In this article we will analyze all types of problems to find
Let's remember geometric meaning of derivative: if a tangent is drawn to the graph of a function at a point, then the slope coefficient 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.
Let's 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 tangent equation, we only need to know the equation of the function and the point at which 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. The tangent slope coefficient is given, that is, the value of the derivative of the function at the point.
3. Given are the coordinates of the point through which the tangent is drawn, but which is not the point of tangency.
Let's look at each type of task.
1 . Write the equation of the tangent to the graph of the function 
at the point .
.
b) Find the value of the derivative at point . First, let's find the derivative of the function
Let's 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 abscissa of the points at which the functions are tangent to the graph parallel to the x-axis.
If the tangent is parallel to the x-axis, therefore the angle between the tangent and the positive direction of the axis is zero, therefore the tangent of the tangent angle is zero. This means that the value of the derivative of the function at the points of tangency is zero.
a) Find the derivative of the function.
b) Let's equate the derivative to zero and find the values in which the tangent is parallel to the axis:
Equating each factor to zero, we get:
Answer: 0;3;5
3. Write equations for tangents to the graph of a function , parallel straight .
A tangent is parallel to a line. The slope of this 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, thereby, derivative value at the point of tangency.
This is the second type of problem to find the tangent equation.
So, we are given the function and the value of the derivative at the point of tangency.
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.
Let's 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 point .
Let's find the value of the function at the point.
(by condition).
Let's substitute these values into the tangent equation:
.
Answer:
4 . Write the equation of the tangent to the curve , passing through a point
First, let's check whether the point is a tangent point. If a point is a tangent point, then it belongs to the graph of the function, and its coordinates must satisfy the equation of the function. Let's substitute the coordinates of the point into the equation of the function.
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This is the last type of problem to find the tangent equation. First thing we need to find the abscissa of the tangent point.
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 a point is 
.
Let's find the value of the derivative of the function at the point.
First, let's find the derivative of the function. This .
The derivative at a point is equal to 
.
Let's substitute the expressions for and into the tangent equation. We get the equation for:
Let's solve this equation.
Reduce the numerator and denominator of the fraction by 2:
Let's bring the right side of the equation to a common denominator. We get:
Let's simplify the numerator of the fraction and multiply both sides by - this expression is strictly greater than zero.
We get the equation
Let's solve it. To do this, let's square both parts and move on to the system.
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Let's solve the first equation.
Let's solve the quadratic equation, we get
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Let's write the equation of the tangent to the curve at the point. To do this, substitute the value into 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 powers, 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 a carefully thought-out system of them. In the broadest sense, a system is understood as a set of interconnected interacting elements that have 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 tangent equation 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 the general tangent equation 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 point of tangency, 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 tangent 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 of the 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 task comes down to finding the abscissa of the tangent points of common tangents, that is, solving key problem 1 in general form, drawing up a system of equations and then solving it (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 problems 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 = 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: 
The article provides a detailed explanation of the definitions, the geometric meaning of the derivative with graphic notations. The equation of a tangent line will be considered with examples, the equations of a tangent to 2nd order curves will be found.
Yandex.RTB R-A-339285-1 Definition 1
The angle of inclination of the straight line y = k x + b is called angle α, which is measured from the positive direction of the x axis to the straight line y = k x + b in the positive direction.
In the figure, the x direction is indicated by a green arrow and a green arc, and the angle of inclination by a red arc. The blue line refers to the straight line.
Definition 2
The slope of the straight line y = k x + b is called the numerical coefficient k.
The angular coefficient is equal to the tangent of the straight line, in other words k = t g α.
- The angle of inclination of a straight line is equal to 0 only if it is parallel about x and the slope is equal to zero, because the tangent of zero is equal to 0. This means that the form of the equation will be y = b.
- If the angle of inclination of the straight line y = k x + b is acute, then the conditions 0 are satisfied< α < π 2 или 0 ° < α < 90 ° . Отсюда имеем, что значение углового коэффициента k считается положительным числом, потому как значение тангенс удовлетворяет условию t g α >0, and there is an increase in the graph.
- If α = π 2, then the location of the line is perpendicular to x. Equality is specified by x = c with the value c being a real number.
- If the angle of inclination of the straight line y = k x + b is obtuse, then it corresponds to the conditions π 2< α < π или 90 ° < α < 180 ° , значение углового коэффициента k принимает отрицательное значение, а график убывает.
A secant is a line that passes through 2 points of the function f (x). In other words, a secant is a straight line that is drawn through any two points on the graph of a given function.
The figure shows that A B is a secant, and f (x) is a black curve, α is a red arc, indicating the angle of inclination of the secant.
When the angular coefficient of a straight line is equal to the tangent of the angle of inclination, it is clear that the tangent of a right triangle A B C can be found by the ratio of the opposite side to the adjacent one.
Definition 4
We get a formula for finding a secant of the form:
k = t g α = B C A C = f (x B) - f x A x B - x A, where the abscissas of points A and B are the values x A, x B, and f (x A), f (x B) are the values functions at these points.
Obviously, the angular coefficient of the secant is determined using the equality k = f (x B) - f (x A) x B - x A or k = f (x A) - f (x B) x A - x B, and the equation must be written as y = f (x B) - f (x A) x B - x A x - x A + f (x A) or
y = f (x A) - f (x B) x A - x B x - x B + f (x B) .
The secant divides the graph visually into 3 parts: to the left of point A, from A to B, to the right of B. The figure below shows that there are three secants that are considered coincident, that is, they are set using a similar equation.
By definition, it is clear that the straight line and its secant in this case coincide.
A secant can intersect the graph of a given function multiple times. If there is an equation of the form y = 0 for a secant, then the number of points of intersection with the sinusoid is infinite.
Definition 5
Tangent to the graph of the function f (x) at point x 0 ; f (x 0) is a straight line passing through a given point x 0; f (x 0), with the presence of a segment that has many x values close to x 0.
Example 1
Let's take a closer look at the example below. Then it is clear that the line defined by the function y = x + 1 is considered tangent to y = 2 x at the point with coordinates (1; 2). For clarity, it is necessary to consider graphs with values close to (1; 2). The function y = 2 x is shown in black, the blue line is the tangent line, and the red dot is the intersection point.
Obviously, y = 2 x merges with the line y = x + 1.
To determine the tangent, we should consider the behavior of the tangent A B as point B approaches point A infinitely. For clarity, we present a drawing.
The secant A B, indicated by the blue line, tends to the position of the tangent itself, and the angle of inclination of the secant α will begin to tend to the angle of inclination of the tangent itself α x.
Definition 6
The tangent to the graph of the function y = f (x) at point A is considered to be the limiting position of the secant A B as B tends to A, that is, B → A.
Now let's move on to consider the geometric meaning of the derivative of a function at a point.
Let's move on to considering the secant A B for the function f (x), where A and B with coordinates x 0, f (x 0) and x 0 + ∆ x, f (x 0 + ∆ x), and ∆ x is denoted as the increment of the argument . Now the function will take the form ∆ y = ∆ f (x) = f (x 0 + ∆ x) - f (∆ x) . For clarity, let's give an example of a drawing.
Consider the resulting right triangle A B C. We use the definition of tangent to solve, that is, we obtain the relation ∆ y ∆ x = t g α . From the definition of a tangent it follows that lim ∆ x → 0 ∆ y ∆ x = t g α x . According to the rule of the derivative at a point, we have that the derivative f (x) at the point x 0 is called the limit of the ratio of the increment of the function to the increment of the argument, where ∆ x → 0, then we denote it as f (x 0) = lim ∆ x → 0 ∆ y ∆ x .
It follows that f " (x 0) = lim ∆ x → 0 ∆ y ∆ x = t g α x = k x, where k x is denoted as the slope of the tangent.
That is, we find that f ' (x) can exist at point x 0, and like the tangent to a given graph of the function at the point of tangency equal to x 0, f 0 (x 0), where the value of the slope of the tangent at the point is equal to the derivative at point x 0 . Then we get that k x = f " (x 0) .
The geometric meaning of the derivative of a function at a point is that it gives the concept of the existence of a tangent to the graph at the same point.
To write the equation of any straight line on a plane, it is necessary to have an angular coefficient with the point through which it passes. Its notation is taken to be x 0 at intersection.
The tangent equation to the graph of the function y = f (x) at the point x 0, f 0 (x 0) takes the form y = f "(x 0) x - x 0 + f (x 0).
This means that the final value of the derivative f "(x 0) can determine the position of the tangent, that is, vertically, provided lim x → x 0 + 0 f "(x) = ∞ and lim x → x 0 - 0 f "(x ) = ∞ or absence at all under the condition lim x → x 0 + 0 f " (x) ≠ lim x → x 0 - 0 f " (x) .
The location of the tangent depends on the value of its angular coefficient k x = f "(x 0). When parallel to the o x axis, we obtain that k k = 0, when parallel to o y - k x = ∞, and the form of the tangent equation x = x 0 increases with k x > 0, decreases as k x< 0 .
Example 2
Compile an equation for the tangent to the graph of the function y = e x + 1 + x 3 3 - 6 - 3 3 x - 17 - 3 3 at the point with coordinates (1; 3) and determine the angle of inclination.
Solution
By condition, we have that the function is defined for all real numbers. We find that the point with the coordinates specified by the condition, (1; 3) is a point of tangency, then x 0 = - 1, f (x 0) = - 3.
It is necessary to find the derivative at the point with value - 1. We get that
y " = e x + 1 + x 3 3 - 6 - 3 3 x - 17 - 3 3 " = = e x + 1 " + x 3 3 " - 6 - 3 3 x " - 17 - 3 3 " = e x + 1 + x 2 - 6 - 3 3 y " (x 0) = y " (- 1) = e - 1 + 1 + - 1 2 - 6 - 3 3 = 3 3
The value of f' (x) at the point of tangency is the slope of the tangent, which is equal to the tangent of the slope.
Then k x = t g α x = y " (x 0) = 3 3
It follows that α x = a r c t g 3 3 = π 6
Answer: the tangent equation takes the form
y = f " (x 0) x - x 0 + f (x 0) y = 3 3 (x + 1) - 3 y = 3 3 x - 9 - 3 3
For clarity, we give an example in a graphic illustration.
Black color is used for the graph of the original function, blue color is the image of the tangent, and the red dot is the point of tangency. The figure on the right shows an enlarged view.
Example 3
Determine the existence of a tangent to the graph of a given function
y = 3 · x - 1 5 + 1 at the point with coordinates (1 ; 1) . Write an equation and determine the angle of inclination.
Solution
By condition, we have that the domain of definition of a given function is considered to be the set of all real numbers.
Let's move on to finding the derivative
y " = 3 x - 1 5 + 1 " = 3 1 5 (x - 1) 1 5 - 1 = 3 5 1 (x - 1) 4 5
If x 0 = 1, then f' (x) is undefined, but the limits are written as lim x → 1 + 0 3 5 1 (x - 1) 4 5 = 3 5 1 (+ 0) 4 5 = 3 5 · 1 + 0 = + ∞ and lim x → 1 - 0 3 5 · 1 (x - 1) 4 5 = 3 5 · 1 (- 0) 4 5 = 3 5 · 1 + 0 = + ∞ , which means the existence vertical tangent at point (1; 1).
Answer: the equation will take the form x = 1, where the angle of inclination will be equal to π 2.
For clarity, let's depict it graphically.
Example 4
Find the points on the graph of the function y = 1 15 x + 2 3 - 4 5 x 2 - 16 5 x - 26 5 + 3 x + 2, where
- There is no tangent;
- The tangent is parallel to x;
- The tangent is parallel to the line y = 8 5 x + 4.
Solution
It is necessary to pay attention to the scope of definition. By condition, we have that the function is defined on the set of all real numbers. We expand the module and solve the system with intervals x ∈ - ∞ ; 2 and [ - 2 ; + ∞) . We get that
y = - 1 15 x 3 + 18 x 2 + 105 x + 176 , x ∈ - ∞ ; - 2 1 15 x 3 - 6 x 2 + 9 x + 12 , x ∈ [ - 2 ; + ∞)
It is necessary to differentiate the function. We have that
y " = - 1 15 x 3 + 18 x 2 + 105 x + 176 " , x ∈ - ∞ ; - 2 1 15 x 3 - 6 x 2 + 9 x + 12 ", x ∈ [ - 2 ; + ∞) ⇔ y " = - 1 5 (x 2 + 12 x + 35) , x ∈ - ∞ ; - 2 1 5 x 2 - 4 x + 3 , x ∈ [ - 2 ; + ∞)
When x = − 2, then the derivative does not exist because the one-sided limits are not equal at that point:
lim x → - 2 - 0 y " (x) = lim x → - 2 - 0 - 1 5 (x 2 + 12 x + 35 = - 1 5 (- 2) 2 + 12 (- 2) + 35 = - 3 lim x → - 2 + 0 y " (x) = lim x → - 2 + 0 1 5 (x 2 - 4 x + 3) = 1 5 - 2 2 - 4 - 2 + 3 = 3
We calculate the value of the function at the point x = - 2, where we get that
- y (- 2) = 1 15 - 2 + 2 3 - 4 5 (- 2) 2 - 16 5 (- 2) - 26 5 + 3 - 2 + 2 = - 2, that is, the tangent at the point (- 2; - 2) will not exist.
- The tangent is parallel to x when the slope is zero. Then k x = t g α x = f "(x 0). That is, it is necessary to find the values of such x when the derivative of the function turns it to zero. That is, the values of f ' (x) will be the points of tangency, where the tangent is parallel to x .
When x ∈ - ∞ ; - 2, then - 1 5 (x 2 + 12 x + 35) = 0, and for x ∈ (- 2; + ∞) we get 1 5 (x 2 - 4 x + 3) = 0.
1 5 (x 2 + 12 x + 35) = 0 D = 12 2 - 4 35 = 144 - 140 = 4 x 1 = - 12 + 4 2 = - 5 ∈ - ∞ ; - 2 x 2 = - 12 - 4 2 = - 7 ∈ - ∞ ; - 2 1 5 (x 2 - 4 x + 3) = 0 D = 4 2 - 4 · 3 = 4 x 3 = 4 - 4 2 = 1 ∈ - 2 ; + ∞ x 4 = 4 + 4 2 = 3 ∈ - 2 ; +∞
Calculate the corresponding function values
y 1 = y - 5 = 1 15 - 5 + 2 3 - 4 5 - 5 2 - 16 5 - 5 - 26 5 + 3 - 5 + 2 = 8 5 y 2 = y (- 7) = 1 15 - 7 + 2 3 - 4 5 (- 7) 2 - 16 5 - 7 - 26 5 + 3 - 7 + 2 = 4 3 y 3 = y (1) = 1 15 1 + 2 3 - 4 5 1 2 - 16 5 1 - 26 5 + 3 1 + 2 = 8 5 y 4 = y (3) = 1 15 3 + 2 3 - 4 5 3 2 - 16 5 3 - 26 5 + 3 3 + 2 = 4 3
Hence - 5; 8 5, - 4; 4 3, 1; 8 5, 3; 4 3 are considered to be the required points of the function graph.
Let's look at a graphical representation of the solution.
The black line is the graph of the function, the red dots are the tangency points.
- When the lines are parallel, the angular coefficients are equal. Then it is necessary to search for points on the function graph where the slope will be equal to the value 8 5. To do this, you need to solve an equation of the form y "(x) = 8 5. Then, if x ∈ - ∞; - 2, we obtain that - 1 5 (x 2 + 12 x + 35) = 8 5, and if x ∈ ( - 2 ; + ∞), then 1 5 (x 2 - 4 x + 3) = 8 5.
The first equation has no roots since the discriminant is less than zero. Let's write down that
1 5 x 2 + 12 x + 35 = 8 5 x 2 + 12 x + 43 = 0 D = 12 2 - 4 43 = - 28< 0
Another equation has two real roots, then
1 5 (x 2 - 4 x + 3) = 8 5 x 2 - 4 x - 5 = 0 D = 4 2 - 4 · (- 5) = 36 x 1 = 4 - 36 2 = - 1 ∈ - 2 ; + ∞ x 2 = 4 + 36 2 = 5 ∈ - 2 ; +∞
Let's move on to finding the values of the function. We get that
y 1 = y (- 1) = 1 15 - 1 + 2 3 - 4 5 (- 1) 2 - 16 5 (- 1) - 26 5 + 3 - 1 + 2 = 4 15 y 2 = y (5) = 1 15 5 + 2 3 - 4 5 5 2 - 16 5 5 - 26 5 + 3 5 + 2 = 8 3
Points with values - 1; 4 15, 5; 8 3 are the points at which the tangents are parallel to the line y = 8 5 x + 4.
Answer: black line – graph of the function, red line – graph of y = 8 5 x + 4, blue line – tangents at points - 1; 4 15, 5; 8 3.
There may be an infinite number of tangents for given functions.
Example 5
Write the equations of all available tangents of the function y = 3 cos 3 2 x - π 4 - 1 3, which are located perpendicular to the straight line y = - 2 x + 1 2.
Solution
To compile the tangent equation, it is necessary to find the coefficient and coordinates of the tangent point, based on the condition of perpendicularity of the lines. The definition is as follows: the product of angular coefficients that are perpendicular to straight lines is equal to - 1, that is, written as k x · k ⊥ = - 1. From the condition we have that the angular coefficient is located perpendicular to the line and is equal to k ⊥ = - 2, then k x = - 1 k ⊥ = - 1 - 2 = 1 2.
Now you need to find the coordinates of the touch points. You need to find x and then its value for a given function. Note that from the geometric meaning of the derivative at the point
x 0 we obtain that k x = y "(x 0). From this equality we find the values of x for the points of contact.
We get that
y " (x 0) = 3 cos 3 2 x 0 - π 4 - 1 3 " = 3 - sin 3 2 x 0 - π 4 3 2 x 0 - π 4 " = = - 3 sin 3 2 x 0 - π 4 3 2 = - 9 2 sin 3 2 x 0 - π 4 ⇒ k x = y " (x 0) ⇔ - 9 2 sin 3 2 x 0 - π 4 = 1 2 ⇒ sin 3 2 x 0 - π 4 = - 1 9
This trigonometric equation will be used to calculate the ordinates of the tangent points.
3 2 x 0 - π 4 = a r c sin - 1 9 + 2 πk or 3 2 x 0 - π 4 = π - a r c sin - 1 9 + 2 πk
3 2 x 0 - π 4 = - a r c sin 1 9 + 2 πk or 3 2 x 0 - π 4 = π + a r c sin 1 9 + 2 πk
x 0 = 2 3 π 4 - a r c sin 1 9 + 2 πk or x 0 = 2 3 5 π 4 + a r c sin 1 9 + 2 πk , k ∈ Z
Z is a set of integers.
x points of contact have been found. Now you need to move on to searching for the values of y:
y 0 = 3 cos 3 2 x 0 - π 4 - 1 3
y 0 = 3 1 - sin 2 3 2 x 0 - π 4 - 1 3 or y 0 = 3 - 1 - sin 2 3 2 x 0 - π 4 - 1 3
y 0 = 3 1 - - 1 9 2 - 1 3 or y 0 = 3 - 1 - - 1 9 2 - 1 3
y 0 = 4 5 - 1 3 or y 0 = - 4 5 + 1 3
From this we obtain that 2 3 π 4 - a r c sin 1 9 + 2 πk ; 4 5 - 1 3 , 2 3 5 π 4 + a r c sin 1 9 + 2 πk ; - 4 5 + 1 3 are the points of tangency.
Answer: the necessary equations will be written as
y = 1 2 x - 2 3 π 4 - a r c sin 1 9 + 2 πk + 4 5 - 1 3 , y = 1 2 x - 2 3 5 π 4 + a r c sin 1 9 + 2 πk - 4 5 + 1 3 , k ∈ Z
For a visual representation, consider a function and a tangent on a coordinate line.
The figure shows that the function is located on the interval [ - 10 ; 10 ], where the black line is the graph of the function, the blue lines are tangents, which are located perpendicular to the given line of the form y = - 2 x + 1 2. Red dots are touch points.
The canonical equations of 2nd order curves are not single-valued functions. Tangent equations for them are compiled according to known schemes.
Tangent to a circle
To define a circle with center at point x c e n t e r ; y c e n t e r and radius R, apply the formula x - x c e n t e r 2 + y - y c e n t e r 2 = R 2 .
This equality can be written as a union of two functions:
y = R 2 - x - x c e n t e r 2 + y c e n t e r y = - R 2 - x - x c e n t e r 2 + y c e n t e r
The first function is located at the top, and the second at the bottom, as shown in the figure.
To compile the equation of a circle at the point x 0; y 0 , which is located in the upper or lower semicircle, you should find the equation of the graph of a function of the form y = R 2 - x - x c e n t e r 2 + y c e n t e r or y = - R 2 - x - x c e n t e r 2 + y c e n t e r at the indicated point.
When at points x c e n t e r ; y c e n t e r + R and x c e n t e r ; y c e n t e r - R tangents can be given by the equations y = y c e n t e r + R and y = y c e n t e r - R , and at points x c e n t e r + R ; y c e n t e r and
x c e n t e r - R ; y c e n t e r will be parallel to o y, then we obtain equations of the form x = x c e n t e r + R and x = x c e n t e r - R .
Tangent to an ellipse
When the ellipse has a center at x c e n t e r ; y c e n t e r with semi-axes a and b, then it can be specified using the equation x - x c e n t e r 2 a 2 + y - y c e n t e r 2 b 2 = 1.
An ellipse and a circle can be denoted by combining two functions, namely the upper and lower half-ellipse. Then we get that
y = b a · a 2 - (x - x c e n t e r) 2 + y c e n t e r y = - b a · a 2 - (x - x c e n t e r) 2 + y c e n t e r
If the tangents are located at the vertices of the ellipse, then they are parallel about x or about y. Below, for clarity, consider the figure.
Example 6
Write the equation of the tangent to the ellipse x - 3 2 4 + y - 5 2 25 = 1 at points with values of x equal to x = 2.
Solution
It is necessary to find the tangent points that correspond to the value x = 2. We substitute into the existing equation of the ellipse and find that
x - 3 2 4 x = 2 + y - 5 2 25 = 1 1 4 + y - 5 2 25 = 1 ⇒ y - 5 2 = 3 4 25 ⇒ y = ± 5 3 2 + 5
Then 2 ; 5 3 2 + 5 and 2; - 5 3 2 + 5 are the tangent points that belong to the upper and lower half-ellipse.
Let's move on to finding and solving the equation of the ellipse with respect to y. We get that
x - 3 2 4 + y - 5 2 25 = 1 y - 5 2 25 = 1 - x - 3 2 4 (y - 5) 2 = 25 1 - x - 3 2 4 y - 5 = ± 5 1 - x - 3 2 4 y = 5 ± 5 2 4 - x - 3 2
Obviously, the upper half-ellipse is specified using a function of the form y = 5 + 5 2 4 - x - 3 2, and the lower half ellipse y = 5 - 5 2 4 - x - 3 2.
Let's apply a standard algorithm to create an equation for a tangent to the graph of a function at a point. Let us write that the equation for the first tangent at point 2; 5 3 2 + 5 will look like
y " = 5 + 5 2 4 - x - 3 2 " = 5 2 1 2 4 - (x - 3) 2 4 - (x - 3) 2 " = = - 5 2 x - 3 4 - ( x - 3) 2 ⇒ y " (x 0) = y " (2) = - 5 2 2 - 3 4 - (2 - 3) 2 = 5 2 3 ⇒ y = y " (x 0) x - x 0 + y 0 ⇔ y = 5 2 3 (x - 2) + 5 3 2 + 5
We find that the equation of the second tangent with a value at the point
2 ; - 5 3 2 + 5 takes the form
y " = 5 - 5 2 4 - (x - 3) 2 " = - 5 2 1 2 4 - (x - 3) 2 4 - (x - 3) 2 " = = 5 2 x - 3 4 - (x - 3) 2 ⇒ y " (x 0) = y " (2) = 5 2 2 - 3 4 - (2 - 3) 2 = - 5 2 3 ⇒ y = y " (x 0) x - x 0 + y 0 ⇔ y = - 5 2 3 (x - 2) - 5 3 2 + 5
Graphically, tangents are designated as follows:
Tangent to hyperbole
When a hyperbola has a center at x c e n t e r ; y c e n t e r and vertices x c e n t e r + α ; y c e n t e r and x c e n t e r - α ; y c e n t e r , the inequality x - x c e n t e r 2 α 2 - y - y c e n t e r 2 b 2 = 1 takes place, if with vertices x c e n t e r ; y c e n t e r + b and x c e n t e r ; y c e n t e r - b , then is specified using the inequality x - x c e n t e r 2 α 2 - y - y c e n t e r 2 b 2 = - 1 .
A hyperbola can be represented as two combined functions of the form
y = b a · (x - x c e n t e r) 2 - a 2 + y c e n t e r y = - b a · (x - x c e n t e r) 2 - a 2 + y c e n t e r or y = b a · (x - x c e n t e r) 2 + a 2 + y c e n t e r y = - b a · (x - x c e n t e r) 2 + a 2 + y c e n t e r
In the first case we have that the tangents are parallel to y, and in the second they are parallel to x.
It follows that in order to find the equation of the tangent to a hyperbola, it is necessary to find out which function the point of tangency belongs to. To determine this, it is necessary to substitute into the equations and check for identity.
Example 7
Write an equation for the tangent to the hyperbola x - 3 2 4 - y + 3 2 9 = 1 at point 7; - 3 3 - 3 .
Solution
It is necessary to transform the solution record for finding a hyperbola using 2 functions. We get that
x - 3 2 4 - y + 3 2 9 = 1 ⇒ y + 3 2 9 = x - 3 2 4 - 1 ⇒ y + 3 2 = 9 x - 3 2 4 - 1 ⇒ y + 3 = 3 2 x - 3 2 - 4 and y + 3 = - 3 2 x - 3 2 - 4 ⇒ y = 3 2 x - 3 2 - 4 - 3 y = - 3 2 x - 3 2 - 4 - 3
It is necessary to identify which function a given point with coordinates 7 belongs to; - 3 3 - 3 .
Obviously, to check the first function it is necessary y (7) = 3 2 · (7 - 3) 2 - 4 - 3 = 3 3 - 3 ≠ - 3 3 - 3, then the point does not belong to the graph, since the equality does not hold.
For the second function we have that y (7) = - 3 2 · (7 - 3) 2 - 4 - 3 = - 3 3 - 3 ≠ - 3 3 - 3, which means the point belongs to the given graph. From here you should find the slope.
We get that
y " = - 3 2 (x - 3) 2 - 4 - 3 " = - 3 2 x - 3 (x - 3) 2 - 4 ⇒ k x = y " (x 0) = - 3 2 x 0 - 3 x 0 - 3 2 - 4 x 0 = 7 = - 3 2 7 - 3 7 - 3 2 - 4 = - 3
Answer: the tangent equation can be represented as
y = - 3 x - 7 - 3 3 - 3 = - 3 x + 4 3 - 3
It is clearly depicted like this:
Tangent to a parabola
To create an equation for the tangent to the parabola y = a x 2 + b x + c at the point x 0, y (x 0), you must use a standard algorithm, then the equation will take the form y = y "(x 0) x - x 0 + y ( x 0). Such a tangent at the vertex is parallel to x.
You should define the parabola x = a y 2 + b y + c as the union of two functions. Therefore, we need to solve the equation for y. We get that
x = a y 2 + b y + c ⇔ a y 2 + b y + c - x = 0 D = b 2 - 4 a (c - x) y = - b + b 2 - 4 a (c - x) 2 a y = - b - b 2 - 4 a (c - x) 2 a
Let's depict it graphically as:
To find out whether a point x 0, y (x 0) belongs to a function, proceed gently according to the standard algorithm. Such a tangent will be parallel to o y relative to the parabola.
Example 8
Write the equation of the tangent to the graph x - 2 y 2 - 5 y + 3 when we have a tangent angle of 150 °.
Solution
We begin the solution by representing the parabola as two functions. We get that
2 y 2 - 5 y + 3 - x = 0 D = (- 5) 2 - 4 · (- 2) · (3 - x) = 49 - 8 x y = 5 + 49 - 8 x - 4 y = 5 - 49 - 8 x - 4
The value of the slope is equal to the value of the derivative at point x 0 of this function and is equal to the tangent of the angle of inclination.
We get:
k x = y "(x 0) = t g α x = t g 150 ° = - 1 3
From here we determine the x value for the points of contact.
The first function will be written as
y " = 5 + 49 - 8 x - 4 " = 1 49 - 8 x ⇒ y " (x 0) = 1 49 - 8 x 0 = - 1 3 ⇔ 49 - 8 x 0 = - 3
Obviously, there are no real roots, since we got a negative value. We conclude that there is no tangent with an angle of 150° for such a function.
The second function will be written as
y " = 5 - 49 - 8 x - 4 " = - 1 49 - 8 x ⇒ y " (x 0) = - 1 49 - 8 x 0 = - 1 3 ⇔ 49 - 8 x 0 = - 3 x 0 = 23 4 ⇒ y (x 0) = 5 - 49 - 8 23 4 - 4 = - 5 + 3 4
We have that the points of contact are 23 4 ; - 5 + 3 4 .
Answer: the tangent equation takes the form
y = - 1 3 x - 23 4 + - 5 + 3 4
Let's depict it graphically this way:
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Tangent is a straight line passing through a point on the curve and coinciding with it at this point up to first order (Fig. 1).
Another definition: this is the limiting position of the secant at Δ x→0.
Explanation: Take a straight line intersecting the curve at two points: A And b(see picture). This is a secant. We will rotate it clockwise until it finds only one common point with the curve. This will give us a tangent.
Strict definition of tangent:
Tangent to the graph of a function f, differentiable at the point xO, is a straight line passing through the point ( xO; f(xO)) and having a slope f′( xO).
The slope has a straight line of the form y =kx +b. Coefficient k and is slope this straight line.
The angular coefficient is equal to the tangent of the acute angle formed by this straight line with the abscissa axis:
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Here angle α is the angle between the straight line y =kx +b and positive (that is, counterclockwise) direction of the x-axis. It is called angle of inclination of a straight line(Fig. 1 and 2). 
If the angle of inclination is straight y =kx +b acute, then the slope is a positive number. The graph is increasing (Fig. 1).
If the angle of inclination is straight y =kx +b is obtuse, then the slope is a negative number. The graph is decreasing (Fig. 2).
If the straight line is parallel to the x-axis, then the angle of inclination of the straight line is zero. In this case, the slope of the line is also zero (since the tangent of zero is zero). The equation of the straight line 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 abscissa axis, then the straight line is given by the equality x =c, Where c– some real number (Fig. 4).
Equation of the tangent to the graph of a functiony = f(x) at point xO:
Example: 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 xO is equal to 2. Calculate f(xO):
f(xO) = f(2) = 2 3 – 2 ∙ 2 2 + 1 = 8 – 8 + 1 = 1
2) Find f′( x). To do this, we apply 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′( xO):
f′( xO) = f′(2) = 3 ∙ 2 2 – 4 ∙ 2 = 12 – 8 = 4.
3) So, we have all the necessary data: xO = 2, f(xO) = 1, f ′( xO) = 4. Substitute these numbers into the tangent equation and find the final solution:
y = f(xO) + f′( xO) (x – x o) = 1 + 4 ∙ (x – 2) = 1 + 4x – 8 = –7 + 4x = 4x – 7.
Answer: y = 4x – 7.
