Solution of tangent equation. Tangent to the graph of a function at a point. Tangent equation. Geometric meaning of derivative

Equation of the tangent to the graph of a function

P. Romanov, T. Romanova,
Magnitogorsk,
Chelyabinsk region

Equation of the tangent to the graph of a function

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

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

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

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

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

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

y = f(a) + f "(a)(x – a)

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Solution.

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

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

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

y = 9x + 8 – tangent equation;

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

y = 9x – 24 – tangent equation.

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

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

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

y = x – 7 – tangent equation.

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

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

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

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

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

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

The further solution comes down to key task 3.

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

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

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

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

Solution. The task comes down to finding the abscissa of the tangent points of common tangents, that is, solving key problem 1 in general form, drawing up a system of equations and then solving it (Fig. 6).

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

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

Since tangents are general, then

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

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

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

Solution.

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

Let's compose and solve a system of equations

Answer:

Problems to solve independently

1. Write the equations of the tangents drawn to the graph of the function y = 2x 2 – 4x + 3 at the points of intersection of the graph with the line y = x + 3.

Answer: y = – 4x + 3, y = 6x – 9.5.

2. For what values ​​of a does the tangent drawn to the graph of the function y = x 2 – ax at the point of the graph with the abscissa x 0 = 1 pass through the point M(2; 3)?

Answer: a = 0.5.

3. For what values ​​of p does the straight line y = px – 5 touch the curve y = 3x 2 – 4x – 2?

Answer: p 1 = – 10, p 2 = 2.

4. Find all common points of the graph of the function y = 3x – x 3 and the tangent drawn to this graph through the point P(0; 16).

Answer: A(2; – 2), B(– 4; 52).

5. Find the shortest distance between the parabola y = x 2 + 6x + 10 and the straight line

Answer:

6. On the curve y = x 2 – x + 1, find the point at which the tangent to the graph is parallel to the straight line y – 3x + 1 = 0.

Answer: M(2; 3).

7. Write the equation of the tangent to the graph of the function y = x 2 + 2x – | 4x |, which touches it at two points. Make a drawing.

Answer: y = 2x – 4.

8. Prove that the line y = 2x – 1 does not intersect the curve y = x 4 + 3x 2 + 2x. Find the distance between their closest points.

Answer:

9. On the parabola y = x 2, two points are taken with abscissas x 1 = 1, x 2 = 3. A secant is drawn through these points. At what point of the parabola will the tangent to it be parallel to the secant? Write the secant and tangent equations.

Answer: y = 4x – 3 – secant equation; y = 4x – 4 – tangent equation.

10. Find the angle q between the tangents to the graph of the function y = x 3 – 4x 2 + 3x + 1, drawn at the points with abscissas 0 and 1.

Answer: q = 45°.

11. At what points does the tangent to the graph of the function form an angle of 135° with the Ox axis?

Answer: A(0; – 1), B(4; 3).

12. At point A(1; 8) to the curve a tangent is drawn. Find the length of the tangent segment between the coordinate axes.

Answer:

13. Write the equation of all common tangents to the graphs of the functions y = x 2 – x + 1 and y = 2x 2 – x + 0.5.

Answer: y = – 3x and y = x.

14. Find the distance between the tangents to the graph of the function parallel to the x-axis.

Answer:

15. Determine at what angles the parabola y = x 2 + 2x – 8 intersects the x-axis.

Answer: q 1 = arctan 6, q 2 = arctan (– 6).

16. Function graph find all points, the tangent at each of which to this graph intersects the positive semi-axes of coordinates, cutting off equal segments from them.

Answer: A(– 3; 11).

17. The line y = 2x + 7 and the parabola y = x 2 – 1 intersect at points M and N. Find the point K of intersection of the lines tangent to the parabola at points M and N.

Answer: K(1; – 9).

18. For what values ​​of b is the line y = 9x + b tangent to the graph of the function y = x 3 – 3x + 15?

Answer: – 1; 31.

19. For what values ​​of k does the straight line y = kx – 10 have only one common point with the graph of the function y = 2x 2 + 3x – 2? For the found values ​​of k, determine the coordinates of the point.

Answer: k 1 = – 5, A(– 2; 0); k 2 = 11, B(2; 12).

20. For what values ​​of b does the tangent drawn to the graph of the function y = bx 3 – 2x 2 – 4 at the point with the abscissa x 0 = 2 pass through the point M(1; 8)?

Answer: b = – 3.

21. A parabola with a vertex on the Ox axis touches the line passing through points A(1; 2) and B(2; 4) at point B. Find the equation of the parabola.

Answer:

22. At what value of the coefficient k does the parabola y = x 2 + kx + 1 touch the Ox axis?

Answer: k = d 2.

23. Find the angles between the straight line y = x + 2 and the curve y = 2x 2 + 4x – 3.

29. Find the distance between the tangents to the graph of the function and the generators with the positive direction of the Ox axis at an angle of 45°.

Answer:

30. Find the locus of the vertices of all parabolas of the form y = x 2 + ax + b tangent to the line y = 4x – 1.

Answer: straight line y = 4x + 3.

Literature

1. Zvavich L.I., Shlyapochnik L.Ya., Chinkina M.V. Algebra and beginnings of analysis: 3600 problems for schoolchildren and those entering universities. – M., Bustard, 1999.
2. Mordkovich A. Seminar four for young teachers. Topic: Derivative Applications. – M., “Mathematics”, No. 21/94.
3. Formation of knowledge and skills based on the theory of gradual assimilation of mental actions.

/ Ed. P.Ya. Galperina, N.F. Talyzina.– M., Moscow State University, 1968. Example 1.(Given a function) = 3Given a function 2 + 4Given a function f Example 1.(Given a function x Given a function 0 = 1.

Solution.– 5. Let’s write the equation of the tangent to the graph of the function Example 1.(Given a function) at the graph point with the abscissa Derivative of a function ) exists for any x

= (3Given a function 2 + 4Given a function R Given a function + 4.

. Let's find her: Example 1.(Given a function 0) = Example 1.(1) = 2; (Given a function– 5)′ = 6

Then = (Given a function 0) (Given a function – Given a function 0) + Example 1.(Given a function 0),

Then = 10(Given a function – 1) + 2,

Then = 10Given a function – 8.

0) = = 10. The tangent equation has the form: Then = 10Given a function – 8.

y– M., Moscow State University, 1968. Example 1.(Given a function) = Given a function 3 – 3Given a function 2 + 2Given a function Answer. Example 1.(Given a function Example 2. + 5. Let's write the equation of the tangent to the graph of the function = 2Given a function – 11.

Solution.– 5. Let’s write the equation of the tangent to the graph of the function Example 1.(Given a function) at the graph point with the abscissa Derivative of a function ) exists for any x

= (Given a function 3 – 3Given a function 2 + 2Given a function), parallel to the line Given a function 2 – 6Given a function + 2.

y Example 1.(Given a function+ 5)′ = 3 Given a function Since the tangent to the graph of the function + 5. Let's write the equation of the tangent to the graph of the function = 2Given a function) at the abscissa point Given a function 0 is parallel to the line Given a function– 6Given a function– 11, then its slope is equal to 2, i.e. ( Given a function 0) = 2. Let’s find this abscissa from the condition that 3 Given a function 0 + 2 = 2. This equality is valid only when Example 1.(Given a function 0 = 0 and at Then = 2Given a function + 0 = 2. Since in both cases 0) = 5, then straight

b 0 = 2. Since in both cases touches the graph of the function either at the point (0; 5) or at the point (2; 5). 0 = 2. Since in both cases In the first case, the numerical equality 5 = 2×0 + is true 0 = 2. Since in both cases touches the graph of the function either at the point (0; 5) or at the point (2; 5). 0 = 2. Since in both cases = 1.

, where Then = 2Given a function= 5, and in the second case the numerical equality 5 = 2×2 + is true Then = 2Given a function So there are two tangents Example 1.(Given a function+ 5 and + 5. Let's write the equation of the tangent to the graph of the function = 2Given a function – 11.

0) = = 10. The tangent equation has the form: Then = 2Given a function + 5, Then = 2Given a function + 1.

+ 1 to the graph of the function– M., Moscow State University, 1968. Example 1.(Given a function) = Given a function 2 – 6Given a function), parallel to the line Example 1.(Given a function Example 3. + 7. Let's write the equation of the tangent to the graph of the function (2; –5).

Solution.), passing through the point Example 1. A + 7. Let's write the equation of the tangent to the graph of the function Because Example 1.(Given a function(2) –5, then point Given a function does not belong to the graph of the function

). Let Example 1.(Given a function) at the graph point with the abscissa Derivative of a function ) exists for any x

= (Given a function 2 – 6Given a function 0 - abscissa of the tangent point. Given a function – 6.

. Let's find her: Example 1.(Given a function 0) = Given a function– 6Given a function 0 + 7; (Given a function 0) = 2Given a function Derivative of a function

Then = (2Given a function 0 – 6)(Given a function – Given a function 0) + Given a function– 6Given a function+ 7,

Then = (2Given a function 0 – 6)Given a function– Given a function+ 7.

+ 1)′ = 2 + 7. Let's write the equation of the tangent to the graph of the function 0 – 6. The tangent equation has the form:

–5 = (2Given a function Since the point Given a function+ 7,

belongs to the tangent, then the numerical equality is true Given a function 0 – 6)×2– Given a function where + 7. Let's write the equation of the tangent to the graph of the function 0 = 0 or Example 1.(Given a function).

0 = 4. This means that through the point Given a function you can draw two tangents to the graph of the function Then = –6Given a function If Given a function 0 = 0, then the tangent equation has the form Then = 2Given a function – 9.

0) = = 10. The tangent equation has the form: Then = –6Given a function + 7, Then = 2Given a function – 9.

+ 7. If 0 = 4, then the tangent equation has the form Example 1.(Given a function) = Given a function 2 – 2Given a function Example 4. Functions given(Given a function) = –Given a function+ 2 and

Solution. g Given a function 2 – 3. Let’s write the equation of the common tangent to the graphs of these functions. Example 1.(Given a function Let Given a function 1 - abscissa of the point of tangency of the desired line with the graph of the function Functions given(Given a function).

). Let Example 1.(Given a function) at the graph point with the abscissa Derivative of a function ) exists for any x

= (Given a function 2 – 2Given a function), A Given a function – 2.

. Let's find her: Example 1.(Given a function 1) = Given a function– 2Given a function 1 + 2; (Given a function 1) = 22 - abscissa of the point of tangency of the same line with the graph of the function+ 2)′ = 2

Then = (2Given a function 1 – 2)(Given a function – Given a function 1) + Given a function– 2Given a function 1 + 2,

+ 5. Let's write the equation of the tangent to the graph of the function = (2Given a function 1 – 2)Given a function – Given a function+ 2. (1)

x Functions given(Given a function):

= (–Given a function 1 – 2. The tangent equation has the form: Given a function.

Let's find the derivative of the function

2 – 3)′ = –2

If now point P is shifted along the graph to point M, then straight line MR will rotate around point M. In this case, ∆x will tend to zero. From here we can formulate the definition of a tangent to the graph of a function.

Tangent to the graph of a function

The tangent to the graph of a function is the limiting position of the secant as the increment of the argument tends to zero. It should be understood that the existence of the derivative of the function f at the point x0 means that at this point on the graph there is tangent to him.

In this case, the angular coefficient of the tangent will be equal to the derivative of this function at this point f’(x0). This is the geometric meaning of the derivative. The tangent to the graph of a function f differentiable at point x0 is a certain straight line passing through the point (x0;f(x0)) and having an angular coefficient f’(x0).

Tangent equation

Let's try to obtain the equation of the tangent to the graph of some function f at point A(x0; f(x0)). The equation of a straight line with slope k has the following form:

Since our slope coefficient is equal to the derivative f’(x0), then the equation will take the following form: y = f’(x0)*x + b.

Now let's calculate the value of b. To do this, we use the fact that the function passes through point A.

f(x0) = f’(x0)*x0 + b, from here we express b and get b = f(x0) - f’(x0)*x0.

We substitute the resulting value into the tangent equation:

y = f’(x0)*x + b = f’(x0)*x + f(x0) - f’(x0)*x0 = f(x0) + f’(x0)*(x - x0).

y = f(x0) + f’(x0)*(x - x0).

Consider the following example: find the equation of the tangent to the graph of the function f(x) = x 3 - 2*x 2 + 1 at point x = 2.

2. f(x0) = f(2) = 2 2 - 2*2 2 + 1 = 1.

3. f’(x) = 3*x 2 - 4*x.

4. f’(x0) = f’(2) = 3*2 2 - 4*2 = 4.

5. Substitute the obtained values ​​into the tangent formula, we get: y = 1 + 4*(x - 2). Opening the brackets and bringing similar terms we get: y = 4*x - 7.

Answer: y = 4*x - 7.

General scheme for composing the tangent equation to the graph of the function y = f(x):

1. Determine x0.

2. Calculate f(x0).

3. Calculate f’(x)

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

Another definition: this is the limiting position of the secant at Δ Given a function→0.

Explanation: Take a straight line intersecting the curve at two points: A And 0 = 2. Since in both cases(see picture). This is a secant. We will rotate it clockwise until it finds only one common point with the curve. This will give us a tangent.

Strict definition of tangent:

Tangent to the graph of a function Example 1., differentiable at the point Given a functionO, is a straight line passing through the point ( Given a functionO; Example 1.(Given a functionO)) and having a slope Example 1.′( Given a functionO).

The slope has a straight line of the form y=kx +0 = 2. Since in both cases. Coefficient k and is slope this straight line.

The angular coefficient is equal to the tangent of the acute angle formed by this straight line with the abscissa axis:

Here angle α is the angle between the straight line y=kx +0 = 2. Since in both cases and positive (that is, counterclockwise) direction of the x-axis. It is called angle of inclination of a straight line(Fig. 1 and 2).

If the angle of inclination is straight y=kx +0 = 2. Since in both cases acute, then the slope is a positive number. The graph is increasing (Fig. 1).

If the angle of inclination is straight y=kx +0 = 2. Since in both cases is obtuse, then the slope is a negative number. The graph is decreasing (Fig. 2).

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

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

Equation of the tangent to the graph of a functionThen = Example 1.(Given a function) at point Given a functionO:

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

Solution .

We follow the algorithm.

1) Touch point Given a functionO is equal to 2. Calculate Example 1.(Given a functionO):

Example 1.(Given a functionO) = Example 1.(2) = 2 3 – 2 ∙ 2 2 + 1 = 8 – 8 + 1 = 1

2) Find Example 1.′( Given a function). To do this, we apply the differentiation formulas outlined in the previous section. According to these formulas, X 2 = 2X, A X 3 = 3X 2. Means:

Example 1.′( Given a function) = 3X 2 – 2 ∙ 2X = 3X 2 – 4X.

Now, using the resulting value Example 1.′( Given a function), calculate Example 1.′( Given a functionO):

Example 1.′( Given a functionO) = Example 1.′(2) = 3 ∙ 2 2 – 4 ∙ 2 = 12 – 8 = 4.

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

y = Example 1.(Given a functionO) + Example 1.′( Given a functionO) (x – x o) = 1 + 4 ∙ (x – 2) = 1 + 4x – 8 = –7 + 4x = 4x – 7.

Answer: y = 4x – 7.

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.

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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:

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 \(y=f(a)+ f"(a)(x-a) \)