Presentation on the topic "equation of a tangent to the graph of a function." Lesson summary "Physical and geometric meaning of the derivative. Tangent to the graph of a function" IV Study of new material
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 tangent point 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)
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Slide captions:
Tangent to the graph of a function. Grade 10
Tangent to the graph of the function x y 0 A Tangent A straight line passing through the point (x 0 ; f (x 0)), with the segment of which the graph of the function f practically merges for values close to x 0, is called the tangent to the graph of the function f at the point (x 0 ; f (x 0)).
The tangent is the limiting position of the secant at ∆х →0 x y 0 k – the angular coefficient of the line (secant) The angular coefficient of the tangent is equal to f ˈ(x 0). This is the geometric meaning of the derivative. Tangent Secant Automatic display. Click 1 time. Secant k → f’(x 0)
The tangent to the graph of a function f differentiable at a point x o is a straight line passing through the point (x o; f (x o)) and having an angular coefficient f ˈ (x o). Let us derive the equation of the tangent to the graph of the function f at point A (x o; f (x o)). k = f ˈ (x o) => y = fˈ (x o) x + b Find b: f (x o) = f ˈ (x o) x o + b => b = f (x o) - f ˈ (x o) x o y = fˈ (x o) x + f (x o) - f ˈ (x o) x o y = f (x o) – f ˈ (x o)(x - x o)
Lagrange's formula. If the function is differentiable, then on the interval (a; b) there is a point with Є (a; b) such that f' (c) = f (b) – f (a) b - a x y 0 A B a b c l o α C f ' (c) = tg α l o ll AB
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Lesson on learning new material in 10th grade
"Equation of a tangent to the graph of a function"
UMK: Algebra and beginning of mathematical analysis. 10-11 grades
(baseline) 2011
Item: mathematics.
Class: 10
Lesson type: learning new material
Subject: Equation of the tangent to the graph of a function
Target: derive the formula for the equation of a tangent to the graph of a function at a given point, create an algorithm for finding the equation of a tangent, learn how to write an equation for a tangent.
Tasks:
Educational:
practice and systematize skills and abilities on the topic “Tangent, equation of a tangent to the graph of a function.”
Educational:
promote the development of attention;
promote the development of mental calculation skills;
promote the development of logical thinking and mathematical intuition;
promote the development and understanding of interdisciplinary connections among students;
Educational:
develop students’ communicative competencies (communication culture, ability to work in groups, ability to argue their point of view);
create conditions for understanding the need for independent action in solving problems;
realize the great practical and historical significance of the derivative.
Equipment: computer, projector, presentation, textbook, “Living Mathematics” program, drawings of graphs of functions in the “Living Mathematics” program.
Lesson structure and plan:
1.Motivation (self-determination) for educational activities.
2. Updating knowledge and fixing difficulties in activity.
3. Statement of the educational task.
4.Discovery of new knowledge.
Task 9 of the presentation slide: “Make an equation for the tangent to the graph of the function f(x) = x 2 +3x+1at the abscissa x 0 =1" takes you to the next stage of the lesson.
3. Statement of the educational task.
Purpose: discussion of difficulties. Why were there difficulties? What don't we know yet? (1-2 min) Students formulate the goals and objectives of the lesson.
4.Discovery of new knowledge.
Goal: building a project for getting out of a difficulty (5-7 min)
As an additional homework, 2 “strong” students Ivan Shein and Vitaly Konev were asked to use a textbook to understand the derivation of the general formula for the equation of a tangent (textbook page 174) and an example of composing the equation of a tangent to the graph of a function 2 at point x = 1 (textbook page 166, example 2).
Students write their conclusions on the board, and write the rest in their notebooks. After the students leave, the teacher demonstrates drawing 1, made in the “Live Mathematics” program (graph of a function and a tangent to it at a point) and with an equation for the tangent.
5.Primary consolidation in external speech.
Goal: pronouncing new knowledge, recording in the form of a reference signal (5 min).
The class is divided into 4 groups, which are asked to create an algorithm for composing an equation for a tangent to the graph of a function. Students use only the general tangent equation. After discussion, the algorithm is discussed point by point, supplemented, and corrected. As a result, it is demonstrated.
6.Independent work with self-test according to the standard.
Goal: everyone must come to a conclusion for themselves about what they already know how to do (5-6 min).
At this stage, we return to the problem on slide 9 about composing the tangent equation; students solve it independently, followed by self-test. , as well as drawing 2 "Living Mathematics".
7. Inclusion of new knowledge into the knowledge system and repetition.
Purpose: exercises are performed in which new knowledge is used together with previously learned knowledge (10-12 min).
Working with the problem book: page 91, independent choice of number from Nos. 29.12 - 29.16 (answers are in the textbook). Students have the opportunity to choose tasks according to difficulty level.
HOMEWORK will be the same numbers 29.12 – 29.16, work on composing the tangent equation using the algorithm. Solve at least 3 letters, not counting those completed in class.
8.Reflection of activity (lesson summary).
Goal: students’ awareness of their educational activities, self-assessment of the results of their own and the entire class’s activities (2-3 min).
Questions:
What was the task?
Did you manage to solve the problem?
How?
What results did you get?
Where can you apply new knowledge?
And finally, after “all sorts of smart things”, a little humor. The screen shows graphs of the level of your knowledge depending on time, in the interval from the beginning of the lesson to its completion.
Please choose the schedule that you think is closest to you. Are they relevant to the topic of our lesson? From these graphs one can judgeabout the rate of increase your knowledge during the lesson. Graph 1 – we achieved the goal and solved the tasks set at the beginning of the lesson.
Thank you for the lesson!
Literature
Algebra and beginning of mathematical analysis. 10-11 grades. At 2 o'clock. Parts 1,2. Textbook and problem book for students of general education institutions (basic level) / ed. A. G. Mordkovich. - M.: Mnemosyne, 2011.
Living mathematics: a collection of teaching materials. – M.: INT. 176 p.
V. M. Chernyavsky Working with the “Living Mathematics” program.
Various Internet resources for children to find additional information on the topic “Derivatives”.
The video lesson “Equation of a tangent to the graph of a function” demonstrates educational material for mastering the topic. During the video lesson, the theoretical material necessary to formulate the concept of the equation of a tangent to the graph of a function at a given point, an algorithm for finding such a tangent, and examples of solving problems using the studied theoretical material are described.
The video tutorial uses methods that improve the clarity of the material. The presentation contains drawings, diagrams, important voice comments, animation, highlighting and other tools.
The video lesson begins with a presentation of the topic of the lesson and an image of a tangent to the graph of some function y=f(x) at the point M(a;f(a)). It is known that the angular coefficient of the tangent plotted to the graph at a given point is equal to the derivative of the function f΄(a) at this point. Also from the algebra course we know the equation of the straight line y=kx+m. The solution to the problem of finding the tangent equation at a point is schematically presented, which reduces to finding the coefficients k, m. Knowing the coordinates of a point belonging to the graph of the function, we can find m by substituting the coordinate value into the tangent equation f(a)=ka+m. From it we find m=f(a)-ka. Thus, knowing the value of the derivative at a given point and the coordinates of the point, we can represent the tangent equation in this way y=f(a)+f΄(a)(x-a).
The following is an example of composing a tangent equation following the diagram. Given the function y=x 2 , x=-2. Taking a=-2, we find the value of the function at a given point f(a)= f(-2)=(-2) 2 =4. We determine the derivative of the function f΄(x)=2x. At this point the derivative is equal to f΄(a)= f΄(-2)=2·(-2)=-4. To compose the equation, all coefficients a=-2, f(a)=4, f΄(a)=-4 were found, so the tangent equation is y=4+(-4)(x+2). Simplifying the equation, we get y = -4-4x.
The following example suggests constructing an equation for the tangent at the origin to the graph of the function y=tgx. At a given point a=0, f(0)=0, f΄(x)=1/cos 2 x, f΄(0)=1. So the tangent equation looks like y=x.
As a generalization, the process of composing an equation tangent to the graph of a function at a certain point is formalized in the form of an algorithm consisting of 4 steps:
- Enter the designation a for the abscissa of the tangent point;
- f(a) is calculated;
- f΄(x) is determined and f΄(a) is calculated. The found values of a, f(a), f΄(a) are substituted into the tangent equation formula y=f(a)+f΄(a)(x-a).
Example 1 considers composing the tangent equation to the graph of the function y=1/x at point x=1. To solve the problem we use an algorithm. For a given function at point a=1, the value of the function f(a)=-1. Derivative of the function f΄(x)=1/x 2. At point a=1 the derivative f΄(a)= f΄(1)=1. Using the data obtained, the tangent equation y=-1+(x-1), or y=x-2, is drawn up.
In example 2, it is necessary to find the equation of the tangent to the graph of the function y=x 3 +3x 2 -2x-2. The main condition is the parallelism of the tangent and straight line y=-2x+1. First, we find the angular coefficient of the tangent, equal to the angular coefficient of the straight line y=-2x+1. Since f΄(a)=-2 for a given line, then k=-2 for the desired tangent. We find the derivative of the function (x 3 +3x 2 -2x-2)΄=3x 2 +6x-2. Knowing that f΄(a)=-2, we find the coordinates of point 3a 2 +6a-2=-2. Having solved the equation, we get a 1 =0, and 2 =-2. Using the found coordinates, you can find the tangent equation using a well-known algorithm. We find the value of the function at the points f(a 1)=-2, f(a 2)=-18. The value of the derivative at the point f΄(а 1)= f΄(а 2)=-2. Substituting the found values into the tangent equation, we obtain for the first point a 1 =0 y=-2x-2, and for the second point a 2 =-2 the tangent equation y=-2x-22.
Example 3 describes the composition of the tangent equation for drawing it at the point (0;3) to the graph of the function y=√x. The solution is made using a well-known algorithm. The tangent point has coordinates x=a, where a>0. The value of the function at the point f(a)=√x. The derivative of the function f΄(х)=1/2√х, therefore at a given point f΄(а)=1/2√а. Substituting all the obtained values into the tangent equation, we obtain y = √a + (x-a)/2√a. Transforming the equation, we get y=x/2√а+√а/2. Knowing that the tangent passes through the point (0;3), we find the value of a. We find a from 3=√a/2. Hence √a=6, a=36. We find the tangent equation y=x/12+3. The figure shows the graph of the function under consideration and the constructed desired tangent.
Students are reminded of the approximate equalities Δy=≈f΄(x)Δxand f(x+Δx)-f(x)≈f΄(x)Δx. Taking x=a, x+Δx=x, Δx=x-a, we get f(x)- f(a)≈f΄(a)(x-a), hence f(x)≈f(a)+ f΄(a)(x-a).
In example 4, it is necessary to find the approximate value of the expression 2.003 6. Since it is necessary to find the value of the function f(x)=x 6 at the point x=2.003, we can use the well-known formula, taking f(x)=x 6, a=2, f(a)= f(2)=64, f ΄(x)=6x 5. Derivative at the point f΄(2)=192. Therefore, 2.003 6 ≈65-192·0.003. Having calculated the expression, we get 2.003 6 ≈64.576.
The video lesson “Equation of a tangent to the graph of a function” is recommended for use in a traditional mathematics lesson at school. For a teacher teaching remotely, video material will help explain the topic more clearly. The video can be recommended for students to review independently if necessary to deepen their understanding of the subject.
TEXT DECODING:
We know that if a point M (a; f(a)) (em with coordinates a and ef from a) belongs to the graph of the function y = f (x) and if at this point it is possible to draw a tangent to the graph of the function that is not perpendicular to the axis abscissa, then the angular coefficient of the tangent is equal to f"(a) (eff prime from a).
Let a function y = f(x) and a point M (a; f(a)) be given, and it is also 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 that is not parallel to the ordinate axis, has the form y = kx+m (the y is equal to ka x plus em), so the task is to find the values of the coefficients k and m. (ka and em)
Angle coefficient k= f"(a). To calculate the value of m, 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 the straight line, we obtain the correct equality : f(a) = ka+m, from where we find that m = f(a) - ka.
It remains to substitute the found values of the coefficients ki and m into the equation of the straight line:
y = kx+(f(a) -ka);
y = f(a)+k(x-a);
y= f(a)+ f"(a) (x- a). ( y is equal to ef from a plus ef prime from a, multiplied by x minus a).
We have obtained the equation for the tangent to the graph of the function y = f(x) at the point x=a.
If, say, y = x 2 and x = -2 (i.e. a = -2), then f (a) = f (-2) = (-2) 2 = 4; f´(x) = 2x, which means f"(a) = f´(-2) = 2·(-2) = -4. (then the ef of a is equal to four, the ef of the prime of x is equal to two x, which means ef prime from a equals minus four)
Substituting the found values a = -2, f(a) = 4, f"(a) = -4 into the equation, we obtain: y = 4+(-4)(x+2), i.e. y = -4x -4.
(E is equal to minus four x minus four)
Let's create an equation for the tangent to the graph of the function y = tanx (the y is equal to the tangent x) at the origin. We have: a = 0, f(0) = tan0=0;
f"(x)= , which means f"(0) = l. Substituting the found values a=0, f(a)=0, f´(a) = 1 into the equation, we get: y=x.
Let us summarize our steps in finding the equation of the tangent to the graph of a function at point x using an algorithm.
ALGORITHM FOR DEVELOPING AN EQUATION FOR A 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).
Example 1. Create an equation for the tangent to the graph of the function y = - in
point x = 1.
Solution. Let's use the algorithm, taking into account that in this example
2) f(a)=f(1)=- =-1
3) f´(x)=; f´(a)= f´(1)= =1.
4) Substitute the found three numbers: a = 1, f(a) = -1, f"(a) = 1 into the formula. We get: y = -1+(x-1), y = x-2.
Answer: y = x-2.
Example 2. Given the function y = x 3 +3x 2 -2x-2. Write down the equation of the tangent to the graph of the function y = f(x), parallel to the straight line y = -2x +1.
Using the algorithm for composing the tangent equation, we take into account that in this example f(x) = x 3 +3x 2 -2x-2, but the abscissa of the tangent point is not indicated here.
Let's start thinking like this. The desired tangent must be parallel to the straight line y = -2x+1. And parallel lines have equal angular coefficients. This means that the angular coefficient of the tangent is equal to the angular coefficient of the given straight line: k tangent. = -2. Hok cas. = f"(a). Thus, we can find the value of a from the equation f ´(a) = -2.
Let's find the derivative of the function y=f(x):
f"(x)= (x 3 +3x 2 -2x-2)´ =3x 2 +6x-2;f"(a)= 3a 2 +6a-2.
From the equation f"(a) = -2, i.e. 3a 2 +6a-2=-2 we find a 1 =0, a 2 =-2. This means that there are two tangents that satisfy the conditions of the problem: one at the point with abscissa 0, the other at the point with abscissa -2.
Now you can follow the algorithm.
1) a 1 =0, and 2 =-2.
2) f(a 1)= 0 3 +3·0 2 -2∙0-2=-2; f(a 2)= (-2) 3 +3·(-2) 2 -2·(-2)-2=6;
3) f"(a 1) = f"(a 2) = -2.
4) Substituting the values a 1 = 0, f(a 1) = -2, f"(a 1) = -2 into the formula, we get:
y=-2-2(x-0), y=-2x-2.
Substituting the values a 2 = -2, f(a 2) =6, f"(a 2) = -2 into the formula, we get:
y=6-2(x+2), y=-2x+2.
Answer: y=-2x-2, y=-2x+2.
Example 3. From the point (0; 3) draw a tangent to the graph of the function y = . Solution. Let's use the algorithm for composing the tangent equation, taking into account that in this example f(x) = . Note that here, as in example 2, the abscissa of the tangent point is not explicitly indicated. Nevertheless, we follow the algorithm.
1) Let x = a be the abscissa of the point of tangency; it is clear that a >0.
3) f´(x)=()´=; f´(a) =.
4) Substituting the values of a, f(a) = , f"(a) = into the formula
y=f (a) +f "(a) (x-a), we get:
By condition, the tangent passes through the point (0; 3). Substituting the values x = 0, y = 3 into the equation, we get: 3 = , and then =6, a =36.
As you can see, in this example, only at the fourth step of the algorithm we managed to find the abscissa of the tangent point. Substituting the value a =36 into the equation, we get: y=+3
In Fig. Figure 1 shows a geometric illustration of the considered example: a graph of the function y = is constructed, a straight line is drawn y = +3.
Answer: y = +3.
We know that for a function y = f(x), which has a derivative at point x, the approximate equality is valid: Δyf´(x)Δx (delta y is approximately equal to the eff prime of x multiplied by delta x)
or, in more detail, f(x+Δx)-f(x) f´(x) Δx (eff from x plus delta x minus ef from x is approximately equal to eff prime from x by delta x).
For the convenience of further discussion, let us change the notation:
instead of x we will write A,
instead of x+Δx we will write x
Instead of Δx we will write x-a.
Then the approximate equality written above will take the form:
f(x)-f(a)f´(a)(x-a)
f(x)f(a)+f´(a)(x-a). (eff from x is approximately equal to ef from a plus ef prime from a, multiplied by the difference between x and a).
Example 4. Find the approximate value of the numerical expression 2.003 6.
Solution. We are talking about finding the value of the function y = x 6 at the point x = 2.003. Let's use the formula f(x)f(a)+f´(a)(x-a), taking into account that in this example f(x)=x 6, a = 2,f(a) = f(2) = 2 6 =64; x = 2.003, f"(x) = 6x 5 and, therefore, f"(a) = f"(2) = 6 2 5 =192.
As a result we get:
2.003 6 64+192· 0.003, i.e. 2.003 6 =64.576.
If we use a calculator, we get:
2,003 6 = 64,5781643...
As you can see, the approximation accuracy is quite acceptable.
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 objective 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.