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, the geometric meaning of the derivative becomes clear - the calculation of the angular coefficient 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 the general tangent equation, 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 will be found and tangent.

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 the equation of a 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.

First level

Equation of a tangent to the graph of a function. The 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 the point with the 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 increment ratio 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 the 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 a line on a 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 particular 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%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough...

For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

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?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

You can use our tasks (optional) and we, of course, recommend them.

In order to get better at using our tasks, you need to help extend the life of the YouClever textbook you are currently reading.

How? There are two options:

1. Unlock all hidden tasks in this article - 299 rub.
2. Unlock access to all hidden tasks in all 99 articles of the textbook - 999 rub.

Yes, we have 99 such articles in our textbook and access to all tasks and all hidden texts in them can be opened immediately.

In the second case we will give you simulator “6000 problems with solutions and answers, for each topic, at all levels of complexity.” It will definitely be enough to get your hands on solving problems on any topic.

In fact, this is much more than just a simulator - a whole training program. If necessary, you can also use it for FREE.

Access to all texts and programs is provided for the ENTIRE period of the site's existence.

In conclusion...

If you don't like our tasks, find others. Just don't stop at theory.

“Understood” and “I can solve” are completely different skills. You need both.

Find problems and solve them!

This mathematical program finds the equation of the tangent to the graph of the function \(f(x)\) at a user-specified point \(a\).

The program not only displays the tangent equation, but also displays the process of solving the problem.

This online calculator can be useful for high school students in secondary schools when preparing for tests and exams, when testing knowledge before the Unified State Exam, and for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get your math or algebra homework done as quickly as possible? In this case, you can also use our programs with detailed solutions.

In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.

If you need to find the derivative of a function, then for this we have the task Find the derivative.

If you are not familiar with the rules for entering functions, we recommend that you familiarize yourself with them.

Enter the function expression \(f(x)\) and the number \(a\)

f(x)=
a=

Find tangent equation

It was discovered that some scripts necessary to solve this problem were not loaded, and the program may not work.
You may have AdBlock enabled.
In this case, disable it and refresh the page.

JavaScript is disabled in your browser.
For the solution to appear, you need to enable JavaScript.
Here are instructions on how to enable JavaScript in your browser.

Because There are a lot of people willing to solve the problem, your request has been queued.
In a few seconds the solution will appear below.
Please wait sec...

If you noticed an error in the solution, then you can write about this in the Feedback Form.
Do not forget indicate which task you decide what enter in the fields.

Our games, puzzles, emulators:

A little theory.

Direct slope

Recall that the graph of the linear function \(y=kx+b\) is a straight line. The number \(k=tg \alpha \) is called slope of a straight line, and the angle \(\alpha \) is the angle between this line and the Ox axis

If \(k>0\), then \(0 If \(kEquation of the tangent to the graph of the function

If point M(a; f(a)) belongs to the graph of the function y = f(x) and if at this point a tangent can be drawn to the graph of the function that is not perpendicular to the x-axis, then from the geometric meaning of the derivative it follows that the angular coefficient of the tangent is equal to f "(a). Next, we will develop an algorithm for composing an equation for a tangent to the graph of any function.

Let a function y = f(x) and a point M(a; f(a)) be given on the graph of this function; let it be known that f"(a) exists. Let's create an equation for the tangent to the graph of a given function at a given point. This equation, like the equation of any straight line not parallel to the ordinate axis, has the form y = kx + b, so the task is to find the values ​​of the coefficients k and b.

Everything is clear with the angular coefficient k: it is known that k = f"(a). To calculate the value of b, we use the fact that the desired straight line passes through the point M(a; f(a)). This means that if we substitute the coordinates of the point M into the equation of a straight line, we obtain the correct equality: \(f(a)=ka+b\), i.e. \(b = f(a) - ka\).

It remains to substitute the found values ​​of the coefficients k and b into the equation of the straight line:

$$ y=kx+b $$ $$ y=kx+ f(a) - ka $$ $$ y=f(a)+ k(x-a) $$ $$ y=f(a)+ f"(a )(x-a)$$

We received equation of the tangent to the graph of a function\(y = f(x) \) at the point \(x=a \).

Algorithm for finding the equation of the tangent to the graph of the function \(y=f(x)\)
1. Designate the abscissa of the tangent point with the letter \(a\)
2. Calculate \(f(a)\)
3. Find \(f"(x)\) and calculate \(f"(a)\)
4. Substitute the found numbers \(a, f(a), f"(a) \) into the formula \(y=f(a)+ f"(a)(x-a) \)

Books (textbooks) Abstracts of the Unified State Examination and the Unified State Examination tests online Games, puzzles Plotting graphs of functions Spelling dictionary of the Russian language Dictionary of youth slang Catalog of Russian schools Catalog of secondary educational institutions of Russia Catalog of Russian universities Catalog of Russian universities List of problems Finding GCD and LCM Simplifying a polynomial (multiplying polynomials)

Back forward

Attention! Slide previews are for informational purposes only and may not represent all the features of the presentation. If you are interested in this work, please download the full version.

Lesson type: learning new material.

Teaching methods: visual, partly search.

The purpose of the lesson:

1. Introduce the concept of a tangent to the graph of a function at a point, find out what the geometric meaning of the derivative is, derive the tangent equation and teach how to find it for specific functions.
2. Development of logical thinking, research skills, functional thinking, mathematical speech.
3. Developing communication skills in work, promoting the development of independent activities of students.

Equipment: computer, multimedia projector, handouts.

Lesson Plan

IOrganizing time.
<слайд 2, 3>Checking students' readiness for the lesson. State the theme and motto of the lesson.

IIUpdating the material.
(Activate attention, show lack of knowledge about the tangent, formulate the goals and objectives of the lesson.)<слайд 5>

Let's discuss what is a tangent to the graph of a function? Do you agree with the statement that “A tangent is a straight line that has one common point with a given curve”?
There is a discussion going on. Children's statements (yes and why, no and why). During the discussion, we come to the conclusion that this statement is not true.

Examples. <слайд 6>
1) The straight line x = 1 has one common point M(1; 1) with the parabola y = x2, but is not tangent to the parabola. The straight line y = 2x – 1 passing through the same point is tangent to this parabola<рисунок 1>.
2) Similarly, the line x = π is not tangent to the graph y = cosx, although it has a single common point K(π; 1). On the other hand, the line y = - 1 passing through the same point is tangent to the graph, although it has infinitely many common points of the form

(π+2 πk; 1), where k is an integer, in each of which it touches the graph<рисунок 2>.

Picture 1

Figure 2

Setting goals and objectives for children in the lesson: <слайд 7>find out what a tangent to the graph of a function at a point is, how to write an equation for the tangent?
What do we need for this?
Recall the general form of the equation of a line, the conditions for parallelism of lines, the definition of a derivative, the rules of differentiation.

III Preparatory work for studying new material.
Questioning material using cards: (tasks are completed on the board)
1 student: fill out the table of derivatives of elementary functions

Student 2: remember the rules of differentiation

Student 3: write an equation for a straight line y =kx + 4 passing through point A(3; -2).
(y = -2x+4)

4th student: write a straight line equation y=3x+b, passing through point C(4; 2).
(y = 3x – 2).

The rest are front-line work.<слайд 8>

1. State the definition of a derivative.
2. Which of the following lines are parallel? y = 0.5x; y = - 0.5x; y = - 0.5x + 2. Why?

Guess the name of the scientist<слайд 9>:

Answer key

Who this scientist was and what his work was related to, we will find out in the next lesson.
Checking students' answers using cards.<слайд 10>

IV Studying new material.
To set the equation of a straight line on a plane, it is enough for us to know its angular
coefficient and coordinates of one point.

• Let's start with the slope <слайд 11>

Figure 3

Consider the graph of the function y =f(x) differentiable at point A (x 0 ,f(x 0)) <рисунок 3>.
Let's select a point on it M (x 0 + Δх,f(x 0 + Δх)) and draw a secant A.M..
Question: what is the angular coefficient of the secant? (∆f/∆x=tgβ)

We will approach the point along an arc M to the point A. In this case the straight line A.M. will rotate around a point A, approaching (for smooth lines) to some limiting position - straight line AT. In other words< TAM → 0 если длина АМ → 0. Прямую AT, which has this property is called tangent to the graph of the function y =f(x) at point A(x 0 , f(x 0)). <слайд 12>

Angular coefficient of the secant A.M. at A.M.→ 0 tends to the tangent slope AT Δf/Δx → f "(x 0). Derivative value at a point x 0 Let us take the tangent angle as the angular coefficient. They say that the tangent is the limiting position of the secant at ∆х → 0.

Existence of a derivative of a function at point x 0 is equivalent to the existence of a (non-vertical) tangent at the point (x 0 , f(x 0 )) graphics, while the angular coefficient of the tangent is equal to f "(x 0). This is geometric meaning of derivative. <слайд 13>

Tangent Definition: <слайд 14>Tangent to the graph differentiable at a point x 0 functions f is a straight line passing through a point (x 0 ,f(x 0)) and having a slope f "(x 0).
Let's draw tangents to the graph of the function y = f(x) at points x 1, x 2, x 3,<рисунок 4>and note the angles they form with the x-axis. (This is the angle measured in the positive direction from the positive direction of the axis to the straight line.)

Figure 4

We see that angle α 1 is acute, angle α 3 is obtuse, and angle α 2 is equal to zero, since the straight line l parallel to the axis Oh. The tangent of an acute angle is positive, and the tangent of an obtuse angle is negative. That's why f "(x 1)>0 , f "(x 2) = 0 , f "(x 3)< 0 . <слайд 15, 16>

• Let us now derive the tangent equation <слайд 17, 18>to the graph of the function f at point A( x 0 ,f(x 0)).

General view of the equation of a line y =kx +b.

1. Let's find the slope k =f "(x 0), we get y =f "(x0)∙x+b,f(x) =f "(x 0)∙x+b
2. Let's find b. b =f(x 0) -f "(x 0)∙x 0.
3. Let's substitute the obtained values k And b into the equation of a straight line: y = f "(x 0 )∙ x+ f( x 0 ) - f "(x 0 )∙ x 0 or y = f( x 0 ) + f "(x 0 )( x- x 0 )

• Summarizing the lecture material. <слайд 19>

What is the tangent to the graph of a function at a point?
- What is the geometric meaning of the derivative?
- formulate an algorithm for finding the equation of a tangent at a point?

1. Value of the function at the point of contact
2. General derivative of a function
3. The value of the derivative at the point of tangency
4. Substitute the found values ​​into the general tangent equation.

V Consolidation of the studied material.

1. Oral work:
1) <слайд 20>At what points on the graph?<рисунок 5>tangent to it
a) horizontal;
b) forms an acute angle with the abscissa axis;
c) forms an obtuse angle with the x-axis?
2) <слайд 21>At what values ​​of the argument is the derivative of the function specified by the graph<рисунок 6>
a) equal to 0;
b) more than 0;
c) less than 0?

Figure 5

Figure 6

3) <слайд 22>The figure shows the graph of the function f(x) and the tangent to it at the abscissa point x 0. Find the value of the derivative of the function f "(x) at the point x 0<рисунок 7>.

Figure 7

2. Written work.
No. 253 (a, b), No. 254 (a, b). (field work, with commentary)

3. Solving support problems.<слайд 23>
Let's look at four types of problems. Children read the conditions of the problem, propose a solution algorithm, one of the students draws it up on the board, the rest write it down in a notebook.
1. If the touch point is specified
Write an equation for the tangent to the graph of the function f(x) = x 3 – 3x – 1 at point M with abscissa –2.
Solution:

1. Let's calculate the value of the function: f(-2) =(-2) 3 – 3(-2) – 1 = -3 ;
2. let's find the derivative of the function: f "(x) = 3x 2 – 3;
3. Let's calculate the value of the derivative: f "(-2)= - 9.;
4. Let's substitute these values ​​into the tangent equation: y = 9(x + 2) – 3 = 9x + 15.

Answer: y = 9x + 15.

2. Along the ordinate of the tangent point.
Write an equation for the tangent at a point on the graph with y-ordinate 0 = 1.
Solution:

Answer: y = –x + 2.

3. A given direction.
Write tangent equations to the graph y = x 3 – 2x + 7, parallel to the line y = x.
Solution.
The desired tangent is parallel to the line y = x. This means they have the same slope k = 1, y"(x) = 3x2 – 2. Abscissa x 0 points of tangency satisfies the equation 3x 2 – 2 = 1, where x 0 = ±1.
Now we can write tangent equations: y = x + 5 And y = x + 9.
Answer: y = x + 5, y = x + 9.

4. Conditions for tangency between the graph and the straight line.
Task. At what b straight y = 0.5x + b is tangent to the graph of the function f(x) = ?
Solution.
Recall that the slope of a tangent is the value of the derivative at the point of tangency. The slope of this line is k = 0.5. From here we obtain the equation for determining the abscissa x of the tangent point: f "(x) == 0.5. Obviously, its only root is –x = 1. The value of this function at this point is y(1) = 1. So, the coordinates of the tangent point are (1; 1). Now it remains to select a value of the parameter b at which the straight line passes through this point, that is, the coordinates of the point satisfy the equation of the straight line: 1 = 0.5 1 + b, whence b = 0.5.

5. Independent educational work. <слайд 24>

Work in pairs.


Check: the results of the solution are entered into a table on the board (one answer from each pair), discussion of the answers.

6. Finding the angle of intersection of the graph of a function and a straight line. <слайд 25>
Angle of intersection of the graph of the function y = f(x) and straight l is the angle at which the tangent to the graph of the function intersects the line at the same point.
No. 259 (a, b), No. 260 (a) - disassemble at the board.

7. Independent work of a controlling nature. <слайд 26>(the work is differentiated, checked by the teacher for the next lesson)
Option 1.

Option 2.

1. At what points is the tangent to the graph of the function f(x) = 3x 2 - 12x + 7 parallel to the x axis?
2. Write an equation for the tangent to the graph of the function f(x)= x 2 - 4 at the abscissa x 0= - 2. Complete the drawing.
3. Find out if the line is straight y = 12x – 10 tangent to the graph of the function y = 4x 3.

Option 3.

VI Summing up the lesson.<слайд 27>
1. Answers to questions
- what is called a tangent to the graph of a function at a point?
- What is the geometric meaning of the derivative?
- formulate an algorithm for finding the equation of a tangent at a point?
2. Remember the goals and objectives of the lesson, have we achieved this goal?
3. What were the difficulties in the lesson, what parts of the lesson did you like most?
4. Giving marks for the lesson.
VII Homework commentary: paragraph 19 (1, 2), No. 253 (c), No. 255 (d), No. 256 (d), No. 257 (d), No. 259 (d). Prepare a report on Leibniz<слайд 28>.

Literature<слайд 29>

1. Algebra and the beginnings of analysis: Textbook. for 10-11 grades. general education institutions / A.N.Kolmogorov, A.M.Abramov, Yu.P. Dudnitsyn and others; Under. ed. A.N. Kolmogorov. - M.: Education, 2004.
2. Didactic materials on algebra and the beginnings of analysis for grade 10 / B.M. Ivlev, S.M. Sahakyan, S.I. Schwartzburd. - M.: Education, 2003.
3. Multimedia disk from 1C. 1C: Tutor. Mathematics (part 1) + Unified State Exam options. 2006.
4. Open bank of tasks in mathematics / http://mathege.ru/