Find the tangent equation. Online calculator. Equation of a straight tangent to the graph of a function at a given point
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
The article was published with the support of the ITAKA+ Hotel Complex. When staying in the city of shipbuilders Severodvinsk, you will not encounter the problem of finding temporary housing. , Online hotel complex“ITHAKA+” http://itakaplus.ru, you can easily and quickly rent an apartment in the city, for any period of time, with a daily payment.
On modern stage development of education, one of its main tasks is the formation of a creatively thinking personality. The ability for creativity in students can be developed only if they are systematically involved in the basics of research activities. The foundation for students to use their creative powers, abilities and talents is formed full-fledged knowledge and skills. In this regard, the problem of forming a system 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 tangent equation 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 tangent point is given, the first part of the solution is reduced to key problem 1.
1. a = 3 – abscissa of the point of tangency of one of the sides 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.
Consider the following figure:
It depicts a certain function y = f(x), which is differentiable at point a. Point M with coordinates (a; f(a)) is marked. A secant MR is drawn through an arbitrary point P(a + ∆x; f(a + ∆x)) of the graph.
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 of 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 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: f’(x0)
, then the equation will take the following form: y =
*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). Let's consider
next 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. to the graph of the function y = f(x):
1. Determine x0.
2. Calculate f(x0).
3. Calculate f’(x)
Let a function f be given, which at some point x 0 has a finite derivative f (x 0). Then the straight line passing through the point (x 0 ; f (x 0)), having an angular coefficient f ’(x 0), is called a tangent.
What happens if the derivative does not exist at the point x 0? There are two options:
- There is no tangent to the graph either. Classic example- function y = |x | at point (0; 0).
- The tangent becomes vertical. This is true, for example, for the function y = arcsin x at the point (1; π /2).
to him.
Any non-vertical straight line is given by an equation of the form y = kx + b, where k is the slope. The tangent is no exception, and in order to create its equation at some point x 0, it is enough to know the value of the function and the derivative at this point.
So, let a function y = f (x) be given, which has a derivative y = f ’(x) on the segment. Then at any point x 0 ∈ (a ; b) a tangent can be drawn to the graph of this function, which is given by the equation:
y = f ’(x 0) (x − x 0) + f (x 0)
Here f ’(x 0) is the value of the derivative at point x 0, and f (x 0) is the value of the function itself.
Task. Given the function y = x 3 . Write an equation for the tangent to the graph of this function at the point x 0 = 2.
Tangent equation: y = f ’(x 0) · (x − x 0) + f (x 0). The point x 0 = 2 is given to us, but the values f (x 0) and f ’(x 0) will have to be calculated.
First, let's find the value of the function. Everything is easy here: f (x 0) = f (2) = 2 3 = 8;
Now let’s find the derivative: f ’(x) = (x 3)’ = 3x 2;
We substitute x 0 = 2 into the derivative: f ’(x 0) = f ’(2) = 3 2 2 = 12;
In total we get: y = 12 · (x − 2) + 8 = 12x − 24 + 8 = 12x − 16.
This is the tangent equation.
Task. Write an equation for the tangent to the graph of the function f (x) = 2sin x + 5 at point x 0 = π /2.
This time we will not describe each action in detail - we will only indicate the key steps. We have:
f (x 0) = f (π /2) = 2sin (π /2) + 5 = 2 + 5 = 7;
f ’(x) = (2sin x + 5)’ = 2cos x;
f ’(x 0) = f ’(π /2) = 2cos (π /2) = 0;
Tangent equation:
y = 0 · (x − π /2) + 7 ⇒ y = 7
IN the latter case the straight line turned out to be horizontal, because its angular coefficient k = 0. There is nothing wrong with this - we just stumbled upon an extremum point.
Example 1. Given a function f(x) = 3x 2 + 4x– 5. Let’s write the equation of the tangent to the graph of the function f(x) at the graph point with the abscissa x 0 = 1.
Solution. Derivative of a function f(x) exists for any x R . Let's find her:
= (3x 2 + 4x– 5)′ = 6 x + 4.
Then f(x 0) = f(1) = 2; (x 0) = = 10. The tangent equation has the form:
y = (x 0) (x – x 0) + f(x 0),
y = 10(x – 1) + 2,
y = 10x – 8.
Answer. y = 10x – 8.
Example 2. Given a function f(x) = x 3 – 3x 2 + 2x+ 5. Let’s write the equation of the tangent to the graph of the function f(x), parallel to the line y = 2x – 11.
Solution. Derivative of a function f(x) exists for any x R . Let's find her:
= (x 3 – 3x 2 + 2x+ 5)′ = 3 x 2 – 6x + 2.
Since the tangent to the graph of the function f(x) at the abscissa point x 0 is parallel to the line y = 2x– 11, then its slope is equal to 2, i.e. ( x 0) = 2. Let’s find this abscissa from the condition that 3 x– 6x 0 + 2 = 2. This equality is valid only when x 0 = 0 and at x 0 = 2. Since in both cases f(x 0) = 5, then straight y = 2x + b touches the graph of the function either at the point (0; 5) or at the point (2; 5).
In the first case, the numerical equality 5 = 2×0 + is true b, where b= 5, and in the second case the numerical equality 5 = 2×2 + is true b, where b = 1.
So there are two tangents y = 2x+ 5 and y = 2x+ 1 to the graph of the function f(x), parallel to the line y = 2x – 11.
Answer. y = 2x + 5, y = 2x + 1.
Example 3. Given a function f(x) = x 2 – 6x+ 7. Let’s write the equation of the tangent to the graph of the function f(x), passing through the point A (2; –5).
Solution. Because f(2) –5, then point A does not belong to the graph of the function f(x). Let x 0 - abscissa of the tangent point.
Derivative of a function f(x) exists for any x R . Let's find her:
= (x 2 – 6x+ 1)′ = 2 x – 6.
Then f(x 0) = x– 6x 0 + 7; (x 0) = 2x 0 – 6. The tangent equation has the form:
y = (2x 0 – 6)(x – x 0) + x– 6x+ 7,
y = (2x 0 – 6)x– x+ 7.
Since the point A belongs to the tangent, then the numerical equality is true
–5 = (2x 0 – 6)×2– x+ 7,
where x 0 = 0 or x 0 = 4. This means that through the point A you can draw two tangents to the graph of the function f(x).
If x 0 = 0, then the tangent equation has the form y = –6x+ 7. If x 0 = 4, then the tangent equation has the form y = 2x – 9.
Answer. y = –6x + 7, y = 2x – 9.
Example 4. Functions given f(x) = x 2 – 2x+ 2 and g(x) = –x 2 – 3. Let’s write the equation of the common tangent to the graphs of these functions.
Solution. Let x 1 - abscissa of the point of tangency of the desired line with the graph of the function f(x), A x 2 - abscissa of the point of tangency of the same line with the graph of the function g(x).
Derivative of a function f(x) exists for any x R . Let's find her:
= (x 2 – 2x+ 2)′ = 2 x – 2.
Then f(x 1) = x– 2x 1 + 2; (x 1) = 2x 1 – 2. The tangent equation has the form:
y = (2x 1 – 2)(x – x 1) + x– 2x 1 + 2,
y = (2x 1 – 2)x – x+ 2. (1)
Let's find the derivative of the function g(x):
= (–x 2 – 3)′ = –2 x.
Instructions
We determine the angular coefficient of the tangent to the curve at point M.
The curve representing the graph of the function y = f(x) is continuous in a certain neighborhood of the point M (including the point M itself).
If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, the geometric meaning of the derivative becomes clear - calculation slope tangent.
Find the abscissa value of the tangent point, which is denoted by the letter “a”. If it coincides with a given tangent point, then "a" will be its x-coordinate. Determine the value functions f(a) by substituting into the equation functions abscissa value.
Determine the first derivative of the equation functions f’(x) and substitute the value of point “a” into it.
Take the general tangent equation, which is defined as y = f(a) = f (a)(x – a), and substitute the found values of a, f(a), f "(a) into it. As a result, the solution to the graph will be found and tangent.
Solve the problem in a different way if the given tangent point does not coincide with the tangent point. In this case, it is necessary to substitute “a” instead of numbers in the tangent equation. After this, instead of the letters “x” and “y”, substitute the value of the coordinates of the given point. Solve the resulting equation in which “a” is the unknown. Plug the resulting value into the tangent equation.
Write an equation for a tangent with the letter “a” if the problem statement specifies the equation functions and the equation of a parallel line relative to the desired tangent. After this we need the derivative functions, to the coordinate at point “a”. Substitute the appropriate value into the tangent equation and solve the function.
