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On the present stage development of education as 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 forces, abilities and talents is formed full-fledged knowledge and skills. In this regard, the problem of forming a system basic knowledge and skills for each topic school course mathematics is of great importance. At the same time, full-fledged skills should be didactic purpose not individual tasks, but a carefully thought-out system of them. In the very broad 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 problems of finding the tangent equation are reduced to the need to select from the set (sheaf, family) of lines those of them that satisfy certain requirement– are tangent to the graph of some 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) slope factor(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 methodical technique, in our opinion, allows students to quickly and easily realize where in the general tangent equation the coordinates of the current point are written, and where are the touch points.

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 general equation 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 consistent solution each of the key tasks with the help of 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 stage-by-stage formation 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 = 3 - abscissa of the touch point of one of the sides 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 problem is reduced to finding the abscissas of the common tangent points, that is, to solving key problem 1 in general view, compiling a system of equations and its subsequent solution (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 purpose of the considered tasks is to prepare students for self-recognition of the type of key task when solving more challenging tasks requiring certain research skills (the ability to analyze, compare, generalize, put forward a hypothesis, etc.). Such tasks include any task in which key task included as a component. Consider as an example the problem ( inverse problem 1) to find a function by 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:

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. As 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 a 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.

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 general education schools in preparation for control work and exams, when testing knowledge before the exam, 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 soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.

Thus, you can carry out your own training and/or training their younger brothers or sisters, while the level of education in the field of tasks being solved increases.

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 schedule 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 geometric sense 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 formulate the equation of the tangent to the graph given function in given point. This equation, like the equation of any straight line not parallel to the y-axis, has the form y = 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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