a” is the unknown. Plug the resulting value into the tangent equation.

Write an equation for a tangent with the letter “a” if the problem statement specifies the equation functions and equation parallel line relative to the desired tangent. After this we need the derivative functions, to the coordinate at point “a”. Substitute the appropriate value into the tangent equation and solve the function.

In this article we will analyze all types of problems to find

Let's remember geometric meaning of derivative: if a tangent is drawn to the graph of a function at a point, then the slope coefficient of the tangent (equal to the tangent of the angle between the tangent and the positive direction of the axis) is equal to the derivative of the function at the point.

Let's take an arbitrary point on the tangent with coordinates:

And consider right triangle :

In this triangle

From here

This is the equation of the tangent drawn to the graph of the function at the point.

To write the tangent equation, we only need to know the equation of the function and the point at which the tangent is drawn. Then we can find and .

There are three main types of tangent equation problems.

1. Given a point of contact

2. The tangent slope coefficient is given, that is, the value of the derivative of the function at the point.

3. Given are the coordinates of the point through which the tangent is drawn, but which is not the point of tangency.

Let's look at each type of task.

1 . Write the equation of the tangent to the graph of the function at the point .

.

b) Find the value of the derivative at point . First let's find the derivative of the function

Let's substitute the found values ​​into the tangent equation:

Let's open the brackets on the right side of the equation. We get:

Answer: .

2. Find the abscissa of the points at which the functions are tangent to the graph parallel to the x-axis.

If the tangent is parallel to the x-axis, therefore the angle between the tangent and the positive direction of the axis is zero, therefore the tangent of the tangent angle is zero. This means that the value of the derivative of the function at the points of tangency is zero.

a) Find the derivative of the function.

b) Let’s equate the derivative to zero and find the values ​​in which the tangent is parallel to the axis:

Equating each factor to zero, we get:

Answer: 0;3;5

3. Write equations for tangents to the graph of a function , parallel straight .

A tangent is parallel to a line. The slope of this line is -1. Since the tangent is parallel to this line, therefore, the slope of the tangent is also -1. That is we know the slope of the tangent, and, thereby, derivative value at the point of tangency.

This is the second type of problem to find the tangent equation.

So, we are given the function and the value of the derivative at the point of tangency.

a) Find the points at which the derivative of the function is equal to -1.

First, let's find the derivative equation.

Let's equate the derivative to the number -1.

Let's find the value of the function at the point.

(by condition)

.

b) Find the equation of the tangent to the graph of the function at point .

Let's find the value of the function at the point.

(by condition).

Let's substitute these values ​​into the tangent equation:

.

Answer:

4 . Write the equation of the tangent to the curve , passing through a point

First, let's check if the point is a tangent point. If a point is a tangent point, then it belongs to the graph of the function, and its coordinates must satisfy the equation of the function. Let's substitute the coordinates of the point into the equation of the function.

Title="1sqrt(8-3^2)">. Мы получили под корнем !} a negative number, the equality is not true, and the point does not belong to the graph of the function and is not a point of contact.

This is the last type of problem to find the tangent equation. First thing we need to find the abscissa of the tangent point.

Let's find the value.

Let be the point of contact. The point belongs to the tangent to the graph of the function. If we substitute the coordinates of this point into the tangent equation, we get the correct equality:

.

The value of the function at a point is .

Let's find the value of the derivative of the function at the point.

First, let's find the derivative of the function. This .

The derivative at a point is equal to .

Let's substitute the expressions for and into the tangent equation. We get the equation for:

Let's solve this equation.

Reduce the numerator and denominator of the fraction by 2:

Let's bring the right side of the equation to a common denominator. We get:

Let's simplify the numerator of the fraction and multiply both sides by - this expression is strictly greater than zero.

We get the equation

Let's solve it. To do this, let's square both parts and move on to the system.

Title="delim(lbrace)(matrix(2)(1)((64-48(x_0)+9(x_0)^2=8-(x_0)^2) (8-3x_0>=0 ) ))( )">!}

Let's solve the first equation.

Let's decide quadratic equation, we get

The second root does not satisfy the condition title="8-3x_0>=0">, следовательно, у нас только одна точка касания и её абсцисса равна .!}

Let's write the equation of the tangent to the curve at the point. To do this, substitute the value into the equation - We already recorded it.

Answer:
.

On modern stage development of education, one of its main tasks is the formation of a creatively thinking personality. The ability for creativity in students can be developed only if they are systematically involved in the basics of research activities. The foundation for students to use their creative powers, abilities and talents is formed full-fledged knowledge and skills. In this regard, the problem of forming a system basic knowledge and skills for each topic school course mathematics is of no small importance. At the same time, full-fledged skills should be the didactic goal not of individual tasks, but of a carefully thought-out system of them. In the broadest sense, a system is understood as a set of interconnected interacting elements that have integrity and a stable structure.

Let's consider a technique for teaching students how to write an equation for a tangent to the graph of a function. Essentially, all problems of finding the tangent equation come down to the need to select from a set (bundle, family) of lines those that satisfy a certain requirement - they are tangent to the graph of a certain function. In this case, the set of lines from which selection is carried out can be specified in two ways:

a) a point lying on the xOy plane (central pencil of lines);
b) angular coefficient (parallel beam of straight lines).

In this regard, when studying the topic “Tangent to the graph of a function” in order to isolate the elements of the system, we identified two types of problems:

1) tangent problems, given by the point, through which it passes;
2) problems on a tangent given by its slope.

Training in solving tangent problems was carried out using the algorithm proposed by A.G. Mordkovich. His fundamental difference from those already known is that the abscissa of the point of tangency is denoted by the letter a (instead of x0), and therefore the equation of the tangent takes the form

y = f(a) + f "(a)(x – a)

(compare with y = f(x 0) + f "(x 0)(x – x 0)). This methodical technique, in our opinion, allows students to quickly and easily understand where in the general tangent equation the coordinates of the current point are written, and where the tangent points are.

Algorithm for composing the tangent equation to the graph of the function y = f(x)

1. Designate the abscissa of the tangent point with the letter a.
2. Find f(a).
3. Find f "(x) and f "(a).
4. Substitute the found numbers a, f(a), f "(a) into the general tangent equation y = f(a) = f "(a)(x – a).

This algorithm can be compiled on the basis of students’ independent identification of operations and the sequence of their implementation.

Practice has shown that the sequential solution of each of the key problems using an algorithm allows you to develop the skills of writing the equation of a tangent to the graph of a function in stages, and the steps of the algorithm serve as reference points for actions. This approach is consistent with the theory gradual formation mental actions, developed by P.Ya. Galperin and N.F. Talyzina.

In the first type of tasks, two key tasks were identified:

• the tangent passes through a point lying on the curve (problem 1);
• the tangent passes through a point not lying on the curve (problem 2).

Task 1. Write an equation for the tangent to the graph of the function at point M(3; – 2).

Solution. Point M(3; – 2) is a tangent point, since

1. a = 3 – abscissa of the tangent point.
2. f(3) = – 2.
3. f "(x) = x 2 – 4, f "(3) = 5.
y = – 2 + 5(x – 3), y = 5x – 17 – tangent equation.

Problem 2. Write the equations of all tangents to the graph of the function y = – x 2 – 4x + 2 passing through the point M(– 3; 6).

Solution. Point M(– 3; 6) is not a tangent point, since f(– 3) 6 (Fig. 2).

2. f(a) = – a 2 – 4a + 2.
3. f "(x) = – 2x – 4, f "(a) = – 2a – 4.
4. y = – a 2 – 4a + 2 – 2(a + 2)(x – a) – tangent equation.

The tangent passes through the point M(– 3; 6), therefore, its coordinates satisfy the tangent equation.

6 = – a 2 – 4a + 2 – 2(a + 2)(– 3 – a),
a 2 + 6a + 8 = 0 ^ a 1 = – 4, a 2 = – 2.

If a = – 4, then the tangent equation is y = 4x + 18.

If a = – 2, then the tangent equation has the form y = 6.

In the second type, the key tasks will be the following:

• the tangent is parallel to some line (problem 3);
• the tangent passes at a certain angle to the given line (problem 4).

Problem 3. Write the equations of all tangents to the graph of the function y = x 3 – 3x 2 + 3, parallel to the line y = 9x + 1.

1. a – abscissa of the tangent point.
2. f(a) = a 3 – 3a 2 + 3.
3. f "(x) = 3x 2 – 6x, f "(a) = 3a 2 – 6a.

But, on the other hand, f "(a) = 9 (parallelism condition). This means that we need to solve the equation 3a 2 – 6a = 9. Its roots are a = – 1, a = 3 (Fig. 3).

4. 1) a = – 1;
2) f(– 1) = – 1;
3) f "(– 1) = 9;
4) y = – 1 + 9(x + 1);

y = 9x + 8 – tangent equation;

1) a = 3;
2) f(3) = 3;
3) f "(3) = 9;
4) y = 3 + 9(x – 3);

y = 9x – 24 – tangent equation.

Problem 4. Write the equation of the tangent to the graph of the function y = 0.5x 2 – 3x + 1, passing at an angle of 45° to the straight line y = 0 (Fig. 4).

Solution. From the condition f "(a) = tan 45° we find a: a – 3 = 1 ^ a = 4.

1. a = 4 – abscissa of the tangent point.
2. f(4) = 8 – 12 + 1 = – 3.
3. f "(4) = 4 – 3 = 1.
4. y = – 3 + 1(x – 4).

y = x – 7 – tangent equation.

It is easy to show that the solution to any other problem comes down to solving one or more key problems. Consider the following two problems as an example.

1. Write the equations of the tangents to the parabola y = 2x 2 – 5x – 2, if the tangents intersect at right angles and one of them touches the parabola at the point with abscissa 3 (Fig. 5).

Solution. Since the abscissa of the tangent point is given, the first part of the solution is reduced to key problem 1.

1. a = 3 – abscissa of the point of tangency of one of the sides right angle.
2. f(3) = 1.
3. f "(x) = 4x – 5, f "(3) = 7.
4. y = 1 + 7(x – 3), y = 7x – 20 – equation of the first tangent.

Let a be the angle of inclination of the first tangent. Since the tangents are perpendicular, then is the angle of inclination of the second tangent. From the equation y = 7x – 20 of the first tangent we have tg a = 7. Let us find

This means that the slope of the second tangent is equal to .

The further solution comes down to key task 3.

Let B(c; f(c)) be the point of tangency of the second line, then

1. – abscissa of the second point of tangency.
2.
3.
4.
– equation of the second tangent.

Note. The angular coefficient of the tangent can be found more easily if students know the ratio of the coefficients of perpendicular lines k 1 k 2 = – 1.

2. Write the equations of all common tangents to the graphs of functions

Solution. The problem comes down to finding the abscissa of the points of tangency of common tangents, that is, to solving key problem 1 in general view, drawing up a system of equations and its subsequent solution (Fig. 6).

1. Let a be the abscissa of the tangent point lying on the graph of the function y = x 2 + x + 1.
2. f(a) = a 2 + a + 1.
3. f "(a) = 2a + 1.
4. y = a 2 + a + 1 + (2a + 1)(x – a) = (2a + 1)x + 1 – a 2 .

1. Let c be the abscissa of the tangent point lying on the graph of the function
2.
3. f "(c) = c.
4.

Since tangents are general, then

So y = x + 1 and y = – 3x – 3 are common tangents.

The main goal of the considered tasks is to prepare students to independently recognize the type of key problem when solving more complex problems that require certain research skills (the ability to analyze, compare, generalize, put forward a hypothesis, etc.). Such tasks include any task in which the key task is included as a component. Let us consider as an example the problem (inverse to Problem 1) of finding a function from the family of its tangents.

3. For what b and c are the lines y = x and y = – 2x tangent to the graph of the function y = x 2 + bx + c?

Let t be the abscissa of the point of tangency of the straight line y = x with the parabola y = x 2 + bx + c; p is the abscissa of the point of tangency of the straight line y = – 2x with the parabola y = x 2 + bx + c. Then the tangent equation y = x will take the form y = (2t + b)x + c – t 2 , and the tangent equation y = – 2x will take the form y = (2p + b)x + c – p 2 .

Let's compose and solve a system of equations

Answer:

First level

Equation of a tangent to the graph of a function. Comprehensive guide (2019)

Do you already know what a derivative is? If not, read the topic first. So you say you know the derivative. Let's check it now. Find the increment of the function when the increment of the argument is equal to. Did you manage? It should work. Now find the derivative of the function at a point. Answer: . Happened? If you have any difficulties with any of these examples, I strongly recommend that you return to the topic and study it again. I know the topic is very big, but otherwise there is no point in going further. Consider the graph of some function:

Let's select a certain point on the graph line. Let its abscissa, then the ordinate is equal. Then we select a point with an abscissa close to the point; its ordinate is:

Let's draw a straight line through these points. It is called a secant (just like in geometry). Let us denote the angle of inclination of the straight line to the axis as. As in trigonometry, this angle is measured from the positive direction of the x-axis counterclockwise. What values ​​can the angle take? No matter how you tilt this straight line, one half will still stick up. Therefore, the maximum possible angle is , and the minimum possible angle is . Means, . The angle is not included, since the position of the straight line in this case exactly coincides with, and it is more logical to choose a smaller angle. Let’s take a point in the figure such that the straight line is parallel to the abscissa axis and a is the ordinate axis:

From the figure it can be seen that, a. Then the ratio of the increments is:

(since it is rectangular).

Let's reduce it now. Then the point will approach the point. When it becomes infinitesimal, the ratio becomes equal to the derivative of the function at the point. What will happen to the secant? The point will be infinitely close to the point, so that they can be considered the same point. But a straight line that has only one common point with a curve is nothing more than tangent(in this case, this condition is met only in a small area - near the point, but this is enough). They say that in this case the secant takes limit position.

Let's call the angle of inclination of the secant to the axis. Then it turns out that the derivative

that is the derivative is equal to the tangent of the angle of inclination of the tangent to the graph of the function at a given point.

Since a tangent is a line, let's now remember the equation of a line:

What is the coefficient responsible for? For the slope of the straight line. This is what it's called: slope. What does it mean? And the fact that it is equal to the tangent of the angle between the straight line and the axis! So this is what happens:

But we got this rule by considering an increasing function. What will change if the function is decreasing? Let's see:
Now the angles are obtuse. And the increment of the function is negative. Let's consider again: . On the other side, . We get: , that is, everything is the same as last time. Let us again direct the point to the point, and the secant will take a limiting position, that is, it will turn into a tangent to the graph of the function at the point. So, let’s formulate the final rule:
The derivative of a function at a given point is equal to the tangent of the angle of inclination of the tangent to the graph of the function at this point, or (which is the same) the slope of this tangent:

That's what it is geometric meaning of derivative. Okay, all this is interesting, but why do we need it? Here example:
The figure shows a graph of a function and a tangent to it at the abscissa point. Find the value of the derivative of the function at a point.
Solution.
As we recently found out, the value of the derivative at the point of tangency is equal to the slope of the tangent, which in turn is equal to the tangent of the angle of inclination of this tangent to the abscissa axis: . This means that to find the value of the derivative we need to find the tangent of the tangent angle. In the figure we have marked two points lying on the tangent, the coordinates of which are known to us. So let's complete the construction of a right triangle passing through these points and find the tangent of the tangent angle!

The angle of inclination of the tangent to the axis is. Let's find the tangent of this angle: . Thus, the derivative of the function at a point is equal to.
Answer:. Now try it yourself:

Answers:

Knowing geometric meaning of derivative, we can very simply explain the rule that the derivative at the point of a local maximum or minimum is equal to zero. Indeed, the tangent to the graph at these points is “horizontal”, that is, parallel to the x-axis:

What is the angle between parallel lines? Of course, zero! And the tangent of zero is also zero. So the derivative is equal to zero:

Read more about this in the topic “Monotonicity of functions. Extremum points."

Now let's focus on arbitrary tangents. Let's say we have some function, for example, . We have drawn its graph and want to draw a tangent to it at some point. For example, at a point. We take a ruler, attach it to the graph and draw:

What do we know about this line? What is the most important thing to know about direct to coordinate plane? Since a straight line is an image of a linear function, it would be very convenient to know its equation. That is, the coefficients in the equation

But we already know! This is the slope of the tangent, which is equal to the derivative of the function at that point:

In our example it will be like this:

Now all that remains is to find it. It's as simple as shelling pears: after all - the value of. Graphically, this is the coordinate of the intersection of the line with the ordinate axis (after all, at all points of the axis):

Let's draw it (so it's rectangular). Then (to the same angle between the tangent and the x-axis). What are and equal to? The figure clearly shows that, a. Then we get:

We combine all the obtained formulas into the equation of a straight line:

Now decide for yourself:

1. Find tangent equation to a function at a point.
2. The tangent to a parabola intersects the axis at an angle. Find the equation of this tangent.
3. The line is parallel to the tangent to the graph of the function. Find the abscissa of the tangent point.
4. The line is parallel to the tangent to the graph of the function. Find the abscissa of the tangent point.

Solutions and answers:

EQUATION OF A TANGENT TO THE GRAPH OF A FUNCTION. BRIEF DESCRIPTION AND BASIC FORMULAS

The derivative of a function at a specific point is equal to the tangent of the tangent to the graph of the function at this point, or the slope of this tangent:

Equation of the tangent to the graph of a function at a point:

Algorithm for finding the tangent equation:

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

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But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

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