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 point of the graph with the abscissa x 0 = 1.

Decision. Function derivative f(x) exists for any x R . Let's find it:

= (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.

Decision. Function derivative f(x) exists for any x R . Let's find it:

= (x 3 – 3x 2 + 2x+ 5)′ = 3 x 2 – 6x + 2.

Since the tangent to the graph of the function f(x) at the point with the abscissa x 0 is parallel to the line y = 2x– 11, then its slope is 2, i.e. ( x 0) = 2. Find this abscissa from the condition that 3 x– 6x 0 + 2 = 2. This equality is valid only for x 0 = 0 and x 0 = 2. Since in both cases f(x 0) = 5, then the straight line 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 is true 5 = 2×0 + b, where b= 5, and in the second case, the numerical equality is true 5 = 2 × 2 + 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).

Decision. Because f(2) –5, then the point A does not belong to the graph of the function f(x). Let be x 0 - abscissa of the touch point.

Function derivative f(x) exists for any x R . Let's find it:

= (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 it is possible to 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 Given functions 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.

Decision. Let be x 1 - abscissa of the point of contact of the desired line with the graph of the function f(x), a x 2 - abscissa of the point of contact of the same line with the graph of the function g(x).

Function derivative f(x) exists for any x R . Let's find it:

= (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.

In this article, we will analyze all types of problems for finding

Let's remember geometric meaning of the derivative: if a tangent is drawn to the graph of a function at a point, then the slope of the tangent (equal to the tangent of the angle between the tangent and the positive direction of the axis) is equal to the derivative of the function at the point.

Take an arbitrary point on the tangent with coordinates :

And consider a right triangle:

In this triangle

From here

This is the equation of the tangent drawn to the graph of the function at the point.

To write the equation of the tangent, we only need to know the equation of the function and the point where the tangent is drawn. Then we can find and .

There are three main types of tangent equation problems.

1. Given a point of contact

2. Given the coefficient of slope of the tangent, that is, the value of the derivative of the function at the point.

3. Given the coordinates of the point through which the tangent is drawn, but which is not a tangent point.

Let's look at each type of problem.

one . Write the equation of the tangent to the graph of the function at the point .

.

b) Find the value of the derivative at the point . First we find the derivative of the function

Substitute the found values ​​into the tangent equation:

Let's open the brackets on the right side of the equation. We get:

Answer: .

2. Find the abscissas of the points at which the functions tangent to the graph parallel to the x-axis.

If the tangent is parallel to the x-axis, then the angle between the tangent and the positive direction of the axis is zero, so the tangent of the slope of the tangent is zero. So the value of the derivative of the function at the points of contact is zero.

a) Find the derivative of the function .

b) Equate the derivative to zero and find the values ​​in which the tangent is parallel to the axis:

We equate each factor to zero, we get:

Answer: 0;3;5

3 . Write equations of tangents to the graph of a function , parallel straight .

The tangent is parallel to the line. The slope of this straight line is -1. Since the tangent is parallel to this line, therefore, the slope of the tangent is also -1. That is we know the slope of the tangent, and thus the value of the derivative at the point of contact.

This is the second type of problem for finding the tangent equation.

So, we are given a function and the value of the derivative at the point of contact.

a) Find the points at which the derivative of the function is equal to -1.

First, let's find the derivative equation.

Let's equate the derivative to the number -1.

Find the value of the function at the point .

(by condition)

.

b) Find the equation of the tangent to the graph of the function at the point .

Find the value of the function at the point .

(by condition).

Substitute these values ​​into the tangent equation:

.

Answer:

4 . Write an equation for a tangent to a curve , passing through a point

First, check if the point is not a touch point. If the point is a tangent point, then it belongs to the graph of the function, and its coordinates must satisfy the equation of the function. Substitute the coordinates of the point in the equation of the function.

Title="(!LANG:1sqrt(8-3^2)">. Мы получили под корнем отрицательное число, равенство не верно, и точка не принадлежит графику функции и !} is not a point of contact.

This is the last type of problem for finding the tangent equation. First thing we need to find the abscissa of the point of contact.

Let's find the value.

Let be the point of contact. The point belongs to the tangent to the graph of the function . If we substitute the coordinates of this point into the tangent equation, we get the correct equality:

.

The value of the function at the point is .

Find the value of the derivative of the function at the point .

Let's find the derivative of the function first. This .

The derivative at a point is .

Let us substitute the expressions for and into the equation of the tangent. We get the equation for:

Let's solve this equation.

Reduce the numerator and denominator of the fraction by 2:

We bring the right side of the equation to a common denominator. We get:

Simplify the numerator of the fraction and multiply both parts by - this expression is strictly greater than zero.

We get the equation

Let's solve it. To do this, we square both parts and go to the system.

Title="(!LANG:delim(lbrace)(matrix(2)(1)((64-48(x_0)+9(x_0)^2=8-(x_0)^2) (8-3x_0>=0 ) ))( )">!}

Let's solve the first equation.

We solve the quadratic equation, we get

The second root does not satisfy the condition title="(!LANG:8-3x_0>=0">, следовательно, у нас только одна точка касания и её абсцисса равна .!}

Let's write the equation of the tangent to the curve at the point . To do this, we substitute the value in the equation We already recorded it.

Answer:
.

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 forces, 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 their carefully thought-out system. In the broadest sense, a system is understood as a set of interrelated interacting elements that has integrity and a stable structure.

Consider a methodology for teaching students how to draw up an equation of a tangent to a function graph. In essence, all tasks for finding the tangent equation are reduced to the need to select from the set (sheaf, family) of lines those of them 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 bundle of 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 tasks:

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

Learning to solve problems on a tangent 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), in connection with which the tangent equation takes the form

y \u003d f (a) + f "(a) (x - a)

(compare with y \u003d f (x 0) + f "(x 0) (x - x 0)). This methodological technique, in our opinion, allows students to quickly and easily realize where the coordinates of the current point are written in the general tangent equation, and where are the points of contact.

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

1. Designate with the letter a the abscissa of the point of contact.
2. Find f(a).
3. Find f "(x) and f "(a).
4. Substitute the found numbers a, f (a), f "(a) into the general equation of the tangent y \u003d f (a) \u003d f "(a) (x - a).

This algorithm can be compiled on the basis of students' independent selection of operations and the sequence of their execution.

Practice has shown that the consistent solution of each of the key tasks using the algorithm allows you to form the ability to write the equation of the tangent to the graph of the function in stages, and the steps of the algorithm serve as strong 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. Equate the tangent to the graph of the function at the point M(3; – 2).

Decision. The point M(3; – 2) is the point of contact, since

1. a = 3 - abscissa of the touch point.
2. f(3) = – 2.
3. f "(x) \u003d x 2 - 4, f "(3) \u003d 5.
y \u003d - 2 + 5 (x - 3), y \u003d 5x - 17 is the tangent equation.

Task 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).

Decision. The 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) \u003d - 2x - 4, f "(a) \u003d - 2a - 4.
4. y \u003d - 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 \u003d - 2, then the tangent equation has the form y \u003d 6.

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

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

Task 3. Write the equations of all tangents to the graph of the function y \u003d x 3 - 3x 2 + 3, parallel to the line y \u003d 9x + 1.

1. a - abscissa of the touch point.
2. f(a) = a 3 - 3a 2 + 3.
3. f "(x) \u003d 3x 2 - 6x, f "(a) \u003d 3a 2 - 6a.

But, on the other hand, f "(a) \u003d 9 (parallelism condition). So, we need to solve the equation 3a 2 - 6a \u003d 9. Its roots a \u003d - 1, a \u003d 3 (Fig. 3).

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

y = 9x + 8 is the tangent equation;

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

y = 9x – 24 is the tangent equation.

Task 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).

Decision. From the condition f "(a) \u003d tg 45 ° we find a: a - 3 \u003d 1 ^ a \u003d 4.

1. a = 4 - abscissa of the touch point.
2. f(4) = 8 - 12 + 1 = - 3.
3. f "(4) \u003d 4 - 3 \u003d 1.
4. y \u003d - 3 + 1 (x - 4).

y \u003d x - 7 - the equation of the tangent.

It is easy to show that the solution of any other problem is reduced to the solution of one or several key problems. Consider the following two problems as an example.

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

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

1. a \u003d 3 - the abscissa of the point of contact of one of the sides of the right angle.
2. f(3) = 1.
3. f "(x) \u003d 4x - 5, f "(3) \u003d 7.
4. y \u003d 1 + 7 (x - 3), y \u003d 7x - 20 - the equation of the first tangent.

Let a be the slope 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. Find

This means that the slope of the second tangent is .

The further solution is reduced to the key task 3.

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

1. - abscissa of the second point of contact.
2.
3.
4.
is the equation of the second tangent.

Note. The angular coefficient of the tangent can be found easier 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 function graphs

Decision. The task is reduced to finding the abscissas of the points of contact of the common tangents, that is, to solving the key problem 1 in a general form, compiling a system of equations and then solving it (Fig. 6).

1. Let a be the abscissa of the touch 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 \u003d a 2 + a + 1 + (2a + 1) (x - a) \u003d (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 the tangents are common, then

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

The main goal of the tasks considered is to prepare students for self-recognition of the type of key task when solving more complex tasks 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 \u003d x and y \u003d - 2x tangent to the graph of the function y \u003d x 2 + bx + c?

Let t be the abscissa of the point of contact of the line y = x with the parabola y = x 2 + bx + c; p is the abscissa of the point of contact of the 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 .

Compose and solve a system of equations

Answer:

Consider the following figure:

It shows some function y = f(x) that is differentiable at the point a. Marked point M with coordinates (a; f(a)). Through an arbitrary point P(a + ∆x; f(a + ∆x)) of the graph, a secant MP is drawn.

If now the point P is shifted along the graph to the point M, then the straight line MP will rotate around the 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 function graph

The tangent to the graph of the function is the limiting position of the secant when 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 slope 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 the function f differentiable at the point x0 is some straight line passing through the point (x0;f(x0)) and having a slope f’(x0).

Tangent equation

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

Since our slope 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) \u003d x 3 - 2 * x 2 + 1 at the point x \u003d 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 like terms, we get: y = 4*x - 7.

Answer: y = 4*x - 7.

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

1. Determine x0.

2. Calculate f(x0).

3. Calculate f'(x)

This math 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 preparing for tests and exams, when testing knowledge before the Unified State Examination, and 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 your math or algebra homework done as quickly as possible? In this case, you can also use our programs with a detailed solution.

In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.

If you need to find the derivative of a function, then for this we have the Find Derivative task.

If you are not familiar with the rules for introducing functions, we recommend that you familiarize yourself with them.

Enter the function expression \(f(x)\) and the number \(a\)

f(x)=
a=

Find Tangent Equation

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A bit of theory.

Slope of a straight line

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 \(kThe equation of the tangent to the graph of the function

If the point M (a; f (a)) belongs to the graph of the function y \u003d 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 x-axis, then from the geometric meaning of the derivative it follows that the slope of the tangent is equal to f "(a). Next, we will develop an algorithm for compiling the equation of the tangent to the graph of any function.

Let the function y \u003d f (x) and the point M (a; f (a)) on the graph of this function be given; let it be known that f "(a) exists. Let's compose the equation of 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 y-axis, has the form y \u003d kx + b, so the problem is to find the values ​​of the coefficients k and b.

Everything is clear with the slope k: it is known that k \u003d 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 get the correct equality: \ (f (a) \u003d ka + b \), i.e. \ (b \u003d f (a) - ka \).

It remains to substitute the found values ​​of the coefficients k and b into the equation of a straight line:

$$ y=kx+b $$ $$ y=kx+ f(a) - ka $$ $$ y=f(a)+ k(x-a) $$ $$ y=f(a)+ f"(a )(x-a) $$

We received the equation of the tangent to the graph of the 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 point of contact 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 \u003d f (a) + f "(a) (x-a) \)

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