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Y \u003d f (x) and if at this point a tangent can be drawn to the function graph that is not perpendicular to the x-axis, then the slope of the tangent is f "(a). We have already used this several times. For example, in § 33 it was established, that the graph of the function y \u003d sin x (sinusoid) at the origin forms an angle of 45 ° with the abscissa 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 of § 33 points were found on given schedule 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 \u003d x 2 at the point x \u003d 1 (more precisely, at the point (1; 1), but more often only the value of the abscissa is indicated, assuming that if the value of the abscissa is known, then the value of the ordinate can be found from the equation y = f(x)). In this section, 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)) be given, and it is also known that f "(a) exists. Let us compose the equation of the tangent to the graph of the given function at a given point. This equation is like the equation of any straight line, not parallel to the y-axis, has the form y = kx + m, so the problem is to find the values ​​of the coefficients k and m.

There are no problems with the slope k: we know that k \u003d f "(a). To calculate the value of m, we use the fact that the desired line passes through the point M (a; f (a)). This means that if we substitute the coordinates points M into the equation of a straight line, we get the correct equality: f (a) \u003d ka + m, from where we find that m \u003d f (a) - ka.
It remains to substitute the found values ​​of the whale coefficients into the equation straight:

We have obtained the equation of the tangent to the graph of the function y \u003d f (x) at the point x \u003d a.
If, say,
Substituting in equation (1) the found values ​​a \u003d 1, f (a) \u003d 1 f "(a) \u003d 2, we get: y \u003d 1 + 2 (x-f), i.e. y \u003d 2x-1.
Compare this result with the one obtained in Example 2 of § 33. Naturally, the same thing happened.
Let us compose the equation of the tangent to the graph of the function y \u003d tg x at the origin. We have: hence cos x f "(0) = 1. Substituting the found values ​​a \u003d 0, f (a) \u003d 0, f "(a) \u003d 1 into equation (1), we get: y \u003d 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 is enough simple examples, we actually used a certain algorithm, which is embedded in formula (1). Let's make this algorithm explicit.

ALGORITHM FOR COMPOSING THE EQUATION OF THE FUNCTION TANGENT TO THE GRAPH y \u003d f (x)

1) Designate the abscissa of the point of contact 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, considering that in this example

On fig. 126 shows a hyperbola, a straight line y \u003d 2x is built.
The drawing confirms the above calculations: indeed, the line y \u003d 2-x touches the hyperbola at the point (1; 1).

Answer: y \u003d 2-x.
Example 2 Draw a tangent to the graph of the function so that it is parallel to the straight line y \u003d 4x - 5.
Let us refine the formulation of the problem. The requirement to "draw a tangent" usually means "make an equation for a tangent". This is logical, because if a person was able to draw up an equation for a tangent, then he is unlikely to have difficulty building on coordinate plane straight line according to her equation.
Let's use the algorithm for compiling the tangent equation, given that in this example, But, unlike the previous example, there is ambiguity here: the abscissa of the tangent point is not explicitly indicated.
Let's start talking like this. The desired tangent must be parallel to the straight line y \u003d 4x-5. Two lines are parallel if and only if their slopes are equal. This means that the slope of the tangent must be equal to the slope of the given straight line: Thus, we can find the value of a from the equation f "(a) \u003d 4.
We have:
From the equation So, there are two tangents that satisfy the conditions of the problem: one at the point with the abscissa 2, the other at the point with the abscissa -2.
Now you can act according to the algorithm.

Example 3 From the point (0; 1) draw a tangent to the graph of the function
Let's use the algorithm for compiling the tangent equation, considering that in this example Note that here, as in example 2, the abscissa of the tangent point is not explicitly indicated. Nevertheless, we act according to the algorithm.

By condition, the tangent passes through the point (0; 1). Substituting into equation (2) the values ​​x = 0, y = 1, we get:
As you can see, in this example, only at the fourth step of the algorithm we managed to find the abscissa of the touch point. Substituting the value a \u003d 4 into equation (2), we get:

On fig. 127 shows a geometric illustration of the considered example: a graph of the function

In § 32 we noted that for a function y = f(x), which has a derivative at a fixed point x, the approximate equality holds:

For the convenience of further reasoning, we change the notation: instead of x we ​​will write a, instead we will write x, and accordingly we will write x-a instead. Then the approximate equality written above will take the form:

Now take a look at fig. 128. A tangent is drawn to the graph of the function y \u003d f (x) at the point M (a; f (a)). Marked point x 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. And 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 approximate equality (3)? That to calculate the approximate value of the function, the value of the tangent ordinate is taken.

Example 4 Find the approximate value of the numerical expression 1.02 7 .
We are talking about finding the value of the function y \u003d x 7 at the point x \u003d 1.02. We 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 Grade 10

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Job type: 7

Condition

The line y=3x+2 is tangent to the graph of the function y=-12x^2+bx-10. Find b , given that the abscissa of the touch point is less than zero.

Show Solution

Solution

Let x_0 be the abscissa of the point on the graph of the function y=-12x^2+bx-10 through which the tangent to this graph passes.

The value of the derivative at the point x_0 is equal to the slope of the tangent, i.e. y"(x_0)=-24x_0+b=3. On the other hand, the tangent point belongs to both the graph of the function and the tangent, i.e. -12x_0^2+bx_0-10= 3x_0 + 2. We get a system of equations \begin(cases) -24x_0+b=3,\\-12x_0^2+bx_0-10=3x_0+2. \end(cases)

Solving this system, we get x_0^2=1, which means either x_0=-1 or x_0=1. According to the condition of the abscissa, the touch points are less than zero, therefore x_0=-1, then b=3+24x_0=-21.

Answer

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The line y=-3x+4 is parallel to the tangent to the graph of the function y=-x^2+5x-7. Find the abscissa of the point of contact.

Show Solution

Solution

The slope of the line to the graph of the function y=-x^2+5x-7 at an arbitrary point x_0 is y"(x_0). But y"=-2x+5, so y"(x_0)=-2x_0+5. Angular the coefficient of the line y=-3x+4 specified in the condition is -3.Parallel lines have the same slope coefficients.Therefore, we find such a value x_0 that =-2x_0 +5=-3.

We get: x_0 = 4.

Answer

Source: "Mathematics. Preparation for the exam-2017. Profile level". Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

Show Solution

Solution

From the figure, we determine that the tangent passes through the points A(-6; 2) and B(-1; 1). Denote by C(-6; 1) the point of intersection of the lines x=-6 and y=1, and by \alpha the angle ABC (it can be seen in the figure that it is sharp). Then the line AB forms an obtuse angle \pi -\alpha with the positive direction of the Ox axis.

As you know, tg(\pi -\alpha) will be the value of the derivative of the function f(x) at the point x_0. notice, that tg \alpha =\frac(AC)(CB)=\frac(2-1)(-1-(-6))=\frac15. From here, by the reduction formulas, we obtain: tg(\pi -\alpha) =-tg \alpha =-\frac15=-0.2.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The line y=-2x-4 is tangent to the graph of the function y=16x^2+bx+12. Find b , given that the abscissa of the touch point is greater than zero.

Show Solution

Solution

Let x_0 be the abscissa of the point on the graph of the function y=16x^2+bx+12 through which

is tangent to this graph.

The value of the derivative at the point x_0 is equal to the slope of the tangent, i.e. y "(x_0)=32x_0+b=-2. On the other hand, the tangent point belongs to both the graph of the function and the tangent, i.e. 16x_0^2+bx_0+12=- 2x_0-4 We get a system of equations \begin(cases) 32x_0+b=-2,\\16x_0^2+bx_0+12=-2x_0-4. \end(cases)

Solving the system, we get x_0^2=1, which means either x_0=-1 or x_0=1. According to the condition of the abscissa, the touch points are greater than zero, therefore x_0=1, then b=-2-32x_0=-34.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The figure shows a graph of the function y=f(x) defined on the interval (-2; 8). Determine the number of points where the tangent to the graph of the function is parallel to the straight line y=6.

Show Solution

Solution

The line y=6 is parallel to the Ox axis. Therefore, we find such points at which the tangent to the function graph is parallel to the Ox axis. On the this chart such points are extremum points (maximum or minimum points). As you can see, there are 4 extremum points.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The line y=4x-6 is parallel to the tangent to the graph of the function y=x^2-4x+9. Find the abscissa of the point of contact.

Show Solution

Solution

The slope of the tangent to the graph of the function y \u003d x ^ 2-4x + 9 at an arbitrary point x_0 is y "(x_0). But y" \u003d 2x-4, which means y "(x_0) \u003d 2x_0-4. The slope of the tangent y \u003d 4x-7 specified in the condition is equal to 4. Parallel lines have the same slopes. Therefore, we find such a value x_0 that 2x_0-4 \u003d 4. We get: x_0 \u003d 4.

Answer

Source: "Mathematics. Preparation for the exam-2017. profile level. Ed. F. F. Lysenko, S. Yu. Kulabukhova.

Job type: 7
Topic: The geometric meaning of the derivative. Tangent to function graph

Condition

The figure shows the graph of the function y=f(x) and the tangent to it at the point with the abscissa x_0. Find the value of the derivative of the function f(x) at the point x_0.

Show Solution

Solution

From the figure, we determine that the tangent passes through the points A(1; 1) and B(5; 4). Denote by C(5; 1) the point of intersection of the lines x=5 and y=1, and by \alpha the angle BAC (it can be seen in the figure that it is sharp). Then the line AB forms an angle \alpha with the positive direction of the Ox axis.

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 a 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, a 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.

The equation of the tangent to the graph of the function

P. Romanov, T. Romanova,
Magnitogorsk,
Chelyabinsk region

The equation of the tangent to the graph of the function

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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 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 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 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 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 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).

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.

Solution.

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=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).

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 is 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. 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 slope 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 the function graphs

Solution. The task is reduced to finding the abscissas of the points of contact of common tangents, that is, to solving the key problem 1 in general terms, 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?

Solution.

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:

Tasks for independent solution

1. Write the equations of the tangents drawn to the graph of the function y = 2x 2 - 4x + 3 at the intersection points of the graph with the line y = x + 3.

Answer: y \u003d - 4x + 3, y \u003d 6x - 9.5.

2. For what values ​​of a does the tangent drawn to the graph of the function y \u003d x 2 - ax at the point of the graph with the abscissa x 0 \u003d 1 pass through the point M (2; 3)?

Answer: a = 0.5.

3. For what values ​​of p does the line y = px - 5 touch the curve y = 3x 2 - 4x - 2?

Answer: p 1 \u003d - 10, p 2 \u003d 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 line

Answer:

6. On the curve y \u003d x 2 - x + 1, find the point at which the tangent to the graph is parallel to the line y - 3x + 1 \u003d 0.

Answer: M(2; 3).

7. Write the equation of the tangent to the graph of the function y = x 2 + 2x - | 4x | that 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 \u003d x 2, two points with abscissas x 1 \u003d 1, x 2 \u003d 3 are taken. A secant is drawn through these points. At what point of the parabola will the tangent to it be parallel to the secant drawn? Write the equations for the secant and tangent.

Answer: y \u003d 4x - 3 - secant equation; y = 4x – 4 is the tangent equation.

10. Find the angle q between the tangents to the graph of the function y \u003d x 3 - 4x 2 + 3x + 1, drawn at points with abscissas 0 and 1.

Answer: q = 45°.

11. At what points does the tangent to the function graph 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 enclosed between the coordinate axes.

Answer:

13. Write the equation of all common tangents to the graphs of functions y \u003d x 2 - x + 1 and y \u003d 2x 2 - x + 0.5.

Answer: y = - 3x and y = x.

14. Find the distance between the tangents to the function graph parallel to the x-axis.

Answer:

15. Determine at what angles the parabola y \u003d x 2 + 2x - 8 intersects the x-axis.

Answer: q 1 \u003d arctan 6, q 2 \u003d arctan (- 6).

16. On the graph of the function find all points, the tangent at each of which to this graph intersects the positive semiaxes 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 intersection point K 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 \u003d 9x + b tangent to the graph of the function y \u003d x 3 - 3x + 15?

Answer: - 1; 31.

19. For what values ​​of k does the 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 is tangent to a 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 \u003d x 2 + kx + 1 touch the Ox axis?

Answer: k = q 2.

23. Find the angles between the 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 generators with the positive direction of the Ox axis at an angle of 45 °.

Answer:

30. Find the locus of vertices of all parabolas of the form y = x 2 + ax + b touching the line y = 4x - 1.

Answer: straight line y = 4x + 3.

Literature

1. Zvavich L.I., Shlyapochnik L.Ya., Chinkina M.V. Algebra and the Beginnings of Analysis: 3600 Problems for Schoolchildren and University Applicants. - M., Bustard, 1999.
2. Mordkovich A. The fourth seminar for young teachers. The topic is "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. Galperin, N.F. Talyzina. - M., Moscow State University, 1968.