Now we get the equation of the tangent to the graph of the function. Presentation on the topic "equation of a tangent to the graph of a function" Derivation of the equation of a tangent

Sections: Mathematics

Class: 10

The purpose of the lesson. Generalization, systematization and deepening of knowledge on the topic “Geometric meaning of derivatives.”.

Lesson objectives.

Develop the ability to apply theoretical knowledge when solving tasks of varying complexity.
Preparation for the Unified State Exam
Develop the ability to manage lesson time and evaluate your learning activities.

Equipment: Interactive whiteboard, presentation, drawing tools, chalk, textbooks, notebooks. Everyone has a crossword puzzle on their desk.

Lesson type. A lesson in systematizing and deepening knowledge on the topic (preparation for the Unified State Exam.).

During the classes

1. Repetition of theoretical material. Crossword solution (Slide - 3)

2. Repeat the algorithm for composing the tangent equation. (Slide - 6.7)

To create an equation for the tangent to the graph of the function y=f(x) at point x 0, you need to find

2) y"(x0) =f"(x 0)

3) y(x0) =f(x 0)

4) Substitute the found numbers into the formula

3. Solving examples. Peer review. Self-test. Write an equation for the tangent to the graph of the function y=f(x) at point x 0.

a) , x 0 =1 (Slide - 7.8)

b) y=-x 2 +4, x 0 =-1 (Slide - 9.10)

c)y = x 3, x 0 = 1 (Slide - 12-15)

d) x 0 =4 (Slide - 16.17)

e) y = tgx at point x 0 =0 (Slide - 20-22)

4. Solving complex problems.

The second type of tangent equation. (Slide - 23)

Write the equation of the tangent to the graph of the function y=f(x0), if the tangent is parallel to the line y= kx+b.

Finding algorithm.

1. Let's find the derivative of the function.

2. Since the angular coefficient of the tangent to the graph of the function y= f(x0) is equal to the value of the derivative of the function, i.e. k=f "(x0), then we find the abscissa of the point of tangency by solving the equation f "(x0) = k.

3. Find the value of the function at point x0.

4. Substituting the found values into the formula, we obtain the tangent equation.

The third type of tangent equation. (Slide - 27)

Write the equation of the tangent to the graph of the function y=f(x), if it is known that this tangent passes through the point A(x 0 ,y 0).

Solution algorithm.

Write the equation of the tangent to the graph of the function y=f(x), if it is known that this tangent passes through the point A(x 0 ,y 0).

Y=(x-2) 2 -1 ; A(3;-1) (Slide - 28-30)

The fourth type of tangent equation. (Slide - 31)

Write an equation for the common tangent to the graphs of the functions y= f(X) and y = g (x).

Solution algorithm.

Let us introduce the assumed points of tangency x1 - for the function y= f(x) and x2 - for the function y= g(x).
Let's find the derivatives of these functions.
Let's find the values of the derivatives at these points f "(x1) and g" (x2).
Let's find the values of the functions at these points y = f(x1) and y = g(x2).
Let us compose tangent equations for each function respectively.
Let's write down the angular coefficients k1, k2 and b1, b2.
Since the tangent is common, the angular coefficients are equal and the values of b are equal. k1 = k2 and b1= b2
Let's create a system of equations and solve it, find the values of x1 and x2
We substitute the found values into the general tangent equations.
The equations turned out to be the same. We obtained the equation of the common tangent to the graphs

Write an equation for the common tangent to the graphs of the functions y=f(x) and y= g(x).
Y-(x-+2) 2 - 3 and y=x 2 (Slide - 32-36)

Solving tasks in the Unified State Exam format (Slide - 37-40)

Lesson plan for 10th grade

"Equation of a tangent to the graph of a function"

Lesson type: A lesson in the initial presentation of new knowledge and the formation of initial subject skills, mastery of subject skills.

Didactic task of the lesson: Ensuring awareness and assimilation of concepts, rules, algorithms; formation of skills in applying theoretical principles in the context of solving educational problems.

Lesson objectives: withdraw equation of a tangent to a graph of a function, teach how to construct an equation of a tangent for a given function at a given point.

Planned results:

ZUNs. Students must

know: equation of the tangent to the graph of a function at point x 0 ;

be able to: compose an equation for a tangent to the graph of a given function at a given point.

developing the skill of drawing up an equation for a tangent to the graph of a given function at a given point.

Equipment: board, computer, projector, screen, textbooks, student notebooks, writing materials.

Teacher: Nesterova Svetlana Yurievna

Hello guys! Is everyone ready for class? You can sit down.

1 slide. "Tangent to the graph of a function"

Oral work aimed at preparing students to perceive a new topic (repetition of previously studied material)

10.01 – 10.03

Frontal

Oral work

In order to thoroughly understand the topic of today’s lesson, we need to remember what we previously studied.

Answer the following questions.

2 slide.

The graph of which function is a straight line?(linear)

What equation defines a linear function?(y = k x + b )

What is the name of the number before "X »? ( direct slope)

In a different way, the equationy = k x + b called the equation of a straight line with an angular coefficient.

3 slide.

What is the slope of the line?(the tangent of the angle of inclination of the straight line that this straight line forms with the positive direction of the Ox axis).

Formulate the definition of a tangent:(straight line passing through the point (x O ; f (X O )), with the segment of which the graph practically merges differentiable at point x O functions f for values of x close to x O ).

4 slide.

If at point x o exists derivative , That exists tangent (non-vertical) to the graph of the function in point x o .

5 slide.

If f ’ ( x 0 ) does not exist, then the tangent is either

does not exist (like the function y = |x|),

or vertical (like the graph y = 3 √x).

6 slide.

Let us remember what the relative position of the tangent with the abscissa axis can be?

Direct increasing => slopek >0, tg> 0 => acute angle.

Straight line // OX axis => slopek=0, tg= 0 => angle = 0 0

Declining line => slopek <0, tg < 0 =>obtuse angle.

Slide 7

Geometric meaning of derivative:

The slope of the tangent is equal to the value of the derivative of the function at the point where the tangent is drawn k = f `( x o ).

Okay, well done, repetition is over.

Lesson topic. Setting a lesson goal

10.03-10.05

Discussion, conversation

Complete the following task:

Given a function y = x 3 . Write tangent equation to the graph of this function at point x 0 = 1.

PROBLEM? Yes. How to solve it? What are your options? Where can you find help with this problem? In what sources? But is the problem solvable? So what do you think the topic of our lesson will be?

Topic of today's lesson"Tangent Equation" .

Well, now formulate the goals of our lesson (CHILDREN):

1. Derive equations for the tangent to the graph of the function at the pointX O .

2. Learn to write a tangent equation for a given function.

We open the notebooks, write down the number, “class work”, and the topic of the lesson in the margins.

Primary perception and assimilation of new theoretical educational material

10.06- 10.12

Frontal

Search and research

8 slide.

Let's solve this practical problem. I write on the board - you look and reason with me.

Given a function y = x 3 . It is necessary to write the equation of the tangent to the graph of this function at point x 0 = 1.

Let's reason: the equation of a straight line with an angle coefficient has the form:y = k x + b .

In order to write it, we need to know the meaningk And b .

We'll find k (from the geometric meaning of the derivative):

k = f `( x o ) = f `(1) = 3 * 1 2 = 3, i.e. k = 3 .

Our equation takes the form: y= 3x + b .

Remember: if a line passes through a given point, then when substituting the coordinates of this point into the equation of the line, the correct equality should be obtained. This means that we need to find the ordinate of the point - the value of the function at point x 0 = 1: f (1) =1 3 =1. The tangent point has coordinates (1; 1).

We substitute the found values into the equation of the straight line, we get:

1 = 3 . 1+ b ; Means b = - 2 .

Let's substitute the found valuesk = 3 And b = - 2 into the equation of a straight line:y = 3x - 2.

The problem is solved.

Slide 9

Now let’s solve the same problem in general form.

Given a function y = f ( x ), it is necessary to write the equation of the tangent to the graph of this function at point x 0 .

We reason according to the same scheme: the equation of a straight line with an angle coefficient has the form:y = k x + b .

From the geometric meaning of the derivative: k = f `( x o )=> y = f `( x o ) * x + b .

Function value at point x 0 yes f ( x o ), this means the tangent passes through the point with coordinates( X 0 ; f ( x o ))=> f ( x o )= f `( x o ) * x o + b .

Let us express from this record b : b = f ( x o ) - f `( x o ) * x o .

Let's substitute all the expressions into the equation of the straight line:

y = f `( x o ) * x + b = f `( x o ) * x + f ( x o ) - f `( x o ) * x o = f `( x o ) * ( X - x o )+ f ( x o ).

COMPARE WITH THE TEXTBOOK (p. 131)

Please find the entry for the tangent equation in the text of the textbook and compare it with what we got.

The recording is slightly different (by what?), but it is correct.

It is customary to write the tangent equation in the following form:

y = f ( x o ) + f `( x o )( X - x o )

Write this formula in your notebook and highlight it - you must know it!

Slide 9

Now let's create an algorithm for finding the tangent equation. All the “hints” are in our formula.

Find the value of a function at a pointX O

Calculate the derivative of a function

Find the value of the derivative of a function at a pointX O

Substitute the resulting numbers into the formula

y = f ( x o ) + f `( x o )( x – x o )

Reduce the equation to standard form

Practicing primary skills

10.12-10.14

Frontal

Written + joint discussion

How does this formula work? Let's look at an example. Write the example in your notebook.

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

We carry out the derivation of the equation with writing on the board and in notebooks.

Answer: y = 4x – 7.

Working with a source of information

10.14-10.15

Individual

Reading text, discussion

Look at the textbook on p. 131, example 2. Read up to paragraph 3. What is this example about? (you can create an equation for a given function in general form and then find the tangent equation for any value of x 0 , and you can also find the point of intersection of the tangent to the standard parabola with the Ox axis

Dynamic pause

10.15-10.16

Rest

A moment of rest.

Slide – exercise for the body, exercise for the eyes.

Application of theoretical principles in conditions of performing exercises and solving problems

10.16- 10.30

Frontal, individual

Written (board + notebook)

Well, now let’s get down to practical work, the purpose of which is to develop the skill of composing a tangent equation.

Write down numbers 255(a, b), 256(a, b) on the board.reserve 257 (a, b),* .

* – a task of the next level of difficulty for the most prepared students: On a parabola y = 3x 2 - 4x + 6 find the point at which the tangent to it // line y = 2x + 4 and write the equation of the tangent to the parabola at this point.

Students are invited to work at the board (one by one).

Answers:

№255

a) y = - 3x – 6, y = - 3x + 6 b) y = 2x, y = - 2x +4

№256

a) y = 3, y = - 3x + 3π b) y = 2x + 1 – π/ 2, y = 4x + √3 - 4 π/ 3

№257 (reserve)

a) x = 1, y = 1, in t. (1; 1) tangent // Ox

b) x = - 2, y = - 24, in t. (-2; -24) tangent // Oh

Assignment *answers:

A (1; 5), tangent equation y = 2x + 3.

Independent use of skills

10.30-10.35

Group, individual, independent

Written (notebook), discussion of work in pairs

So what did we do? Who understood the material? Who has any questions? We will conduct a self-monitoring of our understanding of the topic of the lesson.

You will work in pairs - you have cards with tasks on your tables. Read the task carefully; 4-5 minutes are given to complete the work.

Assignment: Write an equation for the tangent to the given functionf(x) at a point with a given abscissa.

I: f( x) = x 2 – 2х – 8, at the point with abscissa -1. Answer: y = -4x – 9.

II: f( x) = 2x 2 – 4x + 12, at the abscissa 2. Answer: y = 4x + 4.

III: f( x) = 3x 2 – x – 9, at the point with abscissa 1. Answer: y = 5x –12.

IV: f( x) = 4x 2 + 2x + 3, at the point with abscissa -0.5. Answer: y = -2x + 2.

Checking the completion of independent work

10.35-10.37

Frontal, group

Implementation of self-control according to the model, discussion

Answers on the board (rotated). Students conduct self-control.

Who got the same answers?

Who didn't have the same answers?

Where did you go wrong?

Questions for students to consolidate the geometric meaning of the derivative:

Name the lines that intersect the Ox axis at an acute angle.

Name the straight lines that // are the Ox axes.

Name the straight lines that form an angle with the Ox axis whose tangent is a negative number.

Reflection of activity

10.37-10.39

Frontal

Conversation

Summing up the lesson.

What a PROBLEMappeared before us during the lesson? (we needed to write the tangent equation, but we didn’t know how to do it)

What goals did we set for this lesson? (derive the tangent equation, learn to construct the tangent equation for a given function at a given point)

Did you achieve the goal of the lesson?

How many of you can say with confidence that you have learned how to write a tangent equation?

Who else has questions? We will definitely continue to work on this topic and, I hope, your problems will be resolved 100%!

Homework

10.39-10.40

Write down your homework - No. 255 (vg), 256 (vg), 257 (vg),*, formula!!!

Look in your textbook for your homework assignments.

№№ 255(vg), 256(vg) - continuation of class work on developing the skill of writing a tangent equation.

* – a task of the next level of difficulty for those who want to test themselves:

On a parabola y = x 2 + 5x – 16 find the point at which the tangent to it // line is 5x+y+4 =0.

Thanks for the work. The lesson is over.

Class: 10

Presentation for the lesson

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Lesson type: learning new material.

Teaching methods: visual, partially search.

The purpose of the lesson.

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.
Develop logical thinking and mathematical speech.
Cultivate the will and perseverance to achieve final results.

Equipment: interactive whiteboard, computer.

Lesson Plan

I. Organizational moment

Checking students' readiness for the lesson. Communicate the lesson topic and goals.

II. Updating knowledge.

(Remember with students the geometric definition of a tangent to the graph of a function. Give examples showing that this statement is not complete.)

Let's remember what a tangent is?

“A tangent is a straight line that has one common point with a given curve.” (Slide No. 2)

Discussion of the correctness of this definition. (After discussion, students come to the conclusion that this definition is incorrect.) To clearly prove their conclusion, we give the following example.

Let's look at an example. (Slide No. 3)

Let a parabola and two straight lines be given , which has one common point M (1;1) with a given parabola. There is a discussion of why the first line is not tangent to this parabola (Fig. 1), but the second is (Fig. 2).

In this lesson, you and I need to find out what a tangent to the graph of a function at a point is, how to create an equation for the tangent?

Consider the main tasks for composing the tangent equation.

To do this, recall the general form of the equation of a line, the conditions for parallelism of lines, the definition of a derivative and the rules of differentiation. (Slide No. 4)

III. Preparatory work for learning new material.

Formulate the definition of a derivative. (Slide No. 5)
Fill in the table of arbitrary elementary functions. (Slide No. 6)
Remember the rules of differentiation. (Slide No. 7)
Which of the following lines are parallel and why? (See clearly) (Slide No. 8)

IV Studying new material.

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

Let the graph of the function be given. A point is selected on it, at this point a tangent is drawn to the graph of the function (we assume that it exists). Find the slope of the tangent.

Let's give the argument an increment and consider on the graph (Fig. 3) point P with abscissa. The angular coefficient of the secant MP, i.e. the tangent of the angle between the secant and the x-axis is calculated by the formula.

If we now tend to zero, then point P will begin to approach point M along a curve. We characterized the tangent as the limiting position of the secant during this approach. This means that it is natural to assume that the angular coefficient of the tangent will be calculated using the formula.

Hence, .

If to the graph of the function y = f (x) at the point x = a you can draw a tangent that is not parallel to the axis at, then expresses the slope of the tangent. (Slide number 10)

Or differently. Derivative at a point x = a equal to the slope of the tangent to the graph of the function y = f(x) at this point.

This is the geometric meaning of the derivative. (Slide No. 11)

Moreover, if:

Let us find out the general form of the tangent equation.

Let the line be given by the equation . We know that . To calculate m, we use the fact that the line passes through the point. Let's plug it into the equation. We get, i.e. . Let's substitute the found values k And m into the equation of a straight line:

– equation of the tangent to the graph of the function. (Slide No. 12)

Let's look at examples:

Let's create an equation for the tangent:

(Slide No. 14)

When solving these examples, we used a very simple algorithm, which is as follows: (Slide No. 15)

Let's look at typical tasks and their solutions.

No. 1 Write an equation for the tangent to the graph of the function at the point.

(Slide No. 16)

Solution. Let's use the algorithm, taking into account that in this example .

3) ;

4) Substitute the found numbers ,, into the formula.

No. 2 Draw a tangent to the graph of the function so that it is parallel to the straight line. (Slide No. 17)

Solution. Let us clarify the formulation of the problem. The requirement to “draw a tangent” usually means “to form an equation for the tangent.” Let's use the algorithm for constructing a tangent, taking into account that in this example .

The desired tangent must be parallel to the line. Two lines are parallel if and only if their slopes are equal. This means that the angular coefficient of the tangent must be equal to the angular coefficient of the given straight line: .But . Hence: ; ., i.e.

V. Problem solving.

1. Solving problems using finished drawings (Slide No. 18 and Slide No. 19)

2. Solving problems from the textbook: No. 29.3 (a, c), No. 29.12 (b, d), No. 29.18, No. 29.23 (a) (Slide No. 20)

VI. Summarizing.

1. Answer the questions:

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

2. What were the difficulties during the lesson, what parts of the lesson did you like the most?

3. Marking.

VII. Comments on homework

No. 29.3 (b,d), No. 29.12 (a,c), No. 29.19, No. 29.23 (b) (Slide No. 22)

Literature. (Slide 23)

Algebra and the beginnings of mathematical analysis: Textbook. For 10-11 grades. for students of general education institutions (basic level) / Edited by A.G. Mordkovich. – M.: Mnemosyne, 2009.
Algebra and the beginnings of mathematical analysis: Problem book, For 10-11 grades. for students of general education institutions (basic level) / Edited by A.G. Mordkovich. – M.: Mnemosyne, 2009.
Algebra and the beginnings of analysis. Independent and test work for grades 10-11. / Ershova A.P., Goloborodko V.V. – M.: ILEKSA, 2010.
Unified State Exam 2010. Mathematics. Problem B8. Workbook / Edited by A.L. Semenov and I.V. Yashchenko - M.: Publishing house MTsNMO, 2010.

Lessons 70-71. Equation of the tangent to the graph of a function

09.07.2015 5132 0

Target: obtain the equation of the tangent to the graph of the function.

I. Communicating the topic and purpose of the lessons

II. Repetition and consolidation of the material covered

1. Answers to questions on homework (analysis of unsolved problems).

2. Monitoring the assimilation of the material (test).

Option 1

1. Find the derivative of the function y = 3x4 – 2 cos x .

Answer:

at point x = π.

Answer:

3. Solve the equation y ’(x) = 0, if

Answer:

Option 2

1. Find the derivative of the function y = 5xb + 3 sinx.

Answer:

2. Calculate the value of the derivative of the function at point x = π.

Answer:

3. Solve the equation y ’(x) = 0, if

Answer:

III. Learning new material

Finally, we will move on to the final stage of studying the derivative and consider the use of the derivative in the remaining lessons. In this lesson we will discuss the tangent to the graph of a function.

The concept of a tangent has already been discussed earlier. It was shown that the graph of a function differentiable at point a f (x) near a practically does not differ from the tangent graph, which means that it is close to the secant passing through the points (a; f (a)) and (a + Δx; f (a + Δx)). Any of these secants passes through the point M(a; f (A)). To write an equation for a tangent, you need to specify its slope. Angular coefficient of the secant Δ f /Δ x at Δх → 0 tends to the number f "(a), which is the angular coefficient of the tangent. Therefore, they say that the tangent is the limiting position of the secant at Δx→ 0.

Now we get the equation of the tangent to the graph of the function f (X). Since the tangent is a straight line and its slope is f "(a), then we can write its equation y = f "(a) x + b . Let's find the coefficient b from the condition that the tangent passes through the point M(a; f (A)). Substitute the coordinates of this point into the tangent equation and get: f (a) = f "(a) a + b, whence b = f (a) - f "(a) · a. Now let's substitute the found value b into the tangent equation and we get: or This is the tangent equation. Let's discuss the application of the tangent equation.

Example 1

At what angle is the sine waveintersects the x-axis at the origin?

The angle at which the graph of a given function intersects the x-axis is equal to the slope a of the tangent drawn to the graph of the function f(x ) at this point. Let's find the derivative:Taking into account the geometric meaning of the derivative, we have: and a = 60°.

Example 2

Let's write the tangent equation to the graph of the function f (x) = -x2 + 4x at point a = 1.

f "(x) and the function itself f (x) at point a = 1 and we get: f "(a) = f "(1) = -2 1 + 4 = 2 and f (a) = f (1) = -12 + 4 · 1 = 3. Let’s substitute these values into the tangent equation. We have: y = 2(x - 1) + 3 or y = 2x + 1.

For clarity, the figure shows a graph of the function f(x ) and tangent to this function. Touch occurs at a point M (1; 3).

Based on examples 1 and 2, we can formulate an algorithm for obtaining 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 )(x - a ) + f (a ).

Note that initially point a may be unknown and must be found from the conditions of the problem. Then in the algorithm in paragraphs 2 and 3, the word “calculate” must be replaced with the word “write” (as illustrated by example 3).

In example 2, the abscissa a of the tangency point was specified directly. In many cases, the tangency point is determined by various additional conditions.

Example 3

Let us write the equations of tangents drawn from the point A (0; 4) to the graph of the function f (x) = - x 2 + 2x.

It is easy to check that point A does not lie on a parabola. At the same time, the tangent points of the parabola and the tangents are unknown, therefore, to find these points, an additional condition will be used - the passage of the tangents through point A.

Let us assume that the contact occurs at point a. Let's find the derivative of the function:Let's calculate the derivative values f "(x ) and the function itself f (x) at the point of tangency a and we get: f ’(a) = -2a + 2 and f (a ) = -a2 + 2a. Let's substitute these quantities into the tangent equation. We have: or This is a tangent equation.

Let's write down the condition for the tangent to pass through point A, substituting the coordinates of this point. We get: 4or 4 = a2, whence a = ±2. Thus, the contact occurs at two points B(-2; -8) and C(2; 0). Therefore, there will be two such tangents. Let's find their equations. Let's substitute the values a = ±2 into the tangent equation. We get: when a = 2 or yx = -2x + 4; at a = -2 or y2 = 6x + 4. So, the tangent equations are y1 = -2x + 4 and y2 = 6x + 4.

Example 4

Let's find the angle between the tangents using the conditions of the previous problem.

The drawn tangents y1 = -2x + 4 and y2 = 6x + 4 make angles a1 and a2 with the positive direction of the abscissa axis (and tg a 1 = -2 and tg a 2 = 6) and between themselves the angle φ = a 1 - a2. Let us find, using the well-known formula,whence φ = arctan 8/11.

Example 5

Let's write the equation of the tangent to the graph of the functionparallel to the straight line y = -x + 2.

Two lines are parallel to each other if they have equal slopes. The angular coefficient of the straight line y = -x + 2 is equal to -1, the angular coefficient of the desired tangent is equal to f ’(a), where a - abscissa of the point of tangency. Therefore, to determine a we have an additional condition f ’(a) = -1.

Using the formula for the derivative of a quotient of functions, we find the derivative:Let's find the value of the derivative at the point a and we get:

We get the equationor (a - 2)2 = 4, or a - 2 = ±2, whence a = 4 and a = 0. Thus, there are two tangents that satisfy the conditions of the problem. Let's substitute the values a = 4 and a = 0 into the tangent equation y = f ’(a)(x - a) + f (A). For a = 4 we have:and tangent y1 = -(x - 4) + 3 or y1 = -x + 7. For a = 0 we get:and the tangent y2 = -(x - 0) – 1 or y2 = -x - 1. So, the equations of the tangents are y1 = -x + 7 and y2 = -x - 1.

Note that if f "(a ) does not exist, then the tangent or does not exist (as with the function f (x) = |x| at point (0; 0) - fig. a, or vertical (as in the functionat point (0; 0) - fig. b.

So, the existence of the derivative of the function f (x) at point a is equivalent to the existence of a non-vertical tangent at point (a; f (a)) graphics. In this case, the angular coefficient of the tangent is equal to f "(a). This is the geometric meaning of the derivative.

The concept of derivative allows for approximate calculations. It has already been repeatedly noted that at Δx→ 0 function values f(x ) and the tangent to it y(x) practically coincide. Therefore, at Δх→ 0 function behavior f (x) in the vicinity of the point x0 can be approximately described by the formula(actually a tangent equation). This formula is successfully used for approximate calculations.

Example 6

Let's calculate the value of the function at point x = 2.03.

Let's find the derivative of this function: f "(x) = 12x2 - 4x + 3. We assume that x = a + Δx, where a = 2 and Δx = 0.03. Let's calculate the values of the function and its derivative at point a and get: And Now we determine the value of the function at a given point x = 2.03. We have:

Of course, the above formula is convenient to use if the values f (a) and f "(a ) is easy to calculate.

Example 7

Let's calculate

Consider the functionLet's find the derivative:We will assume that x = a + Δx, where a = 8 and Δx = 0.03. Let's calculate the values of the function and its derivative at point a and get:Now let's determine the value of the function at a given point x = 8.03. We have:

Example 8

Let us summarize the results obtained. Consider the power function f (x) = x n and we will assume that x = a + Δx and Δx→ 0. Find f "(x) = n x n -1 and calculate the values of the function and its derivative at point a, we get: f (a ) = an and f ’(a ) = nan -1 . Now we have the formula f (x) = a n + nan -1 Δx. Let's use it to calculate the number 0.98-20. We will assume that a = 1, Δx = -0.02 and n = -20. Then we get:

Of course, the above formula can be used for any other functions, in particular trigonometric ones.

Example 9

Let's calculate tan 48°.

Consider the function f (x) = tan x and find the derivative:We will assume that x = a + Δ x, where a = 45° = π/4 and (once again, note that in trigonometry, angles are usually measured in radians). Let's find the values of the function and its derivative at point a and get:Now let's calculate(it is taken into account that π = 3.14).

IV. Control questions

1. Equation of a tangent to the graph of a function.

2. Algorithm for deriving the tangent equation.

3. Geometric meaning of the derivative.

4. Application of the tangent equation for approximate calculations.

V. Lesson assignment

§ 29, no. 1 (a); 2 (b); 5 (a, b); 6 (c, d); 9(a); 10 (b); 12 (g); 14(a); 17; 21(a); 22 (a, c); 24 (a, b); 25(a); 26.

VI. Homework assignment

§ 29, No. 1 (b); 2 (c); 5 (c, d); 6 (a, b); 9 (b); 10(a); 12 (b); 14 (b); 18; 21 (c); 22 (b, d); 24 (c, d); 25 (b); 27.

VII. Creative tasks

1. At what points x are the tangents to the graphs of functions parallel?

Answer: x = -1, x = 3.

2. For what x are the tangents to the graphs of functions y = 3 cos 5 x - 7 and y = 5 cos 3 x + 4 are parallel?

Answer:

3. At what angles do the curves y = x2 and intersect?

Answer: π/2 and arctan 3/5.

4. At what angles do the curves y = intersect? cos x and y = sin x?

Answer:

5. A tangent is drawn to the parabola y = 4 - x2 at the point with abscissa x = 1. Find the point of intersection of this tangent with the ordinate axis.

Answer: (0; 5).

6. A tangent is drawn to the parabola y = 4x - x2 at the point with the abscissa x = 3. Find the point of intersection of this tangent with the x-axis.

Answer: (9/2; 0).

7. Find the angle between two tangents drawn from the point (0; -2) to the parabola y = x2.

Answer:

8. Tangents with angular coefficients are drawn to the graph of the function y = 3x2 + 3x + 2 k 1 = 0 and k 2 = 15. Write the equation of the line passing through the points of tangency.

Answer: y = 12x - 4.

9. Find the equations of lines tangent simultaneously to the parabolas y = x2 + x - 2 and y = -x2 + 7x - 11.

Answer: y = 7x - 11 and y = x - 2.

10. Write the equation of the common tangent to the parabolas y = -3x2 + 4x + 4 and y = -3x2 + 16x - 20.

Answer: y = -2x + 7.

11. The tangent to the graph of the function y = x2 - 4x - 3 is drawn at the point x = 0. Find the area of the triangle formed by the tangent and the coordinate axes.

Answer: 9/8.

12. Find the area of the triangle bounded by the coordinate axes and the tangent to the graph of the functionat point x = 2.

Answer: 1.

VIII. Summing up the lessons

Open algebra lesson in 11th grade 10/19. 2011

Teacher: Gorbunova S.V.

Lesson topic: Equation of a tangent to the graph of a function.

Lesson Objectives

Clarify the concept of “tangent”.
Derive the tangent equation.
Create an algorithm for “composing the equation of a tangent to the graph of a function

y = f (x)".

Begin to develop skills in composing tangent equations in various mathematical situations.
To develop the ability to analyze, generalize, show, use elements of research, and develop mathematical speech.

Equipment: computer, presentation, projector, interactive whiteboard, flashcards, reflection cards.

Lesson structure:

HE. U.
Lesson topic message
Repetition of learned material
Formulation of the problem.
Explanation of new material.
Creation of an algorithm for “composing a tangent equation.”
Historical reference.
Consolidation. Practicing skills in drawing up tangent equations.
Homework.
Independent work with self-test
Summing up the lesson.
Reflection

During the classes

1. O.N.U.

2. Report the topic of the lesson

The topic of today's lesson: “Equation of a tangent to the graph of a function.” Open your notebooks, write down the date and topic of the lesson. (slide 1)

Let the words that you see on the screen become the motto of today's lesson. (slide 2)

There are no bad ideas
Think creatively
Take risks
Don't criticize

To get ready for the lesson, we will repeat the previously studied material. Attention to the screen. Write the solution in your notebook.

2. Repetition of the material studied (slide 3).

Purpose: to test knowledge of the basic rules of differentiation.

Find the derivative of the function:

Who has more than one mistake? Who has one?

3. Update

Purpose: To activate attention, show the lack of knowledge about the tangent, formulate the goals and objectives of the lesson. (Slide 4)

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.

Let's look at specific examples:

Examples.(slide 5)
1) The straight line x = 1 has one common point M(1; 1) with the parabola y = x 2, but is not tangent to the parabola.

The straight line y = 2x – 1, passing through the same point, is tangent to this parabola.

The line x = π is not tangent to the graph y = cos x, 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 concerns the schedule.

^ 4. Setting goals and objectives for children in the lesson: (slide 6)

Try to formulate the purpose of the lesson yourself.

Find out what a tangent to the graph of a function at a point is and derive the tangent equation. Apply formula to solve problems
^ 5. Learning new material

See how the position of the straight line x=1 differs from the position y=2x-1? (slide 7)

Conclude what a tangent is?

The tangent is the limiting position of the secant.

Since a tangent is a straight line and we need to write an equation for the tangent, what do you think we need to remember?

Remember the general form of the equation of a straight line. (y = kx + b)

What is another name for the number k? (angular coefficient or tangent of the angle between this straight line and the positive direction of the Ox axis) k = tan α

What is the geometric meaning of derivative?

Tangent of the angle of inclination between the tangent and the positive direction of the oX axis

That is, I can write tan α = yˈ(x). (slide 8)

Let's illustrate this with a drawing. (slide 9)

Let a function y = f (x) be given and a point M belonging to the graph of this function. Let's define its coordinates as follows: x=a, y= f (a), i.e. M (a, f (a)) and let there be a derivative f "(a), i.e. at a given point the derivative is defined. Let us draw a tangent through the point M. The tangent equation is the equation of a straight line, so it has the form: y = kx + b. Therefore, the task is to find k and b. Pay attention to the board, from what is written there, is it possible to find k? (yes, k = f "(a).)

How to find b now? The desired straight line passes through the point M(a; f(a)), we substitute these coordinates into the equation of the straight line: f(a) = ka + b, hence b = f(a) – ka, since k = tan α= yˈ(x), then b = f(a) – f "(a)a

Let's substitute the values of b and k into the equation y = kx + b.

y = f "(a)x + f(a) – f "(a)a, taking the common factor out of brackets, we get:

y = f(a) + f "(a) · (x-a).

We have obtained the equation for the tangent to the graph of the function y = f(x) at the point x = a.

To confidently solve tangent problems, you need to clearly understand the meaning of each element in this equation. Let's look at this again: (slide 10)

(a, f (a)) – point of contact
f "(a) = tan α = k tangent or slope
(x,y) – any tangent point

And so we derived the tangent equation, analyzed the meaning of each element in this equation, let's now try to derive an algorithm for composing the tangent equation to the graph of the function y = f (x)

6. Drawing up an algorithm (slide 11).

I suggest that students create an algorithm themselves:

Let us denote the abscissa of the point of tangency by the letter a.
Let's calculate f(a).
Let's find f "(x) and calculate f "(a).
Let's substitute the found values of the numbers a, f(a), f "(a) into the tangent equation.
y = f(a) + f "(a) · (x-a).

(I distribute to students the algorithm printed in advance as a reminder for subsequent work.)

Historical background (slide 12).

Attention to the screen. Unscramble the word

1	4/3	9	-4	-1	-3	5

Answer: FLUXION (slide 13).

What is the origin story of this name? (slide 14,15)

The concept of derivative arose in connection with the need to solve a number of problems in physics, mechanics and mathematics. The honor of discovering the fundamental laws of mathematical analysis belongs to the English scientist Newton and the German mathematician Leibniz. Leibniz considered the problem of drawing a tangent to an arbitrary curve.

The famous physicist Isaac Newton, born in the English village of Wolstrop, made significant contributions to mathematics. Solving problems involving drawing tangents to curves and calculating the areas of curvilinear figures, he created a general method for solving such problems - fluxion method (derivatives), and called the derivative itself fluenta .

He calculated the derivative and integral of a power function. He writes about differential and integral calculus in his work “The Method of Fluxions” (1665 – 1666), which served as one of the beginnings of mathematical analysis, differential and integral calculus, which the scientist developed independently of Leibniz.

Many scientists over the years have been interested in tangents. The concept of a tangent was encountered sporadically in the works of the Italian mathematician N. Tartaglia (c. 1500 - 1557) - here the tangent appeared during the study of the issue of the angle of inclination of a gun, at which the greatest degree of flight of a projectile is ensured. I. Keppler considered the tangent while solving the problem of the largest volume of a parallelepiped inscribed in a ball of a given radius.

In the 17th century, based on the teachings of G. Galileo on motion, the kinematic concept of the derivative actively developed. Various versions of the presentation are found in R. Descartes, the French mathematician Roberval, the English scientist D. Gregory, and in the works of I. Barrow.

8. Consolidation (slide 16-18).

1) Create an equation for the tangent to the graph of the function f(x) = x² - 3x + 5 at the point with the abscissa

Solution:

Let's create an equation for the tangent (according to the algorithm). Summon a strong student.

a = -1;
f(a) = f(-1) = 1 + 3 + 5 = 9;
f "(x) = 2x – 3,
f "(a) = f "(-1) = -2 – 3 = -5;
y = 9 – 5 (x + 1),

y = 4 – 5x.

Answer: y = 4 – 5x.

Unified State Exam 2011 tasks B-8

1.The function y = f(x) is defined on the interval (-3; 4). The figure shows its graph and the tangent to this graph at the point with the abscissa a = 1. Calculate the value of the derivative f"(x) at the point a = 1.

Solution: to solve, it is necessary to remember that if the coordinates of any two points A and B lying on a given line are known, then its slope can be calculated using the formula: k = , where (x 1 ; y 1), (x 2 ; y 2) are the coordinates of points A and B, respectively. The graph shows that this tangent passes through points with coordinates (1; -2) and (3; -1), which means k=(-1-(-2))/(3-1)= 0.5.

2. The function y = f(x) is defined on the interval (-3;4). The figure shows its graph and the tangent to this graph at the point with abscissa a = -2. Calculate the value of the derivative f"(x) at point a = -2.

Solution: the graph passes through the points (-2;1) (0;-1) . fˈ(-2)= -2

8.Homework (slide 19).

Preparation for the Unified State Exam B-8 No. 3 - 10

^ 9.Independent work

Write the equation of the tangent to the graph of the function y=f(x) at the point with abscissa a.
option 1 option 2

f(x) = x²+ x+1, a=1 f(x)= x-3x², a=2

answers: Option 1: y=3x; Option 2: y= -11x+12

10. Summing up.