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. AT this case The graph can be either a straight line or a curved line. That is, the derivative characterizes the rate of change of the function at a particular point in time. Remember general rules for 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 there.
Learn to distinguish tasks in which slope it is required to calculate through the derivative of the function. In tasks, it is not always suggested 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 given function. You don't 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 the 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).
- Function derivative:
- f ′ (x) = 4 x + 6 (\displaystyle f"(x)=4x+6)
- Substitute the value of the x-coordinate of the given point:
- f ′ (x) = 4 (4) + 6 (\displaystyle f"(x)=4(4)+6)
- Find the slope:
- Slope of the function f (x) = 2 x 2 + 6 x (\displaystyle f(x)=2x^(2)+6x) at point A(4,2) is 22.
If possible, check your answer on a graph. Keep in mind that the slope factor cannot be calculated at every point. Differential calculus considers complex functions and complex graphs, where the slope can not 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 given to you is correct. Otherwise, draw a tangent to the graph at the given point and consider whether the value of the slope you found corresponds to what you see on the graph.
- The tangent will have the same slope as the function graph at a certain point. To draw a tangent at a given point, move right/left 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 you've given. 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 analytically express the geometric properties characterizing the points of the line under consideration by an equation between the current coordinates. Thus, we get the equation of the line. In this chapter, the equations of straight lines will be considered.
To formulate the equation of a straight line in Cartesian coordinates, you need to somehow set the conditions that determine its position relative to the coordinate axes.
First, we introduce the concept of the slope of a straight line, which is one of the quantities characterizing the position of a straight line on a plane.
Let's call the angle of inclination of the line to the Ox axis the angle by which the Ox axis must be rotated so that it coincides with the given line (or turns out to be 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 by an angle of 180 ° will again combine it with the straight line, the angle of inclination of the straight line to the axis can be chosen ambiguously (up to a multiple of ).
The tangent of this angle is uniquely determined (since changing the angle to does not change its tangent).
The tangent of the angle of inclination of a straight line to the x-axis is called the slope of the straight line.
The slope 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 the line is zero, then the line is parallel to the x-axis. With a positive slope, the angle of inclination of the straight line to the Ox axis will be sharp (we are considering here the smallest positive value of the angle of inclination) (Fig. 39); in this case, the larger the slope, the greater the angle of its inclination to the Ox axis. If the slope is negative, then the angle of inclination of the straight line to the x-axis will be obtuse (Fig. 40). Note that a straight line perpendicular to the x-axis does not have a slope (the tangent of an angle does not exist).
The slope coefficient is straight. In this article, we will consider tasks related to the coordinate plane included in the exam in mathematics. These are assignments for:
- determination of the slope 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 lines on the 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 needs to be understood for the type of tasks under consideration? A bit of theory.
The equation of a straight line on the coordinate plane has the form:
where k – this is the slope of the straight line.
Next moment! The slope of a straight line is equal to the tangent of the slope of the straight line. This is the angle between the given line and the axisoh.
It lies between 0 and 180 degrees.
That is, if we reduce the equation of a straight line to the form y = kx + b, then further we can always determine the coefficient k (slope coefficient).
Also, if we can determine the tangent of the slope of the straight line based on the condition, then we will thereby find its slope.
The next theoretical moment!Equation of a straight line passing through two given points.The formula looks like:
Let's consider tasks (similar to tasks from the open bank of tasks):
Find the slope of the straight line passing through the points with coordinates (–6; 0) and (0; 6).
In this problem, the most rational way to solve this 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 angular coefficient. Consider a right triangle formed by a straight line and the x and y axes:
The tangent of an angle in right triangle is the ratio of the opposite leg to the adjacent:
* Both legs are equal to six (these are their lengths).
Of course, this task can be solved using the formula for finding the equation of a straight line passing through two given points. But it will be a longer solution path.
Answer: 1
Find the slope of the straight 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 got that the angular coefficient 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 ox.
In this problem, you can find the equation of a straight line a, determine the slope for it. Straight line b the slope will be the same since they are parallel. Next, you can find the equation of a straight line b. And then, substituting the value y = 0 into it, find the abscissa. BUT!
In this case, it is easier to use the triangle similarity property.
The right triangles formed by the given (parallel) lines of coordinates are similar, which means that the ratios of their respective sides are equal.
The desired 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 ox.
For this problem, the most rational way to solve it is to use the similarity property of triangles. But we will solve it in a different way.
We know the points through which the line passes a. We can write the equation of a straight line. The formula for the equation of a straight line passing through two given points is:
By condition, the points have coordinates (0;8) and (–12;0). Means,
Let's bring 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.
We know that parallel lines have equal slopes. So the equation of a straight line passing through the point (0;-12) has the form:
Find value b we can substitute the abscissa and ordinate into the equation:
So the 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 \u003d 0:
Answer: 18
Find the ordinate of the point of intersection of the axis oy and a straight 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 the points with coordinates (0;0) and (10;24).
The formula for the equation of a straight line passing through two given points is:
Our points have coordinates (0;0) and (10;24). Means,
Let's bring to mind y = kx + b
The slopes of the parallel lines are equal. Hence, the equation of a straight line passing through the point B (10; 12) has the form:
Meaning b we find by substituting the coordinates of the point B (10; 12) into this equation:
We got the equation of a straight line:
To find the ordinate of the point of intersection of this line with the axis OU must be substituted into the found equation X= 0:
*Easiest solution. With the help of parallel translation, we shift this line down along the axis OU to the point (10;12). The shift occurs by 12 units, that is, point A(10;24) "passed" to point B(10;12), and point O(0;0) "passed" to point (0;–12). So the resulting line will intersect the axis OU at the point (0;–12).
The desired ordinate is -12.
Answer: -12
Find the ordinate of the point of intersection of the line given by the equation
3x + 2y = 6, with axis Oy.
Coordinate of the point of intersection of the given line with the axis OU has the form (0; at). Substitute the abscissa into the equation X= 0, and find the ordinate:
Ordinate of point of intersection of a line with an axis OU equals 3.
* The system is being 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, the system of these equations is solved:
In the first equation, we substitute - X instead of at:
The ordinate is minus six.
Answer: – 6
Find the slope of the straight line passing through the points with coordinates (–2; 0) and (0; 2).
Find the slope of the straight line passing through the points with coordinates (2;0) and (0;2).
The line a passes through the 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 x-axis.
Find the ordinate of the point of intersection of the y-axis and the line passing through point B (6;4) and the parallel line passing through the origin and point A (6;8).
1. It is necessary to clearly understand that the slope of the straight line is equal to the tangent of the slope 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 can always find the equation of a straight line if the coordinates of two of its 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 sign of similarity of triangles. Problems are solved practically orally.
5. Tasks in which two lines are given and it is required to find the abscissa or ordinate of their intersection point can be solved graphically. That is, build them on the coordinate plane (on a sheet in a cell) and determine the intersection point visually. *But this method is not always applicable.
6. And the last. 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 slope through finding the tangent of the angle in the formed right triangle. How to "see" this triangle for various arrangements of lines on the plane is schematically shown below:
>> Line inclination angle from 0 to 90 degrees<<
>> Straight line angle from 90 to 180 degrees<<
That's all. Good luck to you!
Sincerely, Alexander.
P.S: I would be grateful if you tell about the site in social networks.
The continuation of the topic of the equation of a straight line on a plane is based on the study of a straight line from algebra lessons. This article gives generalized information on the topic of the equation of a straight line with a slope. Consider the definitions, get the equation itself, reveal the connection with other types of equations. Everything will be discussed on examples of problem solving.
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Before writing such an equation, it is necessary to define the angle of inclination of a straight line to the O x axis with their slope. Let us assume that a Cartesian coordinate system O x is given on the plane.
Definition 1
The angle of inclination of the straight line to the axis O x, 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 a line is parallel to Ox or coincidence occurs in it, the angle of inclination is 0. Then the angle of inclination of the given straight line α is defined on the interval [ 0 , π) .
Definition 2
Slope of a straight line is the tangent of the slope of the given line.
The standard notation is k. From the definition we get that k = t g α . When the line is parallel to Ox, the slope is said to not exist because it goes to infinity.
The slope is positive when the graph of the function is increasing and vice versa. The figure shows various variations of 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 slope coefficient and calculate the tangent of the inclination angle in the plane.
Solution
From the condition we have that α = 120 °. By definition, you need to calculate the slope. Let's find it from the formula k = t g α = 120 = - 3 .
Answer: k = - 3 .
If the angular coefficient is known, but it is necessary to find the angle of inclination to the x-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 a slope equal to 3.
Solution
From the condition we have that the slope is positive, which means that the angle of inclination to O x is less than 90 degrees. Calculations are made according to 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 slope, then α is the angle of inclination to the 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 pi 6 .
An equation of the form y \u003d k x + b, where k is a slope, and b is some real number, is called the equation of a straight 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 given by an equation with a slope that looks like y \u003d k x + b. In this case, it means that the coordinates of any point on the line correspond to the equation. If we substitute the coordinates of the point M, M 1 (x 1, y 1), into the equation y \u003d 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
Given a straight line with slope y = 1 3 x - 1 . 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, so 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 the 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 straight line is defined by the equation y = k · x + b passing through M 1 (0 , b) , substitution yielded an equality of the form b = k · 0 + b ⇔ b = b . From this we can conclude that the equation of a straight line with a slope 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 .
Consider, for example, a straight line defined using a slope given by the form y = 3 · x - 1 . We get 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 along the positive direction of the O x axis. From this it can be seen that the coefficient is 3.
The 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 coefficient 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
Compose the equation of a straight line passing through the point M 1 with coordinates (4, - 1), with a slope equal to - 2.
Solution
By condition, we have that x 1 \u003d 4, y 1 \u003d - 1, k \u003d - 2. From here, the equation of the straight line will be written in this way 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 a slope that passes through the point M 1 with coordinates (3, 5) parallel to the straight line y \u003d 2 x - 2.
Solution
By condition, we have that parallel lines have coinciding angles of inclination, hence the slope coefficients are equal. To find the slope from this equation, you need to remember its basic formula y \u003d 2 x - 2, which implies that k \u003d 2. We compose an equation with a 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 .
The transition from the equation of a straight line with a slope to other types of equations of a straight line and vice versa
Such an equation is not always applicable for solving problems, since it has a not very convenient notation. To do this, it must be presented in a different form. For example, an equation of the form y = k · x + b does not allow you to write down the coordinates of the direction vector of the straight line or the coordinates of the normal vector. To do this, you need to learn how to represent equations of a different kind.
We can get canonical equation a straight line in a plane using the equation of a straight line with a slope. We get x - x 1 a x = y - y 1 a y . It is necessary to transfer 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 straight line with a slope has become the canonical equation of a given straight line.
Example 7
Bring the equation of a straight line with slope y = - 3 x + 12 to canonical form.
Solution
We calculate and represent 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 this requires transformations: y = k x + b ⇔ k x - y + b = 0. A transition is made from the general equation of a straight line to equations of another type.
Example 8
An equation of a straight line of the form y = 1 7 x - 2 is given. Find out if the vector with coordinates a → = (- 1 , 7) is a normal straight line vector?
Solution
To solve it, it is necessary to switch 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 straight 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 a 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 that it is considered a normal vector for the line y = 1 7 x - 2 .
Answer: Is
Let's solve the problem inverse to this one.
Need to move from general view equation A x + B y + C = 0 , where B ≠ 0 , to the slope equation. 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
An equation of a straight line of the form 2 3 x - 4 y + 1 = 0 is given. Get the equation of a given line with a slope.
Solution
Based on the condition, it is necessary to solve for y, then we get 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 .
In a similar way, an equation of the form x a + y b \u003d 1 is solved, which is called the equation of a straight line in segments, or the canonical form x - x 1 a x \u003d y - y 1 a y. It is necessary to solve it with respect to y, only then we get an equation with a 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 a slope. 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 . Bring to the form of an equation with a slope.
Solution.
Based on the condition, it is necessary to transform, then we get an equation of the form _formula_. Both sides of the equation should be multiplied by -3 to get 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
The straight line equation of the form x - 2 2 \u003d y + 1 5 is brought to the form with a slope.
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 fully enable it, for 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 tasks, parametric equations of the straight line of the form x \u003d x 1 + a x λ y \u003d y 1 + a y λ should be reduced to the canonical equation of the straight line, only after that you can proceed to the equation with the slope.
Example 12
Find the slope of the straight line if it is given by parametric equations x = λ y = - 1 + 2 · λ .
Solution
You need to transition from parametric view to 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 a slope. To do this, we write this way:
x 1 = y + 1 2 ⇔ 2 x = 1 (y + 1) ⇔ y = 2 x - 1
It follows that the slope of the straight line is equal to 2. This is written as k = 2 .
Answer: k = 2 .
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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.
This online calculator can be useful for high school students general education schools in preparation for control work and exams, when testing knowledge before the exam, 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 soon as possible? homework math or algebra? In this case, you can also use our programs with a detailed solution.
In this way, you can conduct your own training and/or the training of your younger brothers or sisters, while the level of education in the field of tasks to be solved is increased.
If you need to find the derivative of a function, then for this we have the Find Derivative task.
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A bit of theory.
Slope of a straight line
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 \(kThe equation of the tangent to the graph of the function
If the point M (a; f (a)) belongs to the graph of the function y \u003d f (x) and if at this point it is possible to draw a tangent to the graph of the function that is not perpendicular to the x-axis, then from the geometric meaning of the derivative it follows that the slope of the tangent is equal to f "(a). Next, we will develop an algorithm for compiling the equation of the tangent to the graph of any function.
Let the function y \u003d f (x) and the point M (a; f (a)) on the graph of this function be given; let it be known that f "(a) exists. Let us compose the equation of the tangent to the graph of the given function in given point. This equation, like the equation of any straight line not parallel to the y-axis, has the form y = kx + b, so the problem is to find the values of the coefficients k and b.
Everything is clear with the slope k: it is known that k \u003d f "(a). To calculate the value of b, we use the fact that the desired straight line passes through the point M (a; f (a)). This means that if we substitute the coordinates of the point M into the equation of a straight line, we get the correct equality: \ (f (a) \u003d ka + b \), i.e. \ (b \u003d f (a) - ka \).
It remains to substitute the found values of the coefficients k and b into the equation of a straight line:
We received the equation of the tangent to the graph of the 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 point of contact 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 \u003d f (a) + f "(a) (x-a) \)
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