Calculator is the equation of the tangent to the graph of a given function. Equation of the tangent to the graph of a function
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 in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
In this way, you can conduct your own training and/or training of your younger brothers or sisters, while the level of education in the field of solving problems increases.
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Enter the function expression \(f(x)\) and the number \(a\) Find tangent equation It was discovered that some scripts necessary to solve this problem were not loaded, and the program may not work.
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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 \(y=f(a)+ f"(a)(x-a) \)
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.
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 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 basic knowledge and skills on 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 very in a broad sense A system is understood as a set of interconnected interacting elements with 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 general equation tangent 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.
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 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 be the 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 us 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 problem comes down to finding the abscissa of the points of tangency of common tangents, that is, to solving key problem 1 in general view, drawing up a system of equations and its subsequent solution (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 tasks, requiring 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 problem 1) to find 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?
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: 
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 abscissa axis, then the angular coefficient of the tangent is equal to f"(a). We have already used this several times. For example, in § 33 it was established that that the graph of the function y = sin x (sinusoid) at the origin forms an angle of 45° with the x-axis (more precisely, the tangent to the graph at the origin makes an angle of 45° with the positive direction of the x-axis), and in example 5 § 33 points were found on schedule given functions, in which the tangent is parallel to the x-axis. In example 2 of § 33, an equation was drawn up for the tangent to the graph of the function y = x 2 at point x = 1 (more precisely, at point (1; 1), but more often only the abscissa value is indicated, believing that if the abscissa value is known, then the ordinate value can be found from the equation y = f(x)). In this section we will develop an algorithm for composing a tangent equation to the graph of any function.
Let the function y = f(x) and the point M (a; f(a)) be given, and it is also known that f"(a) exists. Let us compose an equation for the tangent to the graph of a given function at a given point. This equation is like the equation of any a straight line not parallel to the ordinate axis has the form y = kx+m, so the task is to find the values of the coefficients k and m.
There are no problems with the angular coefficient k: we know that 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 point M into the equation of the straight line, we obtain the correct equality: f(a) = ka+m, from which we find that m = f(a) - ka.
It remains to substitute the found values of the kit coefficients into equation direct:
We have obtained the equation for the tangent to the graph of the function y = f(x) at the point x=a.
If, say,
Substituting the found values a = 1, f(a) = 1 f"(a) = 2 into equation (1), we obtain: y = 1+2(x-f), i.e. y = 2x-1.
Compare this result with that obtained in example 2 from § 33. Naturally, the same thing happened.
Let's create an equation for the tangent to the graph of the function y = tan x at the origin. We have: 
this means cos x f"(0) = 1. Substituting the found values a = 0, f(a) = 0, f"(a) = 1 into equation (1), we obtain: y = x.
That is why we drew the tangentoid in § 15 (see Fig. 62) through the origin of coordinates at an angle of 45° to the abscissa axis.
Solving these enough simple examples, we actually used a certain algorithm, which is contained in formula (1). Let's make this algorithm explicit.
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 1 (a).
3) Find f"(x) and calculate f"(a).
4) Substitute the found numbers a, f(a), (a) into formula (1).
Example 1. Write an equation for the tangent to the graph of the function at the point x = 1.
Let's use the algorithm, taking into account that in this example
In Fig. 126 a hyperbola is depicted, a straight line y = 2 is constructed.
The drawing confirms the above calculations: indeed, the line y = 2 touches the hyperbola at the point (1; 1).
Answer: y = 2- x.
Example 2. Draw a tangent to the graph of the function so that it is parallel to the line y = 4x - 5.
Let us clarify the formulation of the problem. The requirement to "draw a tangent" usually means "to form an equation for the tangent." This is logical, because if a person was able to create an equation for a tangent, then he is unlikely to have difficulty constructing on coordinate plane straight line according to her equation.
Let's use the algorithm for composing the tangent equation, taking into account that in this example But, unlike the previous example, there is ambiguity: the abscissa of the tangent point is not explicitly indicated.
Let's start thinking like this. The desired tangent must be parallel to the straight line y = 4x-5. 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: 
Thus, we can find the value of a from the equation f"(a) = 4.
We have: 
From the equation This means that there are two tangents that satisfy the conditions of the problem: one at the point with abscissa 2, the other at the point with abscissa -2.
Now you can follow the algorithm.
Example 3. From point (0; 1) draw a tangent to the graph of the function
Let's use the algorithm for composing the tangent equation, taking into account that in this example, Note that here, as in example 2, the abscissa of the tangent point is not explicitly indicated. Nevertheless, we follow the algorithm.
By condition, the tangent passes through the point (0; 1). Substituting the values x = 0, y = 1 into equation (2), we obtain: 
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 =4 into equation (2), we obtain:
In Fig. 127 presents a geometric illustration of the considered example: a graph of the function is plotted
In § 32 we noted that for a function y = f(x) having a derivative at a fixed point x, the approximate equality is valid:
For the convenience of further reasoning, let us change the notation: instead of x we will write a, instead of we will write x and, accordingly, instead of we will write x-a. Then the approximate equality written above will take the form:
Now look at fig. 128. A tangent is drawn to the graph of the function y = f(x) at point M (a; f (a)). The point x is marked on the x-axis close to a. It is clear that f(x) is the ordinate of the graph of the function at the specified point x. What is f(a) + f"(a) (x-a)? This is the ordinate of the tangent corresponding to the same point x - see formula (1). What is the meaning of the approximate equality (3)? The fact that To calculate the approximate value of the function, take the ordinate value of the tangent.
Example 4. Find approximate value numerical expression 1,02 7 .
We are talking about finding the value of the function y = x 7 at the point x = 1.02. Let us use formula (3), taking into account that in this example
As a result we get:
If we use a calculator, we get: 1.02 7 = 1.148685667...
As you can see, the approximation accuracy is quite acceptable.
Answer: 1,02 7 =1,14.
A.G. Mordkovich Algebra 10th grade
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