Derivative and tangent to the graph of a function. Tangent to the graph of a function at a point
This math program finds the equation of the tangent to the graph of the function \(f(x)\) at a user-specified point \(a\).
The program not only displays the tangent equation, but also displays the process of solving the problem.
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If you need to find the derivative of a function, then for this we have the task Find the derivative.
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Enter the function expression \(f(x)\) and the number \(a\) Find tangent equation It was discovered that some scripts necessary to solve this problem were not loaded, and the program may not work.
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A little theory.
Direct slope
Let us remember that the schedule linear function\(y=kx+b\) is a straight line. The number \(k=tg \alpha \) is called slope of a straight line, and the angle \(\alpha \) is the angle between this line and the Ox axis
If \(k>0\), then \(0 If \(kEquation of the tangent to the graph of the function
If point M(a; f(a)) belongs to the graph of the function 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 x-axis, then from the geometric meaning of the derivative it follows that the angular coefficient of the tangent is equal to f "(a). Next, we will develop an algorithm for composing an equation for a tangent to the graph of any function.
Let a function y = f(x) and a point M(a; f(a)) be given on the graph of this function; let it be known that f"(a) exists. Let's create an equation for the tangent to the graph of the given function in given point. This equation, like the equation of any straight line that is not parallel to the ordinate axis, has the form y = kx + b, so the task is to find the values of the coefficients k and b.
Everything is clear with the angular coefficient k: it is known that k = f"(a). To calculate the value of b, 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 a straight line, we obtain the correct equality: \(f(a)=ka+b\), i.e. \(b = f(a) - ka\).
It remains to substitute the found values of the coefficients k and b into the equation of the straight line:
We received equation of the tangent to the graph of a function\(y = f(x) \) at the point \(x=a \).
Algorithm for finding the equation of the 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 \(y=f(a)+ f"(a)(x-a) \)
First level
Equation of a tangent to the graph of a function. Comprehensive guide (2019)
Do you already know what a derivative is? If not, read the topic first. So you say you know the derivative. Let's check it now. Find the increment of the function when the increment of the argument is equal to. Did you manage? It should work. Now find the derivative of the function at a point. Answer: . Happened? If you have any difficulties with any of these examples, I strongly recommend that you return to the topic and study it again. I know the topic is very big, but otherwise there is no point in going further. Consider the graph of some function:
Let's select a certain point on the graph line. Let its abscissa, then the ordinate is equal. Then we select the point with the abscissa close to the point; its ordinate is:
Let's draw a straight line through these points. It is called a secant (just like in geometry). Let us denote the angle of inclination of the straight line to the axis as. As in trigonometry, this angle is measured from the positive direction of the x-axis counterclockwise. What values can the angle take? No matter how you tilt this straight line, one half will still stick up. Therefore, the maximum possible angle is , and the minimum possible angle is . Means, . The angle is not included, since the position of the straight line in this case exactly coincides with, and it is more logical to choose a smaller angle. Let’s take a point in the figure such that the straight line is parallel to the abscissa axis and a is the ordinate axis:
From the figure it can be seen that, a. Then the increment ratio is:
(since it is rectangular).
Let's reduce it now. Then the point will approach the point. When it becomes infinitesimal, the ratio becomes equal to the derivative of the function at the point. What will happen to the secant? The point will be infinitely close to the point, so that they can be considered the same point. But a straight line that has only one common point with a curve is nothing more than tangent(V in this case this condition is fulfilled only in a small area - near the point, but this is enough). They say that in this case the secant takes limit position.
Let's call the angle of inclination of the secant to the axis. Then it turns out that the derivative
that is the derivative is equal to the tangent of the angle of inclination of the tangent to the graph of the function at a given point.
Since a tangent is a line, let's now remember the equation of a line:
What is the coefficient responsible for? For the slope of the straight line. This is what it's called: slope . What does it mean? And the fact that it is equal to the tangent of the angle between the straight line and the axis! So this is what happens:
But we got this rule by considering an increasing function. What will change if the function is decreasing? Let's see: 
Now the angles are obtuse. And the increment of the function is negative. Let's consider again: . On the other side, . We get: , that is, everything is the same as last time. Let us again direct the point to the point, and the secant will take a limiting position, that is, it will turn into a tangent to the graph of the function at the point. So, let’s formulate the final rule:
The derivative of a function at a given point is equal to the tangent of the angle of inclination of the tangent to the graph of the function at this point, or (which is the same) the slope of this tangent:
That's what it is geometric meaning of derivative. Okay, all this is interesting, but why do we need it? Here example:
The figure shows a graph of a function and a tangent to it at the abscissa point. Find the value of the derivative of the function at a point. 
Solution.
As we recently found out, the value of the derivative at the point of tangency is equal to the slope of the tangent, which in turn is equal to the tangent of the angle of inclination of this tangent to the abscissa axis: . This means that to find the value of the derivative we need to find the tangent of the tangent angle. In the figure we have marked two points lying on the tangent, the coordinates of which are known to us. So let's finish it right triangle, passing through these points, and find the tangent of the tangent angle! 
The angle of inclination of the tangent to the axis is. Let's find the tangent of this angle: . Thus, the derivative of the function at a point is equal to.
Answer:. Now try it yourself:
Answers:
Knowing geometric meaning of derivative, we can very simply explain the rule that the derivative at the point of a local maximum or minimum is equal to zero. Indeed, the tangent to the graph at these points is “horizontal”, that is, parallel to the x-axis: 
Why equal to the angle between parallel lines? Of course, zero! And the tangent of zero is also zero. So the derivative is equal to zero:
Read more about this in the topic “Monotonicity of functions. Extremum points."
Now let's focus on arbitrary tangents. Let's say we have some function, for example, . We have drawn its graph and want to draw a tangent to it at some point. For example, at a point. We take a ruler, attach it to the graph and draw:
What do we know about this line? What is the most important thing to know about a line on a coordinate plane? Since a straight line is an image of a linear function, it would be very convenient to know its equation. That is, the coefficients in the equation
But we already know! This is the slope of the tangent, which is equal to the derivative of the function at that point:
In our example it will be like this:
Now all that remains is to find it. It's as simple as shelling pears: after all - the value of. Graphically, this is the coordinate of the intersection of the line with the ordinate axis (after all, at all points of the axis):
Let's draw it (so it's rectangular). Then (to the same angle between the tangent and the x-axis). What are and equal to? The figure clearly shows that, a. Then we get:
We combine all the obtained formulas into the equation of a straight line:
Now decide for yourself:
- Find tangent equation to a function at a point.
- The tangent to a parabola intersects the axis at an angle. Find the equation of this tangent.
- The line is parallel to the tangent to the graph of the function. Find the abscissa of the tangent point.
- The line is parallel to the tangent to the graph of the function. Find the abscissa of the tangent point.
Solutions and answers:
EQUATION OF A TANGENT TO THE GRAPH OF A FUNCTION. BRIEF DESCRIPTION AND BASIC FORMULAS
The derivative of a function at a particular point is equal to the tangent of the tangent to the graph of the function at this point, or the slope of this tangent:
Equation of the tangent to the graph of a function at a point:
Algorithm for finding the tangent equation:
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Tangent is a straight line passing through a point on the curve and coinciding with it at this point up to first order (Fig. 1).
Another definition: this is the limiting position of the secant at Δ x→0.
Explanation: Take a straight line intersecting the curve at two points: A And b(see picture). This is a secant. We will rotate it clockwise until it finds only one common point with the curve. This will give us a tangent.
Strict definition of tangent:
Tangent to the graph of a function f, differentiable at the point xO, is a straight line passing through the point ( xO; f(xO)) and having a slope f′( xO).
The slope has a straight line of the form y =kx +b. Coefficient k and is slope this straight line.
The angular coefficient is equal to the tangent of the acute angle formed by this straight line with the abscissa axis:
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Here angle α is the angle between the straight line y =kx +b and positive (that is, counterclockwise) direction of the x-axis. It is called angle of inclination of a straight line(Fig. 1 and 2). 
If the angle of inclination is straight y =kx +b acute, then the slope is positive number. The graph is increasing (Fig. 1).
If the angle of inclination is straight y =kx +b is obtuse, then the slope is 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.
The topic “The angular coefficient of a tangent as the tangent of the angle of inclination” is given several tasks in the certification exam. Depending on their condition, the graduate may be required to provide either a full answer or a short answer. When preparing to take the Unified State Examination in mathematics, the student should definitely repeat the tasks that require calculating the slope of a tangent.
It will help you do this educational portal"Shkolkovo". Our specialists prepared and presented theoretical and practical material in the most accessible way possible. Having become familiar with it, graduates with any level of training will be able to successfully solve problems related to derivatives in which it is necessary to find the tangent of the tangent angle.
Basic moments
To find the correct and rational decision For similar tasks in the Unified State Exam, you need to remember the basic definition: the derivative represents the rate of change of a function; it is equal to the tangent of the tangent angle drawn to the graph of the function at a certain point. It is equally important to complete the drawing. It will allow you to find correct solution Unified State Examination problems on the derivative, in which it is necessary to calculate the tangent of the angle of inclination of the tangent. For clarity, it is best to plot the graph on the OXY plane.
If you have already familiarized yourself with the basic material on the topic of derivatives and are ready to begin solving problems on calculating the tangent of the tangent angle, such as Unified State Exam assignments, you can do this online. For each task, for example, problems on the topic “Relationship of the derivative with the speed and acceleration of the body”, we have written down the correct answer and solution algorithm. At the same time, students can practice performing tasks of varying levels of complexity. If necessary, the exercise can be saved in the “Favorites” section so that you can discuss the solution with the teacher later.
