The coefficient of the slope of a straight line. How to find the slope? Transition from a line equation with a slope to other types of line equation and back

Learn to take derivatives of functions. The derivative characterizes the rate of change of a function at a certain point lying on the graph of this function. IN in this case The graph can be either a straight or curved line. That is, the derivative characterizes the rate of change of a function at a specific point in time. Remember general rules, by which derivatives are taken, and only then proceed to the next step.

• Read the article.
• How to take the simplest derivatives, for example, derivative exponential equation, described. The calculations presented in the following steps will be based on the methods described therein.

Learn to distinguish between tasks in which slope needs to be calculated through the derivative of the function. Problems do not always ask you to find the slope or derivative of a function. For example, you may be asked to find the rate of change of a function at point A(x,y). You may also be asked to find the slope of the tangent at point A(x,y). In both cases it is necessary to take the derivative of the function.

The slope is straight. In this article we will look at problems related to the coordinate plane included in the Unified State Examination in mathematics. These are tasks for:

— determination of the angular coefficient of a straight line when two points through which it passes are known;
— determination of the abscissa or ordinate of the point of intersection of two straight lines on a plane.

What is the abscissa and ordinate of a point was described in this section. In it we have already considered several problems related to the coordinate plane. What do you need to understand for the type of problem under consideration? A little theory.

Equation of a line on coordinate plane has the form:

Where k – this is the slope of the line.

Next moment! Direct slope equal to tangent angle of inclination of a straight line. This is the angle between a given line and the axisOh.

It ranges from 0 to 180 degrees.

That is, if we reduce the equation of a straight line to the form y = kx + b, then we can always determine the coefficient k (slope coefficient).

Also, if based on the condition we can determine the tangent of the angle of inclination of the straight line, then we will thereby find its angular coefficient.

Next theoretical point!Equation of a straight line passing through two given points.The formula looks like:

Let us consider problems (similar to problems from open bank tasks):

Find the slope of the line passing through the points with coordinates (–6;0) and (0;6).

In this problem the most rational way The solution is to find the tangent of the angle between the x axis and the given straight line. It is known that it is equal to the slope. Consider a right triangle formed by a straight line and the axes x and oy:

Tangent of the angle in right triangle is the relation opposite leg to adjacent:

*Both legs are equal to six (these are their lengths).

Certainly, this task can be solved using the formula for finding the equation of a straight line passing through two given points. But this will be a longer solution.

Answer: 1

Find the slope of the line passing through the points with coordinates (5;0) and (0;5).

Our points have coordinates (5;0) and (0;5). Means,

Let's bring the formula to the form y = kx + b

We found that the slope k = – 1.

Answer: –1

Straight a passes through points with coordinates (0;6) and (8;0). Straight b passes through the point with coordinates (0;10) and is parallel to the line a b with axle oh.

In this problem you can find the equation of the line a, determine the slope for it. At the straight line b the slope will be the same since they are parallel. Next you can find the equation of the line b. And then, substituting the value y = 0 into it, find the abscissa. BUT!

In this case, it is easier to use the property of similarity of triangles.

Right triangles formed by these (parallel) lines and coordinate axes are similar, which means that the ratios of their corresponding sides are equal.

The required abscissa is 40/3.

Answer: 40/3

Straight a passes through points with coordinates (0;8) and (–12;0). Straight b passes through the point with coordinates (0; –12) and is parallel to the line a. Find the abscissa of the point of intersection of the line b with axle oh.

For this problem, the most rational way to solve it is to use the property of similarity of triangles. But we will solve it in a different way.

We know the points through which the line passes A. We can write an equation for a straight line. The formula for the equation of a straight line passing through two given points has the form:

By condition, the points have coordinates (0;8) and (–12;0). Means,

Let's bring it to mind y = kx + b:

Got that corner k = 2/3.

*The angular coefficient could be found through the tangent of the angle in a right triangle with legs 8 and 12.

It is known that parallel lines have equal angle coefficients. This means that the equation of the straight line passing through the point (0;-12) has the form:

Find the value b we can substitute the abscissa and ordinate into the equation:

Thus, the straight line looks like:

Now, to find the desired abscissa of the point of intersection of the line with the x axis, you need to substitute y = 0:

Answer: 18

Find the ordinate of the axis intersection point oh and a line passing through point B(10;12) and parallel to a line passing through the origin and point A(10;24).

Let's find the equation of a straight line passing through points with coordinates (0;0) and (10;24).

The formula for the equation of a straight line passing through two given points has the form:

Our points have coordinates (0;0) and (10;24). Means,

Let's bring it to mind y = kx + b

The angle coefficients of parallel lines are equal. This means that the equation of the straight line passing through point B(10;12) has the form:

Meaning b Let’s find by substituting the coordinates of point B(10;12) into this equation:

We got the equation of the straight line:

To find the ordinate of the point of intersection of this line with the axis OU need to be substituted into the found equation X= 0:

*The simplest solution. With help parallel transfer move this line down along the axis OU to point (10;12). The shift occurs by 12 units, that is, point A(10;24) “moved” to point B(10;12), and point O(0;0) “moved” to point (0;–12). This means that the resulting straight line will intersect the axis OU at point (0;–12).

The required ordinate is –12.

Answer: –12

Find the ordinate of the point of intersection of the line given by the equation

3x + 2у = 6, with axis Oy.

Coordinate of the point of intersection of a given line with an axis OU has the form (0; at). Let's substitute the abscissa into the equation X= 0, and find the ordinate:

The ordinate of the point of intersection of the line and the axis OU equals 3.

*The system is solved:

Answer: 3

Find the ordinate of the point of intersection of the lines given by the equations

3x + 2y = 6 And y = – x.

When two lines are given, and the question is about finding the coordinates of the point of intersection of these lines, a system of these equations is solved:

In the first equation we substitute - X instead of at:

The ordinate is equal to minus six.

Answer: – 6

Find the slope of the line passing through the points with coordinates (–2;0) and (0;2).

Find the slope of the line passing through the points with coordinates (2;0) and (0;2).

Line a passes through points with coordinates (0;4) and (6;0). Line b passes through the point with coordinates (0;8) and is parallel to line a. Find the abscissa of the point of intersection of line b with the Ox axis.

Find the ordinate of the point of intersection of the oy axis and the line passing through point B (6;4) and parallel to the line passing through the origin and point A (6;8).

1. It is necessary to clearly understand that the angular coefficient of a straight line is equal to the tangent of the angle of inclination of the straight line. This will help you in solving many problems of this type.

2. The formula for finding a straight line passing through two given points must be understood. With its help, you will always find the equation of a line if the coordinates of its two points are given.

3. Remember that the slopes of parallel lines are equal.

4. As you understand, in some problems it is convenient to use the triangle similarity feature. Problems are solved practically orally.

5. Problems in which two lines are given and it is required to find the abscissa or ordinate of the point of their intersection can be solved graphically. That is, build them on a coordinate plane (on a sheet of paper in a square) and determine the intersection point visually. *But this method is not always applicable.

6. And lastly. If a straight line and the coordinates of the points of its intersection with the coordinate axes are given, then in such problems it is convenient to find the angular coefficient by finding the tangent of the angle in the formed right triangle. How to “see” this triangle when various locations straight lines on a plane are shown schematically below:

>> Straight angle from 0 to 90 degrees<<

>> Straight angle from 90 to 180 degrees<<

That's all. Good luck to you!

Sincerely, Alexander.

P.S: I would be grateful if you tell me about the site on social networks.

In the previous chapter it was shown that, by choosing a certain coordinate system on the plane, we can express the geometric properties characterizing the points of the line under consideration analytically by an equation between the current coordinates. Thus we get the equation of the line. This chapter will look at straight line equations.

To create an equation for a straight line in Cartesian coordinates, you need to somehow set the conditions that determine its position relative to the coordinate axes.

First, we will introduce the concept of the angular coefficient of a line, which is one of the quantities characterizing the position of a line on a plane.

Let's call the angle of inclination of the straight line to the Ox axis the angle by which the Ox axis needs to be rotated so that it coincides with the given line (or is parallel to it). As usual, we will consider the angle taking into account the sign (the sign is determined by the direction of rotation: counterclockwise or clockwise). Since an additional rotation of the Ox axis through an angle of 180° will again align it with the straight line, the angle of inclination of the straight line to the axis can not be chosen unambiguously (to within a term, a multiple of ).

The tangent of this angle is determined uniquely (since changing the angle does not change its tangent).

The tangent of the angle of inclination of the straight line to the Ox axis is called the angular coefficient of the straight line.

The angular coefficient characterizes the direction of the straight line (we do not distinguish here between two mutually opposite directions of the straight line). If the slope of a line is zero, then the line is parallel to the x-axis. With a positive angular coefficient, the angle of inclination of the straight line to the Ox axis will be acute (we are considering here the smallest positive value of the inclination angle) (Fig. 39); Moreover, the greater the angular coefficient, the greater the angle of its inclination to the Ox axis. If the angular coefficient is negative, then the angle of inclination of the straight line to the Ox axis will be obtuse (Fig. 40). Note that a straight line perpendicular to the Ox axis does not have an angular coefficient (the tangent of the angle does not exist).

The straight line y=f(x) will be tangent to the graph shown in the figure at point x0 if it passes through the point with coordinates (x0; f(x0)) and has an angular coefficient f"(x0). Find such a coefficient, Knowing the features of a tangent, it’s not difficult.

You will need

• - mathematical reference book;
• - a simple pencil;
• - notebook;
• - protractor;
• - compass;
• - pen.

Instructions

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.

Draw additional tangents that would be in contact with the graph of the function at points x1, x2 and x3, and also mark the angles formed by these tangents with the x-axis (this angle is counted in the positive direction from the axis to the tangent line). For example, the angle, that is, α1, will be acute, the second (α2) will be obtuse, and the third (α3) will be zero, since the tangent line is parallel to the OX axis. In this case, the tangent of an obtuse angle is negative, the tangent of an acute angle is positive, and at tg0 the result is zero.

note

Correctly determine the angle formed by the tangent. To do this, use a protractor.

Helpful advice

Two inclined lines will be parallel if their angular coefficients are equal to each other; perpendicular if the product of the angular coefficients of these tangents is equal to -1.

Sources:

• Tangent to the graph of a function

Cosine, like sine, is classified as a “direct” trigonometric function. Tangent (together with cotangent) is classified as another pair called “derivatives”. There are several definitions of these functions that make it possible to find the tangent given by a known cosine value of the same value.

Instructions

Subtract the quotient of unity by the value raised to the cosine of the given angle, and extract the square root from the result - this will be the tangent value of the angle, expressed by its cosine: tan(α)=√(1-1/(cos(α))²) . Please note that in the formula the cosine is in the denominator of the fraction. The impossibility of dividing by zero precludes the use of this expression for angles equal to 90°, as well as those differing from this value by numbers that are multiples of 180° (270°, 450°, -90°, etc.).

There is an alternative way to calculate the tangent from a known cosine value. It can be used if there is no restriction on the use of others. To implement this method, first determine the angle value from a known cosine value - this can be done using the arc cosine function. Then simply calculate the tangent for the angle of the resulting value. In general, this algorithm can be written as follows: tg(α)=tg(arccos(cos(α))).

There is also an exotic option using the definition of cosine and tangent through the acute angles of a right triangle. In this definition, cosine corresponds to the ratio of the length of the leg adjacent to the angle under consideration to the length of the hypotenuse. Knowing the value of the cosine, you can select the corresponding lengths of these two sides. For example, if cos(α) = 0.5, then the adjacent can be taken equal to 10 cm, and the hypotenuse - 20 cm. The specific numbers do not matter here - you will get the same and correct numbers with any values ​​that have the same . Then, using the Pythagorean theorem, determine the length of the missing side - the opposite leg. It will be equal to the square root of the difference between the lengths of the squared hypotenuse and the known leg: √(20²-10²)=√300. By definition, tangent corresponds to the ratio of the lengths of the opposite and adjacent legs (√300/10) - calculate it and get the tangent value found using the classical definition of cosine.

Sources:

• cosine through tangent formula

One of the trigonometric functions, most often denoted by the letters tg, although tan is also used. The easiest way to represent the tangent is as a sine ratio angle to its cosine. This is an odd periodic and non-continuous function, each cycle of which is equal to the number Pi, and the break point corresponds to half of this number.

The derivative of a function is one of the difficult topics in the school curriculum. Not every graduate will answer the question of what a derivative is.

This article explains in a simple and clear way what a derivative is and why it is needed.. We will not now strive for mathematical rigor in the presentation. The most important thing is to understand the meaning.

Let's remember the definition:

The derivative is the rate of change of a function.

The figure shows graphs of three functions. Which one do you think is growing faster?

The answer is obvious - the third. It has the highest rate of change, that is, the largest derivative.

Here's another example.

Kostya, Grisha and Matvey got jobs at the same time. Let's see how their income changed during the year:

The graph shows everything at once, isn’t it? Kostya’s income more than doubled in six months. And Grisha’s income also increased, but just a little. And Matvey’s income decreased to zero. The starting conditions are the same, but the rate of change of the function, that is derivative, - different. As for Matvey, his income derivative is generally negative.

Intuitively, we easily estimate the rate of change of a function. But how do we do this?

What we're really looking at is how steeply the graph of a function goes up (or down). In other words, how quickly does y change as x changes? Obviously, the same function at different points can have different derivative values ​​- that is, it can change faster or slower.

The derivative of a function is denoted .

We'll show you how to find it using a graph.

A graph of some function has been drawn. Let's take a point with an abscissa on it. Let us draw a tangent to the graph of the function at this point. We want to estimate how steeply the function graph goes up. A convenient value for this is tangent of the tangent angle.

The derivative of a function at a point is equal to the tangent of the tangent angle drawn to the graph of the function at this point.

Please note that as the angle of inclination of the tangent we take the angle between the tangent and the positive direction of the axis.

Sometimes students ask what a tangent to the graph of a function is. This is a straight line that has a single common point with the graph in this section, and as shown in our figure. It looks like a tangent to a circle.

Let's find it. We remember that the tangent of an acute angle in a right triangle is equal to the ratio of the opposite side to the adjacent side. From the triangle:

We found the derivative using a graph without even knowing the formula of the function. Such problems are often found in the Unified State Examination in mathematics under the number.

There is another important relationship. Recall that the straight line is given by the equation

The quantity in this equation is called slope of a straight line. It is equal to the tangent of the angle of inclination of the straight line to the axis.

.

We get that

Let's remember this formula. It expresses the geometric meaning of the derivative.

The derivative of a function at a point is equal to the slope of the tangent drawn to the graph of the function at that point.

In other words, the derivative is equal to the tangent of the tangent angle.

We have already said that the same function can have different derivatives at different points. Let's see how the derivative is related to the behavior of the function.

Let's draw a graph of some function. Let this function increase in some areas and decrease in others, and at different rates. And let this function have maximum and minimum points.

At a point the function increases. A tangent to the graph drawn at point forms an acute angle; with positive axis direction. This means that the derivative at the point is positive.

At the point our function decreases. The tangent at this point forms an obtuse angle; with positive axis direction. Since the tangent of an obtuse angle is negative, the derivative at the point is negative.

Here's what happens:

If a function is increasing, its derivative is positive.

If it decreases, its derivative is negative.

What will happen at the maximum and minimum points? We see that at the points (maximum point) and (minimum point) the tangent is horizontal. Therefore, the tangent of the tangent at these points is zero, and the derivative is also zero.

Point - maximum point. At this point, the increase in the function is replaced by a decrease. Consequently, the sign of the derivative changes at the point from “plus” to “minus”.

At the point - the minimum point - the derivative is also zero, but its sign changes from “minus” to “plus”.

Conclusion: using the derivative, we can learn everything that interests us about the behavior of a function.

If the derivative is positive, then the function increases.

If the derivative is negative, then the function decreases.

At the maximum point, the derivative is zero and changes sign from “plus” to “minus”.

At the minimum point, the derivative is also zero and changes sign from “minus” to “plus”.

Let's write these conclusions in the form of a table:

Let's make two small clarifications. You will need one of them when solving the problem. Another - in the first year, with a more serious study of functions and derivatives.

It is possible that the derivative of a function at some point is equal to zero, but the function has neither a maximum nor a minimum at this point. This is the so-called :

At a point, the tangent to the graph is horizontal and the derivative is zero. However, before the point the function increased - and after the point it continues to increase. The sign of the derivative does not change - it remains positive as it was.

It also happens that at the point of maximum or minimum the derivative does not exist. On the graph, this corresponds to a sharp break, when it is impossible to draw a tangent at a given point.

How to find the derivative if the function is given not by a graph, but by a formula? In this case it applies