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, the derivative of an exponential equation, is 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.
Take the derivative of the function given to you. There is no need to build a graph here - you only need the equation of the function. In our example, take the derivative of the function. Take the derivative according to the methods outlined in the article mentioned above:
- Derivative:
Substitute the coordinates of the point given to you into the found derivative to calculate the slope. The derivative of a function is equal to the slope at a certain point. In other words, f"(x) is the slope of the function at any point (x,f(x)). In our example:
- Find the slope of the function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x) at point A(4,2).
- Derivative of a function:
- f ′ (x) = 4 x + 6 (\displaystyle f"(x)=4x+6)
- Substitute the value of the “x” coordinate of this point:
- f ′ (x) = 4 (4) + 6 (\displaystyle f"(x)=4(4)+6)
- Find the slope:
- Slope function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x) at point A(4,2) is equal to 22.
If possible, check your answer on a graph. Remember that the slope cannot be calculated at every point. Differential calculus is considering complex functions and complex graphs, where the slope cannot be calculated at every point, and in some cases the points do not lie on the graphs at all. If possible, use a graphing calculator to check that the slope of the function you are given is correct. Otherwise, draw a tangent to the graph at the point given to you and think about whether the slope value you found matches what you see on the graph.
- The tangent will have the same slope as the graph of the function at a certain point. To draw a tangent at a given point, move left/right on the X axis (in our example, 22 values to the right), and then up one on the Y axis. Mark the point, and then connect it to the point given to you. In our example, connect the points with coordinates (4,2) and (26,3).
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 (here we do not distinguish 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 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.
The equation of a straight line on the coordinate plane has the form:
Where k – this is the slope of the line.
Next moment! The slope of a straight line is equal to the tangent of the angle of inclination of the 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's consider the tasks (similar to the tasks from the open task bank):
Find the slope of the line passing through the points with coordinates (–6;0) and (0;6).
In this problem, the most rational way to solve 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 ratio of the opposite side to the adjacent side:
*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 angle 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 a parallel 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 the 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. Using parallel translation, we shift 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 test. 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 with different positions of straight lines on the plane is 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.
Continuation of the topic, the equation of a line on a plane is based on the study of a straight line from algebra lessons. This article provides general information on the topic of equation of a straight line with a slope. Let's consider the definitions, get the equation itself, and identify the connection with other types of equations. Everything will be discussed using examples of problem solving.
Yandex.RTB R-A-339285-1
Before writing such an equation, it is necessary to define the angle of inclination of the straight line to the O x axis with their angular coefficient. Let us assume that a Cartesian coordinate system O x on the plane is given.
Definition 1
The angle of inclination of the straight line to the O x axis, located in the Cartesian coordinate system O x y on the plane, this is the angle that is measured from the positive direction O x to the straight line counterclockwise.
When the line is parallel to O x or coincides in it, the angle of inclination is 0. Then the angle of inclination of the given straight line α is determined on the interval [ 0 , π) .
Definition 2
Direct slope is the tangent of the angle of inclination of a given straight line.
Standard designation is k. From the definition we obtain that k = t g α . When the line is parallel to Ox, they say that the slope does not exist, since it goes to infinity.
The slope is positive when the graph of the function increases and vice versa. The figure shows various variations in the location of the right angle relative to the coordinate system with the value of the coefficient.
To find this angle, it is necessary to apply the definition of the angular coefficient and calculate the tangent of the angle of inclination in the plane.
Solution
From the condition we have that α = 120°. By definition, the slope must be calculated. Let's find it from the formula k = t g α = 120 = - 3.
Answer: k = - 3 .
If the angular coefficient is known, and it is necessary to find the angle of inclination to the abscissa axis, then the value of the angular coefficient should be taken into account. If k > 0, then the right angle is acute and is found by the formula α = a r c t g k. If k< 0 , тогда угол тупой, что дает право определить его по формуле α = π - a r c t g k .
Example 2
Determine the angle of inclination of the given straight line to O x with an angular coefficient of 3.
Solution
From the condition we have that the angular coefficient is positive, which means that the angle of inclination to O x is less than 90 degrees. Calculations are made using the formula α = a r c t g k = a r c t g 3.
Answer: α = a r c t g 3 .
Example 3
Find the angle of inclination of the straight line to the O x axis if the slope = - 1 3.
Solution
If we take the letter k as the designation of the angular coefficient, then α is the angle of inclination to a given straight line in the positive direction O x. Hence k = - 1 3< 0 , тогда необходимо применить формулу α = π - a r c t g k При подстановке получим выражение:
α = π - a r c t g - 1 3 = π - a r c t g 1 3 = π - π 6 = 5 π 6.
Answer: 5 π 6 .
An equation of the form y = k x + b, where k is the slope and b is some real number, is called the equation of a line with a slope. The equation is typical for any straight line that is not parallel to the O y axis.
If we consider in detail a straight line on a plane in a fixed coordinate system, which is specified by an equation with an angular coefficient that has the form y = k x + b. In this case, it means that the equation corresponds to the coordinates of any point on the line. If we substitute the coordinates of point M, M 1 (x 1, y 1) into the equation y = k x + b, then in this case the line will pass through this point, otherwise the point does not belong to the line.
Example 4
A straight line with slope y = 1 3 x - 1 is given. Calculate whether the points M 1 (3, 0) and M 2 (2, - 2) belong to the given line.
Solution
It is necessary to substitute the coordinates of the point M 1 (3, 0) into the given equation, then we get 0 = 1 3 · 3 - 1 ⇔ 0 = 0. The equality is true, which means the point belongs to the line.
If we substitute the coordinates of the point M 2 (2, - 2), then we get an incorrect equality of the form - 2 = 1 3 · 2 - 1 ⇔ - 2 = - 1 3. We can conclude that point M 2 does not belong to the line.
Answer: M 1 belongs to the line, but M 2 does not.
It is known that the line is defined by the equation y = k · x + b, passing through M 1 (0, b), upon substitution we obtained an equality of the form b = k · 0 + b ⇔ b = b. From this we can conclude that the equation of a straight line with an angular coefficient y = k x + b on the plane defines a straight line that passes through the point 0, b. It forms an angle α with the positive direction of the O x axis, where k = t g α.
Let us consider, as an example, a straight line defined using an angular coefficient specified in the form y = 3 x - 1. We obtain that the straight line will pass through the point with coordinate 0, - 1 with a slope of α = a r c t g 3 = π 3 radians in the positive direction of the O x axis. This shows that the coefficient is 3.
Equation of a straight line with a slope passing through a given point
It is necessary to solve a problem where it is necessary to obtain the equation of a straight line with a given slope passing through the point M 1 (x 1, y 1).
The equality y 1 = k · x + b can be considered valid, since the line passes through the point M 1 (x 1, y 1). To remove the number b, it is necessary to subtract the equation with the slope from the left and right sides. It follows from this that y - y 1 = k · (x - x 1) . This equality is called the equation of a straight line with a given slope k, passing through the coordinates of the point M 1 (x 1, y 1).
Example 5
Write an equation for a straight line passing through point M 1 with coordinates (4, - 1), with an angular coefficient equal to - 2.
Solution
By condition we have that x 1 = 4, y 1 = - 1, k = - 2. From here the equation of the line will be written as follows: y - y 1 = k · (x - x 1) ⇔ y - (- 1) = - 2 · (x - 4) ⇔ y = - 2 x + 7.
Answer: y = - 2 x + 7 .
Example 6
Write the equation of a straight line with an angular coefficient that passes through the point M 1 with coordinates (3, 5), parallel to the straight line y = 2 x - 2.
Solution
By condition, we have that parallel lines have identical angles of inclination, which means that the angular coefficients are equal. To find the slope from this equation, you need to remember its basic formula y = 2 x - 2, it follows that k = 2. We compose an equation with the slope coefficient and get:
y - y 1 = k (x - x 1) ⇔ y - 5 = 2 (x - 3) ⇔ y = 2 x - 1
Answer: y = 2 x - 1 .
Transition from a straight line equation with a slope to other types of straight line equations and back
This equation is not always applicable for solving problems, since it is not very conveniently written. To do this, you need to present it in a different form. For example, an equation of the form y = k x + b does not allow us to write down the coordinates of the direction vector of a straight line or the coordinates of a normal vector. To do this, you need to learn to represent with equations of a different type.
We can get canonical equation line on a plane using the equation of a line with a slope. We get x - x 1 a x = y - y 1 a y . It is necessary to move the term b to left side and divide by the expression of the resulting inequality. Then we get an equation of the form y = k · x + b ⇔ y - b = k · x ⇔ k · x k = y - b k ⇔ x 1 = y - b k.
The equation of a line with a slope has become the canonical equation of this line.
Example 7
Bring the equation of a straight line with an angular coefficient y = - 3 x + 12 to canonical form.
Solution
Let us calculate and present it in the form of a canonical equation of a straight line. We get an equation of the form:
y = - 3 x + 12 ⇔ - 3 x = y - 12 ⇔ - 3 x - 3 = y - 12 - 3 ⇔ x 1 = y - 12 - 3
Answer: x 1 = y - 12 - 3.
The general equation of a straight line is easiest to obtain from y = k · x + b, but for this it is necessary to make transformations: y = k · x + b ⇔ k · x - y + b = 0. A transition is made from the general equation of the line to equations of a different type.
Example 8
Given a straight line equation of the form y = 1 7 x - 2 . Find out whether the vector with coordinates a → = (- 1, 7) is a normal line vector?
Solution
To solve it is necessary to move to another form of this equation, for this we write:
y = 1 7 x - 2 ⇔ 1 7 x - y - 2 = 0
The coefficients in front of the variables are the coordinates of the normal vector of the line. Let's write it like this: n → = 1 7, - 1, hence 1 7 x - y - 2 = 0. It is clear that the vector a → = (- 1, 7) is collinear to the vector n → = 1 7, - 1, since we have the fair relation a → = - 7 · n →. It follows that the original vector a → = - 1, 7 is a normal vector of the line 1 7 x - y - 2 = 0, which means it is considered a normal vector for the line y = 1 7 x - 2.
Answer: Is
Let's solve the inverse problem of this one.
Need to move from general view equations A x + B y + C = 0, where B ≠ 0, to an equation with a slope. To do this, we solve the equation for y. We get A x + B y + C = 0 ⇔ - A B x - C B .
The result is an equation with a slope equal to - A B .
Example 9
A straight line equation of the form 2 3 x - 4 y + 1 = 0 is given. Obtain the equation of a given line with an angular coefficient.
Solution
Based on the condition, it is necessary to solve for y, then we obtain an equation of the form:
2 3 x - 4 y + 1 = 0 ⇔ 4 y = 2 3 x + 1 ⇔ y = 1 4 2 3 x + 1 ⇔ y = 1 6 x + 1 4 .
Answer: y = 1 6 x + 1 4 .
An equation of the form x a + y b = 1 is solved in a similar way, which is called the equation of a straight line in segments, or canonical of the form x - x 1 a x = y - y 1 a y. We need to solve it for y, only then we get an equation with the slope:
x a + y b = 1 ⇔ y b = 1 - x a ⇔ y = - b a · x + b.
The canonical equation can be reduced to a form with an angular coefficient. For this:
x - x 1 a x = y - y 1 a y ⇔ a y · (x - x 1) = a x · (y - y 1) ⇔ ⇔ a x · y = a y · x - a y · x 1 + a x · y 1 ⇔ y = a y a x · x - a y a x · x 1 + y 1
Example 10
There is a straight line given by the equation x 2 + y - 3 = 1. Reduce to the form of an equation with an angular coefficient.
Solution.
Based on the condition, it is necessary to transform, then we obtain an equation of the form _formula_. Both sides of the equation must be multiplied by - 3 to obtain the required slope equation. Transforming, we get:
y - 3 = 1 - x 2 ⇔ - 3 · y - 3 = - 3 · 1 - x 2 ⇔ y = 3 2 x - 3 .
Answer: y = 3 2 x - 3 .
Example 11
Reduce the straight line equation of the form x - 2 2 = y + 1 5 to a form with an angular coefficient.
Solution
It is necessary to calculate the expression x - 2 2 = y + 1 5 as a proportion. We get that 5 · (x - 2) = 2 · (y + 1) . Now you need to completely enable it, to do this:
5 (x - 2) = 2 (y + 1) ⇔ 5 x - 10 = 2 y + 2 ⇔ 2 y = 5 x - 12 ⇔ y = 5 2 x
Answer: y = 5 2 x - 6 .
To solve such problems, parametric equations of the line of the form x = x 1 + a x · λ y = y 1 + a y · λ should be reduced to the canonical equation of the line, only after this can one proceed to the equation with the slope coefficient.
Example 12
Find the slope of the line if it is given by parametric equations x = λ y = - 1 + 2 · λ.
Solution
It is necessary to transition from the parametric view to the slope. To do this, we find the canonical equation from the given parametric one:
x = λ y = - 1 + 2 · λ ⇔ λ = x λ = y + 1 2 ⇔ x 1 = y + 1 2 .
Now it is necessary to resolve this equality with respect to y in order to obtain the equation of a straight line with an angular coefficient. To do this, let's write it this way:
x 1 = y + 1 2 ⇔ 2 x = 1 (y + 1) ⇔ y = 2 x - 1
It follows that the slope of the line is 2. This is written as k = 2.
Answer: k = 2.
If you notice an error in the text, please highlight it and press Ctrl+Enter
This mathematical 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.
This online calculator may be useful for high school students secondary schools in preparation for tests and exams, when testing knowledge before the Unified State Exam, for parents to control the solution of many problems in mathematics and algebra. Or maybe it’s too expensive for you to hire a tutor or buy new textbooks? Or do you just want to get it done as quickly as possible? homework
in mathematics or algebra? In this case, you can also use our programs with detailed solutions.
If you need to find the derivative of a function, then for this we have the task Find the derivative.
If you are not familiar with the rules for entering functions, we recommend that you familiarize yourself with them.
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.
You may have AdBlock enabled.
In this case, disable it and refresh the page.
For the solution to appear, you need to enable JavaScript.
Here are instructions on how to enable JavaScript in your browser.
Because There are a lot of people willing to solve the problem, your request has been queued.
In a few seconds the solution will appear below.
Please wait sec...
If you noticed an error in the solution, then you can write about this in the Feedback Form.
Do not forget indicate which task you decide what enter in the fields.
Our games, puzzles, emulators:
A little theory.
Direct slope
Recall that the graph of the 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 \(y=f(a)+ f"(a)(x-a) \)
Loading...