The linear function y = kx + m when m = 0 takes the form y = kx. In this case, you can notice that:

1. If x = 0, then y = 0. Therefore, the graph linear function y = kx passes through the origin regardless of the value of k.
2. If x = 1 then y = k.

Let's consider different values ​​of k, and how y changes from this.

If k is positive (k > 0), then the straight line (the graph of the function), passing through the origin, will lie in the I and III coordinate quarters. After all, with positive k, when x is positive, then y will also be positive. And when x is negative, y will also be negative. For example, for the function y = 2x, if x = 0.5, then y = 1; if x = –0.5, then y = –1.

Now, assuming k is positive, consider three different linear equations. Let these be: y = 0.5x and y = 2x and y = 3x. How does the value of y change for the same x? Obviously it increases with k: the larger k, the larger y. This means that the straight line (function graph) with a larger value of k will have a larger angle between the x-axis (abscissa axis) and the function graph. Thus, the angle at which the straight axis intersects the x axis depends on k, and hence k is spoken of as slope of linear function.

Now let's study the situation when k x is positive, then y will be negative; and vice versa: if x y > 0. Thus, the graph of the function y = kx for at k

Let's say there are linear equations y = –0.5x, y = –2x, y = –3x. For x = 1 we get y = –0.5, y = –2, y = –3. For x = 2 we get y = –1, y = –2, y = –6. Thus, the larger k, the larger y if x is positive.

However, if x = –1, then y = 0.5, y = 2, y = 3. For x = –2 we get y = 1, y = 4, y = 6. Here, as the value of k decreases, y at x increases

Graph of the function at k

Graphs of functions of the type y = kx + m differ from graphs y = km only in a parallel shift.

Linear function

Linear function is a function that can be specified by the formula y = kx + b,

where x is the independent variable, k and b are some numbers.

The graph of a linear function is a straight line.

The number k is called slope of a straight line– graph of the function y = kx + b.

If k > 0, then the angle of inclination of the straight line y = kx + b to the axis X spicy; if k< 0, то этот угол тупой.

If the slopes of the lines that are graphs of two linear functions are different, then these lines intersect. And if the angular coefficients are the same, then the lines are parallel.

Graph of a function y=kx +b, where k ≠ 0, is a line parallel to the line y = kx.

Direct proportionality.

Direct proportionality is a function that can be specified by the formula y = kx, where x is an independent variable, k is a non-zero number. The number k is called coefficient of direct proportionality.

The graph of direct proportionality is a straight line passing through the origin of coordinates (see figure).

Direct proportionality is a special case of a linear function.

Function Propertiesy=kx:

Inverse proportionality

Inverse proportionality is called a function that can be specified by the formula:

k
y = -
x

Where x is the independent variable, and k– a non-zero number.

The graph of inverse proportionality is a curve called hyperbole(see picture).

For a curve that is the graph of this function, the axis x And y act as asymptotes. Asymptote- this is the straight line to which the points of the curve approach as they move away to infinity.

k
Function Propertiesy = -:
x

Class: 8

Presentation for the lesson

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Lesson type: lesson of discovering new knowledge.

Basic goals:

• form an idea of ​​the function y = kx 2, its properties and graphics;
• repeat and reinforce: function details y = x 2, properties of the function, known from the 7th grade course.

Demo material:

1) algorithm for constructing a graph of a function:

2) The rule for determining the location of the graph depending on the coefficient k:

3) independent work: In Fig. graphs of functions y = kx are shown 2 .

For each graph, indicate the corresponding coefficient value To.

4) sample for self-test independent work.

Handout:

1) card:

1st, 2nd group:

Plot Function Graphs y = 2X 2 , y = 4X

3, 4 group:

Plot Function Graphs y =– 2X 2 , y = – 4X 2 and determine in which coordinate quarters the graphs of these functions are located. Draw a conclusion regarding the coefficient k.

2) card for reflection:

DURING THE CLASSES

1. Motivation to educational activities

Goals:

• organize updating of requirements for the student in terms of educational activities;
• organize student activities to establish thematic frameworks: we continue to work with functions;
• create conditions for the student to develop an internal need for inclusion in educational activities.

Organization educational process at stage 1:

- Hello! What interesting things did you learn in previous lessons? (We studied the function y = | x |, the graph of this function and its properties.)
– Today you will continue to get acquainted with new functions.
– In what mood will you work today? (With nice mood).
- I wish you success!

2. Updating knowledge and fixing difficulties in individual activities

Goals:

• update educational content that is necessary and sufficient for the perception of new material.
• record updated methods of action in speech and signs;
• organize a generalization of updated methods of action;
• motivate to complete an individual task;
• organize independent implementation individual assignment for new knowledge;
• organize recording of individual difficulties in students’ performance of an individual task or in justifying it.

Organization of the educational process at stage 2:

Analyze several slides 2-5 and answer the question:

– What schedule will you work with today? (With a parabola).

– Choose which function is the graph of a parabola at = X + 2, at = 2/X, y = x 2 ?(y = x 2 . We studied this function in 7th grade).

– Name the numerical coefficient of the function y = x 2 . (It is equal to 1)

– In which coordinate quarters does the graph of the function lie? y = x 2 , What is the domain of definition and range of values ​​of this function, the intervals of increase and decrease? (Graph of the function y = x 2 lies in the 1st and 2nd coordinate quarters or in the upper half-plane, the domain of definition is the entire number line, the range of values ​​is the function y = x 2 takes non-negative values; increases with x > 0, decreases with x < 0.)

– Let’s discuss what happens at other values ​​of the coefficient.

– Formulate the topic of the lesson. (Function y = kx 2 , its properties and graph).

1) A table has been prepared on the board. Find the corresponding function values:

y = 2X 2
y = 4X 2
y =– 2X 2
y =– 4X 2

- Fill the table. 4 students are called to the board in succession.

2) Function graph y = kx 2 passes through point A(2;8). Determine the value of the coefficient. Write down the function. (k = 2, y = 2x 2 ).

3) What plan do you usually use to graph functions? Slide 7.

(Necessary -
1. Fill out the table of values
2. Plot points on coordinate plane
3. Connect the constructed points with a smooth line
4. Write the name of the function.)

-What did you repeat?

– And now, using everything you just repeated and learned, I suggest you complete the following task:
Plot Function Graphs y = 2X 2 , y = – 4X 2 and determine in which coordinate quarters the graphs of these functions are located. Conclude how the graph is located depending on the coefficient k.

Students work on graph paper.

– Who doesn’t have results?
– What couldn’t you do? (I could not__________________)
– Show the results of who performed the construction.
– How can you prove that you completed the task correctly? (I must___________)
– What will you use to prove it? (___________.)
– What couldn’t you do?
– What rule did you use when constructing?
- That you can not do?

3. Identifying the causes of the difficulty

Goals:

• organize the correlation of your actions with the standards used (algorithm, concept, etc.);
• on this basis, organize the identification and recording in external speech of the cause of the difficulty - those specific knowledge and skills that are lacking to solve the original problem.

Organization of the educational process at stage 3:

– What task did you have to complete?
– What did you use to complete the task?
– Where did the difficulty arise?
– What is the reason for the difficulty? (We do not have a way to determine how the graph of the function y = kx2 is located depending on the coefficient k.)

4. Problematic explanation of new knowledge

Goals:

• organize setting the lesson goal;
• organize clarification and agreement on the topic of the lesson;
• organize a leading or stimulating dialogue on the problematic introduction of new knowledge;
• organize the use of objective actions with models, diagrams, properties, etc.;
• organize the recording of a new method of action in speech;
• organize the fixation of a new method of action in signs;
• correlating new knowledge with a rule in a textbook, reference book, dictionary, etc.
• organize a record of overcoming the difficulty.

Organization of the educational process at stage 4:

– Formulate the purpose of your activity. (Find a way to determine how the graph of the function y = kx is located 2 depending on the coefficient k.)

– Specify the topic of the lesson. (Function y = kx 2 ,its properties and graph). Slide 6.

– And now you will work in groups: Slide 8.

1st, 2nd group:

Plot Function Graphs y = 2X 2 , y = 4X 2 and determine in which coordinate quarters the graphs of these functions are located. Draw a conclusion regarding the coefficient k.

3, 4 group:

Plot Function Graphs y = – 2X 2 ,y = – 4X 2 and determine in which coordinate quarters the graphs of these functions are located. Draw a conclusion regarding the coefficient k.

Each group is given a card. (If difficulties arise, students can use a textbook or reference book.)

– Present your version of the algorithm.

Each group presents its own version, the others complement and clarify. After agreement, the rule is posted on the board:

The teacher adds:

– Each of the lines you constructed is called a parabola. In this case, the point (0;0) is called the vertex of the parabola, and the axis at– the axis of symmetry of the parabola.
The “speed of movement” of the branches of the parabola upward (downward) and the “degree of steepness” of the parabola depend on the value of the coefficient k.
-What have you discovered just now?
– What should you do now?

5. Primary consolidation in external speech

Target: organize children’s assimilation of a new way of acting with their pronunciation in external speech.

Organization of the educational process at stage 5:

– In which coordinate quarters are the graphs of functions located? at = 1/5X 2 , at = X 2 /2, at = – X 2 /2, at = 3X 2 ?

The task is performed in pairs, one pair works at the board.

6. Independent work with self-test according to the sample

Goals:

• organize independent implementation by students typical tasks on new way actions;
• Based on the results of independent work, organize the identification and correction of mistakes;
• based on the results of independent work, create a situation of success.

Organization of the educational process at stage 6:

For independent work, a task is provided on the card. Slide 9.

In Fig. graphs of functions are shown at = kh 2 .

For each graph, indicate the corresponding value of the coefficient k.

After completing the work, students check it according to the sample: Slide 10.

– What rules did you use when completing the task?
– Who has a problem - how to determine the sign of the coefficient k?
– Who had difficulty determining the value of the coefficient k?
– Who completed the task correctly?

7. Inclusion in the knowledge system and repetition

Goals:

• train skills in using new content in conjunction with previously studied material;
• review the learning content required in the following lessons:

Organization of the educational process at stage 7:

The task from GIA-9 is performed at the board. Slides 11-16.

– Define the term that was repeated many times today in class. (graph)

1. The graph of which of these functions is a parabola located in the lower half-plane?

3. Find the range of values ​​of the function y = – 5x2

A) at = –15X 2 b) at = – 9X 2 V) at = – X 2 G) at = – 5X 2

ts uh f and

5. Indicate the intervals for increasing the function y = – 5x 2

a) when X > 0 b) when X < 0 c) at X< 0 d) at X > 0

h O And T

6. Specify smallest value functions y = – 5x 2

a) 0 b) does not exist at 5 d) 5

s To d V.

Physics problems: Slide 17.

The path traveled by the body during the first t seconds of free fall is calculated by the formula: H = GT 2 /2, where g= 9.8 m/s 2. Find the dependence of H on the graph t:

A) the distance that the falling stone will fly in the first 6 seconds;
B) the time it takes for the stone to fly the first 250 m?

8. Reflection on activities in the lesson

Goals:

• organize fixation new content, studied in the lesson;
• organize recording of the degree of compliance with the set goal and performance results;
• organize verbal recording of steps to achieve the goal;
• based on the results of the analysis of work in the lesson, organize the recording of directions for future activities;
• organize self-assessment of students’ work in class;
• organize a discussion and recording of homework.

Organization of the educational process at stage 8:

– What did you study today?
– What new did you learn in the lesson?
– What goals did you set for yourself?
– Have you achieved your goals?
– What helped you cope with difficulties?
– Analyze your work in class.

Students work with reflection cards (R).

Homework: Slide 18.

• Read paragraph 17 of the textbook
• №17.2,
• №17.3,
• №17.11.

Bibliography:

1. A.G. Mordkovich. Algebra, 8th grade. In two parts. Textbook for students of general education institutions. M.:Mnemosyne.2011.
2. Internet resources.

Algebra lesson in 7th grade using the textbook by Mordkovich Alexander Grigorievich.

Linear function y=kx and its graph.

Goals:

Generalize and deepen knowledge on the topic “Linear function y = kx +m and its graph” Consider the properties of graphs of linear functions y = kx with different coefficients k.

To promote the development of observation, the ability to analyze, compare, generalize.

Arouse in students the need to substantiate their statements, cultivate self-control and mutual control.

During the classes:

Organizing time.

Teacher's opening speech.

You have already studied the linear function y =kx +m and learned how to build graphs of this function, and now, please consider the graphs of the following functions and answer the questions:

SLIDE 2

Linear functions are plotted on the coordinate plane:

y=x,

y =0.5x ;

y=-x;

y=-4x

Will these functions be linear? Why? What do these four functions discussed have in common? How do they differ from previously studied linear functions?

SLIDE 3

Graphs of linear function data.

SLIDE 4 (questions for slide 3)

Answers:

The graphs of these linear functions are either in the 1st and 3rd quarters, or in the 2nd and 4th quarters.

What is the relationship between the coefficient k and the location of the graph on the coordinate plane?

SLIDE 5 (answers to questions on slide 4)

All graphs of these linear functions pass through the origin O(0;0)

SLIDE 6

If coefficient k<0, то линейная функция убывает и расположена во 2 и 4 четвертях.

SLIDE 7

If the coefficient k >0, then the linear function increases and is located in the first and third quarters.

SLIDE 8

Now complete the following tasks in textbook No. 348 (a, b), 355:

Problem No. 348(a; b).
Plot a linear function:
a) y =2x,
b) y = -3x.
On one coordinate plane.
What can you say about the graphs of these linear functions?

(They pass through the origin, the linear function y=2x is increasing and is located in the 1st and 3rd quarters, and the linear function y=-3x is decreasing and is located in the 2nd and 4th quarters).

SLIDE 9

Solution (finding the coordinates of data points of linear functions). How many point coordinates are needed to plot a graph of given linear functions? Why? (One, because the graphs of linear data pass through the origin, that is, the point with coordinate (0;0), and we already know it.)

SLIDE10

If you completed the task correctly, you should end up with a graph like this.

SLIDE11

We construct the graph of the linear function y = -3x in a similar way

What can you say about this function? In which quadrants will the graph of this linear function be located?

If we take the abscissa value to be positive, then the ordinate is negative, and, conversely, if the abscissa value is negative, then the ordinate is positive.

SLIDE12

If you completed the task correctly, then you should get a graph of this linear function y=-3x.

SLIDE13

(Formulation of problem No. 355)

SLIDE14

(Questions that activate the solution to the task).

SLIDE15

Finding the coordinates of points to plot a graph of a given linear function y=0.4x.

SLIDE16

Using the graph of this linear function, we find the ordinate value corresponding to the abscissa value equal to 0; 5; 10; -5.

If x =0, then y =0

If x =5, then y =2

If x =10, then y =4

If x =-5, then y =-2

SLIDE17

Using the graph of this linear function, we find the value x corresponding to the value y equal to 0; 2; 4; -2.

If y =0, then x =0

If y =2, then x =5

If y =4, then x =10

If y =-2, then x =-5

SLIDE18

Solution to the inequality: 0.4x >0. What do we need to know to solve this inequality? Find at what values ​​of the abscissa (x) the graph of this linear function will be above the ox axis.

SLIDE19

Now, using the graph of this linear function, we solve the inequality: -2≤y ≤0.

Let's think about how to solve this inequality?

1. Mark the points y =-2 and y =0 on the axis oy.

2. We obtain a straight line segment that lies within the values ​​-2≤y ≤0:

From the ordinate equal to -2 and the ordinate equal to 0, we lower the perpendicular to the graph of this linear function.

3. From the ends of the straight line segment of the graph, drop perpendiculars to the ox axis.

4. We obtained the abscissa values ​​within which the graph of this straight line lies: -5≤x ≤0. This interval will be the solution to this task.

SLIDE 20

Homework – independent completion No. 356.

Definition of a Linear Function

Let us introduce the definition of a linear function

Definition

A function of the form $y=kx+b$, where $k$ is nonzero, is called a linear function.

The graph of a linear function is a straight line. The number $k$ is called the slope of the line.

When $b=0$ the linear function is called a function of direct proportionality $y=kx$.

Consider Figure 1.

Rice. 1. Geometric meaning of the slope of a straight line

Consider triangle ABC. We see that $ВС=kx_0+b$. Let's find the point of intersection of the line $y=kx+b$ with the axis $Ox$:

\ \

So $AC=x_0+\frac(b)(k)$. Let's find the ratio of these sides:

\[\frac(BC)(AC)=\frac(kx_0+b)(x_0+\frac(b)(k))=\frac(k(kx_0+b))((kx)_0+b)=k \]

On the other hand, $\frac(BC)(AC)=tg\angle A$.

Thus, we can draw the following conclusion:

Conclusion

Geometric meaning coefficient $k$. The angular coefficient of the straight line $k$ is equal to the tangent of the angle of inclination of this straight line to the $Ox$ axis.

Study of the linear function $f\left(x\right)=kx+b$ and its graph

First, consider the function $f\left(x\right)=kx+b$, where $k > 0$.

1. $f"\left(x\right)=(\left(kx+b\right))"=k>0$. Consequently, this function increases over the entire domain of definition. There are no extreme points.
2. $(\mathop(lim)_(x\to -\infty ) kx\ )=-\infty $, $(\mathop(lim)_(x\to +\infty ) kx\ )=+\infty $
3. Graph (Fig. 2).

Rice. 2. Graphs of the function $y=kx+b$, for $k > 0$.

Now consider the function $f\left(x\right)=kx$, where $k

1. The domain of definition is all numbers.
2. The range of values ​​is all numbers.
3. $f\left(-x\right)=-kx+b$. The function is neither even nor odd.
4. For $x=0,f\left(0\right)=b$. When $y=0.0=kx+b,\ x=-\frac(b)(k)$.

Intersection points with coordinate axes: $\left(-\frac(b)(k),0\right)$ and $\left(0,\ b\right)$

1. $f"\left(x\right)=(\left(kx\right))"=k
2. $f^("")\left(x\right)=k"=0$. Therefore, the function has no inflection points.
3. $(\mathop(lim)_(x\to -\infty ) kx\ )=+\infty $, $(\mathop(lim)_(x\to +\infty ) kx\ )=-\infty $
4. Graph (Fig. 3).