The video lesson “Equation of a tangent to the graph of a function” demonstrates educational material to master the topic. During the video lesson, the theoretical material necessary to formulate the concept of the equation of a tangent to the graph of a function at a given point, an algorithm for finding such a tangent, and examples of solving problems using the studied theoretical material are described.
The video tutorial uses methods that improve the clarity of the material. The presentation contains drawings, diagrams, important voice comments, animation, highlighting and other tools.
The video lesson begins with a presentation of the topic of the lesson and an image of a tangent to the graph of some function y=f(x) at the point M(a;f(a)). It is known that the angular coefficient of the tangent plotted to the graph at a given point is equal to the derivative of the function f΄(a) at this point. Also from the algebra course we know the equation of the straight line y=kx+m. The solution to the problem of finding the tangent equation at a point is schematically presented, which reduces to finding the coefficients k, m. Knowing the coordinates of a point belonging to the graph of the function, we can find m by substituting the coordinate value into the tangent equation f(a)=ka+m. From it we find m=f(a)-ka. Thus, knowing the value of the derivative at a given point and the coordinates of the point, we can represent the tangent equation in this way y=f(a)+f΄(a)(x-a).
The following is an example of composing a tangent equation following the diagram. Given the function y=x 2 , x=-2. Taking a=-2, we find the value of the function at a given point f(a)= f(-2)=(-2) 2 =4. We determine the derivative of the function f΄(x)=2x. At this point the derivative is equal to f΄(a)= f΄(-2)=2·(-2)=-4. To compose the equation, all coefficients a=-2, f(a)=4, f΄(a)=-4 were found, so the tangent equation is y=4+(-4)(x+2). Simplifying the equation, we get y = -4-4x.
The following example suggests constructing an equation for the tangent at the origin to the graph of the function y=tgx. At a given point a=0, f(0)=0, f΄(x)=1/cos 2 x, f΄(0)=1. So the tangent equation looks like y=x.
As a generalization, the process of composing an equation tangent to the graph of a function at a certain point is formalized in the form of an algorithm consisting of 4 steps:
- Enter the designation a for the abscissa of the tangent point;
- f(a) is calculated;
- f΄(x) is determined and f΄(a) is calculated. The found values of a, f(a), f΄(a) are substituted into the tangent equation formula y=f(a)+f΄(a)(x-a).
Example 1 considers composing the tangent equation to the graph of the function y=1/x at point x=1. To solve the problem we use an algorithm. For a given function at point a=1, the value of the function f(a)=-1. Derivative of the function f΄(x)=1/x 2. At point a=1 the derivative f΄(a)= f΄(1)=1. Using the data obtained, the tangent equation y=-1+(x-1), or y=x-2, is drawn up.
In example 2, it is necessary to find the equation of the tangent to the graph of the function y=x 3 +3x 2 -2x-2. The main condition is the parallelism of the tangent and straight line y=-2x+1. First, we find the angular coefficient of the tangent, equal to the angular coefficient of the straight line y=-2x+1. Since f΄(a)=-2 for a given line, then k=-2 for the desired tangent. We find the derivative of the function (x 3 +3x 2 -2x-2)΄=3x 2 +6x-2. Knowing that f΄(a)=-2, we find the coordinates of point 3a 2 +6a-2=-2. Having solved the equation, we get a 1 =0, and 2 =-2. Using the found coordinates, you can find the tangent equation using a well-known algorithm. We find the value of the function at the points f(a 1)=-2, f(a 2)=-18. The value of the derivative at the point f΄(а 1)= f΄(а 2)=-2. Substituting the found values into the tangent equation, we obtain for the first point a 1 =0 y=-2x-2, and for the second point a 2 =-2 the tangent equation y=-2x-22.
Example 3 describes the composition of the tangent equation for drawing it at the point (0;3) to the graph of the function y=√x. The solution is made using a well-known algorithm. The tangent point has coordinates x=a, where a>0. The value of the function at the point f(a)=√x. The derivative of the function f΄(х)=1/2√х, therefore at a given point f΄(а)=1/2√а. Substituting all the obtained values into the tangent equation, we obtain y = √a + (x-a)/2√a. Transforming the equation, we get y=x/2√а+√а/2. Knowing that the tangent passes through the point (0;3), we find the value of a. We find a from 3=√a/2. Hence √a=6, a=36. We find the tangent equation y=x/12+3. The figure shows the graph of the function under consideration and the constructed desired tangent.
Students are reminded of the approximate equalities Δy=≈f΄(x)Δxand f(x+Δx)-f(x)≈f΄(x)Δx. Taking x=a, x+Δx=x, Δx=x-a, we get f(x)- f(a)≈f΄(a)(x-a), hence f(x)≈f(a)+ f΄(a)(x-a).
In example 4, it is necessary to find the approximate value of the expression 2.003 6. Since it is necessary to find the value of the function f(x) = x 6 at point x = 2.003, we can use well-known formula, taking f(x)=x 6, a=2, f(a)= f(2)=64, f΄(x)=6x 5. Derivative at the point f΄(2)=192. Therefore, 2.003 6 ≈65-192·0.003. Having calculated the expression, we get 2.003 6 ≈64.576.
The video lesson “Equation of a tangent to the graph of a function” is recommended for use in a traditional mathematics lesson at school. For a teacher teaching remotely, video material will help explain the topic more clearly. The video can be recommended for students to review independently if necessary to deepen their understanding of the subject.
TEXT DECODING:
We know that if a point M (a; f(a)) (em with coordinates a and ef from a) belongs to the graph of the function y = f (x) and if at this point it is possible to draw a tangent to the graph of the function that is not perpendicular to the axis abscissa, then the angular coefficient of the tangent is equal to f"(a) (eff prime from a).
Let a function y = f(x) and a point M (a; f(a)) be given, and it is also known that f´(a) exists. Let's create an equation for the tangent to the graph of a given function at a given point. This equation, like the equation of any straight line that is not parallel to the ordinate axis, has the form y = kx+m (the y is equal to ka x plus em), so the task is to find the values of the coefficients k and m. (ka and em)
Angle coefficient 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 of the point M into the equation of the straight line, we obtain the correct equality : f(a) = ka+m, from where we find that m = f(a) - ka.
It remains to substitute the found values of the coefficients ki and m into the equation of the straight line:
y = kx+(f(a) -ka);
y = f(a)+k(x-a);
y= f(a)+ f"(a) (x- a). ( y is equal to ef from a plus ef prime from a, multiplied by x minus a).
We have obtained the equation for the tangent to the graph of the function y = f(x) at the point x=a.
If, say, y = x 2 and x = -2 (i.e. a = -2), then f (a) = f (-2) = (-2) 2 = 4; f´(x) = 2x, which means f"(a) = f´(-2) = 2·(-2) = -4. (then the ef of a is equal to four, the ef of the prime of x is equal to two x, which means ef prime from a equals minus four)
Substituting the found values a = -2, f(a) = 4, f"(a) = -4 into the equation, we obtain: y = 4+(-4)(x+2), i.e. y = -4x -4.
(E is equal to minus four x minus four)
Let's create an equation for the tangent to the graph of the function y = tgx(Greek equal to tangent x) at the origin. We have: a = 0, f(0) = tan0=0;
f"(x)= , which means f"(0) = l. Substituting the found values a=0, f(a)=0, f´(a) = 1 into the equation, we get: y=x.
Let us summarize our steps in finding the equation of the tangent to the graph of a function at point x using an algorithm.
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 f(a).
3) Find f´(x) and calculate f´(a).
4) Substitute the found numbers a, f(a), f´(a) into the formula y= f(a)+ f"(a) (x- a).
Example 1. Create an equation for the tangent to the graph of the function y = - in
point x = 1.
Solution. Let's use the algorithm, taking into account that in this example
2) f(a)=f(1)=- =-1
3) f´(x)=; f´(a)= f´(1)= =1.
4) Substitute the found three numbers: a = 1, f(a) = -1, f"(a) = 1 into the formula. We get: y = -1+(x-1), y = x-2.
Answer: y = x-2.
Example 2. Given the function y = x 3 +3x 2 -2x-2. Write down the equation of the tangent to the graph of the function y = f(x), parallel to the straight line y = -2x +1.
Using the algorithm for composing the tangent equation, we take into account that in this example f(x) = x 3 +3x 2 -2x-2, but the abscissa of the tangent point is not indicated here.
Let's start thinking like this. The desired tangent must be parallel to the straight line y = -2x+1. And parallel lines have equal angular coefficients. This means that the angular coefficient of the tangent is equal to the angular coefficient of the given straight line: k tangent. = -2. Hok cas. = f"(a). Thus, we can find the value of a from the equation f ´(a) = -2.
Let's find the derivative of the function y=f(x):
f"(x)= (x 3 +3x 2 -2x-2)´ =3x 2 +6x-2;f"(a)= 3a 2 +6a-2.
From the equation f"(a) = -2, i.e. 3a 2 +6a-2=-2 we find a 1 =0, a 2 =-2. This means that there are two tangents that satisfy the conditions of the problem: one at the point with abscissa 0, the other at the point with abscissa -2.
Now you can follow the algorithm.
1) a 1 =0, and 2 =-2.
2) f(a 1)= 0 3 +3·0 2 -2∙0-2=-2; f(a 2)= (-2) 3 +3·(-2) 2 -2·(-2)-2=6;
3) f"(a 1) = f"(a 2) = -2.
4) Substituting the values a 1 = 0, f(a 1) = -2, f"(a 1) = -2 into the formula, we get:
y=-2-2(x-0), y=-2x-2.
Substituting the values a 2 = -2, f(a 2) =6, f"(a 2) = -2 into the formula, we get:
y=6-2(x+2), y=-2x+2.
Answer: y=-2x-2, y=-2x+2.
Example 3. From the point (0; 3) draw a tangent to the graph of the function y = . Solution. Let's use the algorithm for composing the tangent equation, taking into account that in this example f(x) = . Note that here, as in example 2, the abscissa of the tangent point is not explicitly indicated. Nevertheless, we follow the algorithm.
1) Let x = a be the abscissa of the point of tangency; it is clear that a >0.
3) f´(x)=()´=; f´(a) =.
4) Substituting the values of a, f(a) = , f"(a) = into the formula
y=f (a) +f "(a) (x-a), we get:
By condition, the tangent passes through the point (0; 3). Substituting the values x = 0, y = 3 into the equation, we get: 3 = , and then =6, a =36.
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 =36 into the equation, we get: y=+3
In Fig. Figure 1 shows a geometric illustration of the considered example: a graph of the function y = is constructed, a straight line is drawn y = +3.
Answer: y = +3.
We know that for a function y = f(x), which has a derivative at point x, the approximate equality is valid: Δyf´(x)Δx (delta y is approximately equal to the eff prime of x multiplied by delta x)
or, in more detail, f(x+Δx)-f(x) f´(x) Δx (eff from x plus delta x minus ef from x is approximately equal to ef prime from x by delta x).
For the convenience of further discussion, let us change the notation:
instead of x we will write A,
instead of x+Δx we will write x
Instead of Δx we will write x-a.
Then the approximate equality written above will take the form:
f(x)-f(a)f´(a)(x-a)
f(x)f(a)+f´(a)(x-a). (eff from x is approximately equal to ef from a plus ef prime from a, multiplied by the difference between x and a).
Example 4. Find the approximate value of the numerical expression 2.003 6.
Solution. We are talking about finding the value of the function y = x 6 at the point x = 2.003. Let's use the formula f(x)f(a)+f´(a)(x-a), taking into account that in this example f(x)=x 6, a = 2,f(a) = f(2) = 2 6 =64; x = 2.003, f"(x) = 6x 5 and, therefore, f"(a) = f"(2) = 6 2 5 =192.
As a result we get:
2.003 6 64+192· 0.003, i.e. 2.003 6 =64.576.
If we use a calculator, we get:
2,003 6 = 64,5781643...
As you can see, the approximation accuracy is quite acceptable.
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 a straight line and its secant in in this case match up.
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.
Let's 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) .
Geometric meaning derivative of a function at a point is that the concept of the existence of a tangent to the graph at the same point is given.
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 consider graphic image solutions.
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 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
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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We determine the angular coefficient of the tangent to the curve at point M.
The curve representing the graph of the function y = f(x) is continuous in a certain neighborhood of the point M (including the point M itself).
If the value f‘(x0) does not exist, then either there is no tangent, or it runs vertically. In view of this, the presence of a derivative of the function at the point x0 is due to the existence of a non-vertical tangent tangent to the graph of the function at the point (x0, f(x0)). In this case, the angular coefficient of the tangent will be equal to f "(x0). Thus, the geometric meaning of the derivative becomes clear - the calculation of the angular coefficient of the tangent.
Find the abscissa value of the tangent point, which is denoted by the letter “a”. If it coincides with a given tangent point, then "a" will be its x-coordinate. Determine the value functions f(a) by substituting into the equation functions abscissa value.
Determine the first derivative of the equation functions f’(x) and substitute the value of point “a” into it.
Take the general tangent equation, which is defined as y = f(a) = f (a)(x – a), and substitute the found values of a, f(a), f "(a) into it. As a result, the solution to the graph will be found and tangent.
Solve the problem in a different way if the given tangent point does not coincide with the tangent point. In this case, it is necessary to substitute “a” instead of numbers in the tangent equation. After this, instead of the letters “x” and “y”, substitute the value of the coordinates of the given point. Solve the resulting equation in which “a” is the unknown. Plug the resulting value into the tangent equation.
Write an equation for a tangent with the letter “a” if the problem statement specifies the equation functions and the equation of a parallel line relative to the desired tangent. After this we need the derivative functions, to the coordinate at point “a”. Substitute the appropriate value into the tangent equation and solve the function.
Let a function f be given, which at some point x 0 has a finite derivative f (x 0). Then the straight line passing through the point (x 0 ; f (x 0)), having an angular coefficient f ’(x 0), is called a tangent.
What happens if the derivative does not exist at the point x 0? There are two options:
- There is no tangent to the graph either. Classic example- function y = |x | at point (0; 0).
- The tangent becomes vertical. This is true, for example, for the function y = arcsin x at the point (1; π /2).
Tangent equation
Any non-vertical straight line is given by an equation of the form y = kx + b, where k is the slope. The tangent is no exception, and in order to create its equation at some point x 0, it is enough to know the value of the function and the derivative at this point.
So, let a function y = f (x) be given, which has a derivative y = f ’(x) on the segment. Then at any point x 0 ∈ (a ; b) a tangent can be drawn to the graph of this function, which is given by the equation:
y = f ’(x 0) (x − x 0) + f (x 0)
Here f ’(x 0) is the value of the derivative at point x 0, and f (x 0) is the value of the function itself.
Task. Given the function y = x 3 . Write an equation for the tangent to the graph of this function at the point x 0 = 2.
Tangent equation: y = f ’(x 0) · (x − x 0) + f (x 0). The point x 0 = 2 is given to us, but the values f (x 0) and f ’(x 0) will have to be calculated.
First, let's find the value of the function. Everything is easy here: f (x 0) = f (2) = 2 3 = 8;
Now let’s find the derivative: f ’(x) = (x 3)’ = 3x 2;
We substitute x 0 = 2 into the derivative: f ’(x 0) = f ’(2) = 3 2 2 = 12;
In total we get: y = 12 · (x − 2) + 8 = 12x − 24 + 8 = 12x − 16.
This is the tangent equation.
Task. Write an equation for the tangent to the graph of the function f (x) = 2sin x + 5 at point x 0 = π /2.
This time we will not describe each action in detail - we will only indicate the key steps. We have:
f (x 0) = f (π /2) = 2sin (π /2) + 5 = 2 + 5 = 7;
f ’(x) = (2sin x + 5)’ = 2cos x;
f ’(x 0) = f ’(π /2) = 2cos (π /2) = 0;
Tangent equation:
y = 0 · (x − π /2) + 7 ⇒ y = 7
In the latter case, the straight line turned out to be horizontal, because its angular coefficient k = 0. There is nothing wrong with this - we just stumbled upon an extremum point.
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