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 passes 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
Graphically depicted 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:
If you notice an error in the text, please highlight it and press Ctrl+Enter
A tangent is a straight line , which touches the graph of the function at one point and all points of which are at the shortest distance from the graph of the function. Therefore, the tangent passes tangent to the graph of the function at a certain angle, and several tangents at different angles cannot pass through the point of tangency. Tangent equations and normal equations to the graph of a function are constructed using the derivative.
The tangent equation is derived from the line equation .
Let us 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 geometric meaning of derivative .
Thus, we can replace k on f "(x 0 ) and get the following equation of the tangent to the graph of a function :
y - y 0 = f "(x 0 )(x - x 0 ) .
In problems involving composing the equation of a tangent to the graph of a function (and we will move on to them soon), it is required to reduce the equation obtained from the above formula to equation of a straight line in general form. To do this, you need to move 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 - this is a straight line passing through the point of tangency 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, you are asked to solve the first example 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. Create a tangent equation and a normal equation for the graph of a function at a point M (1, 1) .
Example 1. Write a tangent equation and a normal equation for the graph of a function 
, if the abscissa is tangent .
Let's find the derivative of the function:
Now we have everything that needs to be substituted into the entry given in the theoretical help to get the tangent equation. We get
In this example, we were lucky: the slope turned out to be zero, so there was no need to separately reduce the equation to its general form. Now we can create the normal equation:
In the figure below: the graph of the function is burgundy, the tangent is green, the normal is orange.
The next example is also not complicated: the function, as in the previous one, is also a polynomial, but the slope 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 tangent point:
Let's find the derivative of the function:
.
Let's find the value of the derivative at the point of tangency, that is, the slope of the tangent:
We substitute all the obtained data into the “blank formula” and get the tangent equation:
We bring the equation to its general form (we collect all letters and numbers other than zero on the left side, and leave zero on the right):
We compose the normal equation:
Example 3. Write the equation of the tangent and the equation of the normal to the graph of the function if the abscissa is the point of tangency.
Solution. Let's find the ordinate of the tangent point:
Let's find the derivative of the function:
.
Let's find the value of the derivative at the point of tangency, that is, the slope of the tangent:
.
We find the tangent equation:
Before bringing the equation to its general form, you need to “comb it” a little: multiply term by term by 4. We do this and bring the equation to its general form:
We compose the normal equation:
Example 4. Write the equation of the tangent and the equation of the normal to the graph of the function if the abscissa is the point of tangency.
Solution. Let's find the ordinate of the tangent point:
.
Let's find the derivative of the function:
Let's find the value of the derivative at the point of tangency, that is, the slope of the tangent:
.
We get the tangent equation:
We bring the equation to its general form:
We compose the normal equation:
A common mistake when writing tangent and normal equations is not to notice that the function given in the example is complex and to calculate its derivative as the derivative of a simple function. The following examples are already from complex functions(the corresponding lesson will open in a new window).
Example 5. Write the equation of the tangent and the equation of the normal to the graph of the function if the abscissa is the point of tangency.
Solution. Let's find the ordinate of the tangent point:
Attention! This function is complex, since the tangent argument (2 x) is itself a function. Therefore, we find the derivative of a function as the derivative of a complex function.
Equation of the tangent to the graph of a function
P. Romanov, T. Romanova,
Magnitogorsk,
Chelyabinsk region
Equation of the tangent to the graph of a function
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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 tangent point, 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.
Solution.
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 – 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's 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?
Solution.
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: 
Problems to solve independently
1. Write the equations of the tangents drawn to the graph of the function y = 2x 2 – 4x + 3 at the points of intersection of the graph with the line y = x + 3.
Answer: y = – 4x + 3, y = 6x – 9.5.
2. For what values of a does the tangent drawn to the graph of the function y = x 2 – ax at the point of the graph with the abscissa x 0 = 1 pass through the point M(2; 3)?
Answer: a = 0.5.
3. For what values of p does the straight line y = px – 5 touch the curve y = 3x 2 – 4x – 2?
Answer: p 1 = – 10, p 2 = 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 straight line
Answer:
6. On the curve y = x 2 – x + 1, find the point at which the tangent to the graph is parallel to the straight line y – 3x + 1 = 0.
Answer: M(2; 3).
7. Write the equation of the tangent to the graph of the function y = x 2 + 2x – | 4x |, which 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 = x 2, two points are taken with abscissas x 1 = 1, x 2 = 3. A secant is drawn through these points. At what point of the parabola will the tangent to it be parallel to the secant? Write the secant and tangent equations.
Answer: y = 4x – 3 – secant equation; y = 4x – 4 – tangent equation.
10. Find the angle q between the tangents to the graph of the function y = x 3 – 4x 2 + 3x + 1, drawn at the points with abscissas 0 and 1.
Answer: q = 45°.
11. At what points does the tangent to the graph of the function 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 between the coordinate axes.
Answer:
13. Write the equation of all common tangents to the graphs of the functions y = x 2 – x + 1 and y = 2x 2 – x + 0.5.
Answer: y = – 3x and y = x.
14. Find the distance between the tangents to the graph of the function 
parallel to the x-axis.
Answer:
15. Determine at what angles the parabola y = x 2 + 2x – 8 intersects the x-axis.
Answer: q 1 = arctan 6, q 2 = arctan (– 6).
16. Function graph 
find all points, the tangent at each of which to this graph intersects the positive semi-axes 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 point K of intersection 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 = 9x + b tangent to the graph of the function y = x 3 – 3x + 15?
Answer: – 1; 31.
19. For what values of k does the straight 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 touches the 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 = x 2 + kx + 1 touch the Ox axis?
Answer: k = d 2.
23. Find the angles between the straight 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 and the generators with the positive direction of the Ox axis at an angle of 45°.
Answer:
30. Find the locus of the vertices of all parabolas of the form y = x 2 + ax + b tangent to the line y = 4x – 1.
Answer: straight line y = 4x + 3.
Literature
1. Zvavich L.I., Shlyapochnik L.Ya., Chinkina M.V. Algebra and beginnings of analysis: 3600 problems for schoolchildren and those entering universities. – M., Bustard, 1999.
2. Mordkovich A. Seminar four for young teachers. Topic: 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. Galperina, N.F. Talyzina. – M., Moscow State University, 1968.
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:
|
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.
Y = f(x) and if at this point a tangent can be drawn to the graph of the function that is not perpendicular to the abscissa axis, then the angular coefficient of the tangent is equal to f"(a). We have already used this several times. For example, in § 33 it was established that that the graph of the function y = sin x (sinusoid) at the origin forms an angle of 45° with the x-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 § 33 points were found on schedule given 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 = x 2 at point x = 1 (more precisely, at point (1; 1), but more often only the abscissa value is indicated, believing that if the abscissa value is known, then the ordinate value can be found from the equation y = f(x)). In this section we will develop an algorithm for composing a tangent equation to the graph of any function.
Let the function y = f(x) and the point M (a; f(a)) be given, and it is also known that f"(a) exists. Let us compose an equation for the tangent to the graph of a given function at a given point. This equation is like the equation of any a straight line not parallel to the ordinate axis has the form y = kx+m, so the task is to find the values of the coefficients k and m.
There are no problems with the angular coefficient k: we know that k = f "(a). To calculate the value of m, 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 point M into the equation of the straight line, we obtain the correct equality: f(a) = ka+m, from which we find that m = f(a) - ka.
It remains to substitute the found values of the kit coefficients into the equation straight:
We have obtained the equation for the tangent to the graph of the function y = f(x) at the point x=a.
If, say,
Substituting the found values a = 1, f(a) = 1 f"(a) = 2 into equation (1), we obtain: y = 1+2(x-f), i.e. y = 2x-1.
Compare this result with that obtained in example 2 from § 33. Naturally, the same thing happened.
Let's create an equation for the tangent to the graph of the function y = tan x at the origin. We have: 
this means cos x f"(0) = 1. Substituting the found values a = 0, f(a) = 0, f"(a) = 1 into equation (1), we obtain: y = 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.
When solving these fairly simple examples, we actually used a certain algorithm, which is contained in formula (1). Let's make this algorithm explicit.
ALGORITHM FOR DEVELOPING AN EQUATION FOR A TANGENT TO THE GRAPH OF THE FUNCTION y = f(x)
1) Designate the abscissa of the tangent point 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, taking into account that in this example
In Fig. 126 a hyperbola is depicted, a straight line y = 2 is constructed.
The drawing confirms the above calculations: indeed, the line y = 2 touches the hyperbola at the point (1; 1).
Answer: y = 2- x.
Example 2. Draw a tangent to the graph of the function so that it is parallel to the line y = 4x - 5.
Let us clarify the formulation of the problem. The requirement to "draw a tangent" usually means "to form an equation for the tangent." This is logical, because if a person was able to create an equation for a tangent, then he is unlikely to have difficulty constructing a straight line on the coordinate plane using its equation.
Let's use the algorithm for composing the tangent equation, taking into account that in this example But, unlike the previous example, there is ambiguity: the abscissa of the tangent point is not explicitly indicated.
Let's start thinking like this. The desired tangent must be parallel to the straight line y = 4x-5. Two lines are parallel if and only if their slopes are equal. This means that the angular coefficient of the tangent must be equal to the angular coefficient of the given straight line: 
Thus, we can find the value of a from the equation f"(a) = 4.
We have: 
From the equation This means that there are two tangents that satisfy the conditions of the problem: one at the point with abscissa 2, the other at the point with abscissa -2.
Now you can follow the algorithm.
Example 3. From point (0; 1) draw a tangent to the graph of the function
Let's use the algorithm for composing the tangent equation, taking into account that in this example, Note that here, as in example 2, the abscissa of the tangent point is not explicitly indicated. Nevertheless, we follow the algorithm.
By condition, the tangent passes through the point (0; 1). Substituting the values x = 0, y = 1 into equation (2), we obtain: 
As you can see, in this example, only at the fourth step of the algorithm we managed to find the abscissa of the tangent point. Substituting the value a =4 into equation (2), we obtain:
In Fig. 127 presents a geometric illustration of the considered example: a graph of the function is plotted
In § 32 we noted that for a function y = f(x) having a derivative at a fixed point x, the approximate equality is valid:
For the convenience of further reasoning, let us change the notation: instead of x we will write a, instead of we will write x and, accordingly, instead of we will write x-a. Then the approximate equality written above will take the form:
Now look at fig. 128. A tangent is drawn to the graph of the function y = f(x) at point M (a; f (a)). Point x is marked 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. 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 the approximate equality (3)? The fact that To calculate the approximate value of the function, take the ordinate value of the tangent.
Example 4. Find the approximate value of the numerical expression 1.02 7.
We are talking about finding the value of the function y = x 7 at the point x = 1.02. Let us 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 10th grade
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