Write down the equation of the tangent to the graph. The equation of the tangent to the graph of the function

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

Other definition: this is the limit position of the secant at Δ x→0.

Explanation: Take a line that intersects the curve at two points: BUT And b(see picture). This is a secant. We will turn it clockwise until it has only one common point with a curve. So we get a tangent.

Strict definition of a tangent:

Tangent to function graph f, differentiable at a point xabout, is a line passing through the point ( xabout; f(xabout)) and having slope f′( xabout).

The slope has a straight line y=kx +b. Coefficient k and is slope factor this straight line.

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

k = tgα

Here the angle α is the angle between the line y=kx +b and the positive (i.e. counterclockwise) direction of the x-axis. It is called angle of inclination straight(Fig.1 and 2).

If the angle of inclination is straight y=kx +b acute, then the slope is positive number. The graph increases (Fig. 1).

If the angle of inclination is straight y=kx +b obtuse, then the slope is negative number. The graph is decreasing (Fig. 2).

If the line is parallel to the x-axis, then the slope of the line is zero. In this case, the slope of the line is also zero (since the tangent of zero is zero). The straight line equation 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 x-axis, then the straight line is given by the equality x=c, where c- some real number (Fig. 4).

The equation of the tangent to the graph of the functiony = f(x) at the point xabout:

Example : Let's find the equation of the tangent to the graph of the function f(x) = x 3 – 2x 2 + 1 at the point with abscissa 2.

Solution .

We follow the algorithm.

1) Touch point xabout equals 2. Calculate f(xabout):

f(xabout) = f(2) = 2 3 – 2 ∙ 2 2 + 1 = 8 – 8 + 1 = 1

2) Find f′( x). To do this, we use the differentiation formulas outlined in the previous section. According to these formulas, X 2 = 2X, but X 3 = 3X 2. Means:

f′( x) = 3X 2 – 2 ∙ 2X = 3X 2 – 4X.

Now, using the resulting value f′( x), calculate f′( xabout):

f′( xabout) = f′(2) = 3 ∙ 2 2 – 4 ∙ 2 = 12 – 8 = 4.

3) So, we have all the necessary data: xabout = 2, f(xabout) = 1, f ′( xabout) = 4. We substitute these numbers into the tangent equation and find the final solution:

y= f(xabout) + f′( xabout) (x – x o) \u003d 1 + 4 ∙ (x - 2) \u003d 1 + 4x - 8 \u003d -7 + 4x \u003d 4x - 7.

Answer: y \u003d 4x - 7.

Tangent is a straight line , which touches the graph of the function at one point and all points of which are at the smallest distance from the graph of the function. Therefore, the tangent passes tangent to the function graph at a certain angle and several tangents cannot pass through the tangent point under different angles. The tangent equations and the equations of the normal to the graph of the function are compiled using the derivative.

The tangent equation is derived from the straight line equation .

We derive the equation of the tangent, and then the equation of the normal to the graph of the function.

y = kx + b .

In him k- angular coefficient.

From here we get the following entry:

y - y 0 = k(x - x 0 ) .

Derivative value f "(x 0 ) functions y = f(x) at the point x0 equal to the slope k=tg φ tangent to the graph of a function drawn through a point M0 (x 0 , y 0 ) , where y0 = f(x 0 ) . This is what geometric meaning of the derivative .

Thus, we can replace k on the f "(x 0 ) and get the following the equation of the tangent to the graph of the function :

y - y 0 = f "(x 0 )(x - x 0 ) .

In tasks for compiling the equation of a tangent to the graph of a function (and we will soon move on to them), it is required to bring the equation obtained from the above formula to general equation of a straight line. To do this, you need to transfer all the letters and numbers to the left side of the equation, and leave zero on the right side.

Now about the normal equation. Normal is a straight line passing through the tangent point to the graph of the function perpendicular to the tangent. Normal Equation :

(x - x 0 ) + f "(x 0 )(y - y 0 ) = 0

To warm up the first example, you are asked to solve it yourself, and then look at the solution. There is every reason to hope that this task will not be a "cold shower" for our readers.

Example 0. Compose the equation of the tangent and the equation of the normal to the graph of the function at a point M (1, 1) .

Example 1 Compose the equation of the tangent and the equation of the normal to the graph of the function if the abscissa of the touch point is .

Let's find the derivative of the function:

Now we have everything that needs to be substituted into the entry given in the theoretical reference in order to obtain the tangent equation. We get

In this example, we were lucky: the slope turned out to be equal to zero, so there was no need to separately bring the equation to a general form. Now we can write the normal equation:

In the figure below: graph of the burgundy color function, tangent green color, the normal is orange.

The next example is also not complicated: the function, as in the previous one, is also a polynomial, but the angular coefficient will not be equal to zero, so one more step will be added - bringing the equation to a general form.

Example 2

Solution. Let's find the ordinate of the touch point:

Let's find the derivative of the function:

.

Let's find the value of the derivative at the point of contact, that is, the slope of the tangent:

We substitute all the data obtained into the "blank formula" and get the tangent equation:

We bring the equation to a general form (we collect all letters and numbers other than zero on the left side, and leave zero on the right side):

We compose the equation of the normal:

Example 3 Compose the equation of the tangent and the equation of the normal to the graph of the function if the abscissa of the point of contact is .

Solution. Let's find the ordinate of the touch point:

Let's find the derivative of the function:

.

Let's find the value of the derivative at the point of contact, that is, the slope of the tangent:

.

We find the equation of the tangent:

Before bringing the equation to a general form, you need to “combine” it a bit: multiply term by term by 4. We do this and bring the equation to a general form:

We compose the equation of the normal:

Example 4 Compose the equation of the tangent and the equation of the normal to the graph of the function if the abscissa of the point of contact is .

Solution. Let's find the ordinate of the touch point:

.

Let's find the derivative of the function:

Let's find the value of the derivative at the point of contact, that is, the slope of the tangent:

.

We get the tangent equation:

We bring the equation to a general form:

We compose the equation of the normal:

A common mistake when writing tangent and normal equations is not to notice that the function given in the example is complex and calculate its derivative as the derivative of a simple function. The following examples are already complex functions(the corresponding lesson will open in a new window).

Example 5 Compose the equation of the tangent and the equation of the normal to the graph of the function if the abscissa of the point of contact is .

Solution. Let's find the ordinate of the touch point:

Attention! This function is complex, since the argument of the tangent (2 x) is itself a function. Therefore, we find the derivative of a function as the derivative of a complex function.

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 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 interconnected 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, 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 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 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).

Solution. 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).

Solution. 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).

Solution. 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 the 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).

Solution. 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

Solution. 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 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:

Let a function f be given, which at some point x 0 has a finite derivative f (x 0). Then the line passing through the point (x 0; f (x 0)), which has a slope f '(x 0), is called a tangent.

But what happens if the derivative at the point x 0 does not exist? There are two options:

1. The tangent to the graph also does not exist. Classic example- function y = |x | at the point (0; 0).
2. The tangent becomes vertical. This is true, for example, for the function y = arcsin x at the point (1; π /2).

Tangent equation

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 compose 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 be given y \u003d f (x), which has a derivative y \u003d 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 \u003d f '(x 0) (x - x 0) + f (x 0)

Here f ’(x 0) is the value of the derivative at the point x 0, and f (x 0) is the value of the function itself.

A task. Given a 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 \u003d 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) \u003d (x 3) ' \u003d 3x 2;
Substitute in the derivative x 0 = 2: f '(x 0) = f '(2) = 3 2 2 = 12;
So we get: y = 12 (x - 2) + 8 = 12x - 24 + 8 = 12x - 16.
This is the tangent equation.

A task. Compose the equation of the tangent to the graph of the function f (x) \u003d 2sin x + 5 at the point x 0 \u003d π / 2.

This time we will not describe in detail each action - we will only indicate the key steps. We have:

f (x 0) \u003d f (π / 2) \u003d 2sin (π / 2) + 5 \u003d 2 + 5 \u003d 7;
f '(x) \u003d (2sin x + 5) ' \u003d 2cos x;
f '(x 0) \u003d f '(π / 2) \u003d 2cos (π / 2) \u003d 0;

Tangent equation:

y = 0 (x − π /2) + 7 ⇒ y = 7

In the latter case, the line turned out to be horizontal, because its slope k = 0. There is nothing wrong with that - we just stumbled upon an extremum point.