A linear function is a function of the form y=kx+b, where x is an independent variable, k and b are any numbers.
The graph of a linear function is a straight line.

1. To plot a function graph, we need the coordinates of two points belonging to the graph of the function. To find them, you need to take two x values, substitute them into the equation of the function, and calculate the corresponding y values ​​from them.

For example, to plot the function y= x+2, it is convenient to take x=0 and x=3, then the ordinates of these points will be equal to y=2 and y=3. We get points A(0;2) and B(3;3). Let's connect them and get the graph of the function y= x+2:

2. In the formula y=kx+b, the number k is called the proportionality factor:
if k>0, then the function y=kx+b increases
if k
The coefficient b shows the shift of the graph of the function along the OY axis:
if b>0, then the graph of the function y=kx+b is obtained from the graph of the function y=kx by shifting b units up along the OY axis
if b
The figure below shows the graphs of the functions y=2x+3; y= ½x+3; y=x+3

Note that in all these functions the coefficient k Above zero, and functions are increasing. Moreover, the greater the value of k, the greater the angle of inclination of the straight line to the positive direction of the OX axis.

In all functions b=3 - and we see that all graphs intersect the OY axis at the point (0;3)

Now consider the graphs of functions y=-2x+3; y=- ½ x+3; y=-x+3

This time, in all functions, the coefficient k less than zero and features decrease. The coefficient b=3, and the graphs, as in the previous case, cross the OY axis at the point (0;3)

Consider the graphs of functions y=2x+3; y=2x; y=2x-3

Now, in all equations of functions, the coefficients k are equal to 2. And we got three parallel lines.

But the coefficients b are different, and these graphs intersect the OY axis at different points:
The graph of the function y=2x+3 (b=3) crosses the OY axis at the point (0;3)
The graph of the function y=2x (b=0) crosses the OY axis at the point (0;0) - the origin.
The graph of the function y=2x-3 (b=-3) crosses the OY axis at the point (0;-3)

So, if we know the signs of the coefficients k and b, then we can immediately imagine what the graph of the function y=kx+b looks like.
If a k 0

If a k>0 and b>0, then the graph of the function y=kx+b looks like:

If a k>0 and b, then the graph of the function y=kx+b looks like:

If a k, then the graph of the function y=kx+b looks like:

If a k=0, then the function y=kx+b turns into a function y=b and its graph looks like:

The ordinates of all points of the graph of the function y=b are equal to b If b=0, then the graph of the function y=kx (direct proportionality) passes through the origin:

3. Separately, we note the graph of the equation x=a. The graph of this equation is a straight line parallel to the OY axis, all points of which have an abscissa x=a.

For example, the graph of the equation x=3 looks like this:
Attention! The equation x=a is not a function, since one value of the argument corresponds to different values ​​of the function, which does not correspond to the definition of the function.

4. Condition for parallelism of two lines:

The graph of the function y=k 1 x+b 1 is parallel to the graph of the function y=k 2 x+b 2 if k 1 =k 2

5. The condition for two straight lines to be perpendicular:

The graph of the function y=k 1 x+b 1 is perpendicular to the graph of the function y=k 2 x+b 2 if k 1 *k 2 =-1 or k 1 =-1/k 2

6. Intersection points of the graph of the function y=kx+b with the coordinate axes.

with OY axis. The abscissa of any point belonging to the OY axis is equal to zero. Therefore, to find the point of intersection with the OY axis, you need to substitute zero instead of x in the equation of the function. We get y=b. That is, the point of intersection with the OY axis has coordinates (0;b).

With the x-axis: The ordinate of any point belonging to the x-axis is zero. Therefore, to find the point of intersection with the OX axis, you need to substitute zero instead of y in the equation of the function. We get 0=kx+b. Hence x=-b/k. That is, the point of intersection with the OX axis has coordinates (-b / k; 0):

Linear function y = kx + m when m = 0 becomes y = kx . In this case, you can see that:

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

Consider different values ​​of k and how this changes y .

If k is positive (k > 0), then the straight line (function graph), passing through the origin, will lie in the I and III coordinate quarters. After all, for 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 a function y = 2x , if x = 0.5, then y = 1; if x = –0.5, then y = –1.

Now, under the condition of positive k, consider three different linear equations. Let it be: y = 0.5x and y = 2x and y = 3x . How does the value of y change with the same x ? Obviously it increases with k : the more k , the more y . And 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, it depends on k at what angle the straight axis intersects x, and hence k is spoken of as slope of the 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 when k

Suppose 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. At x = –2, we get y = 1, y = 4, y = 6. Here, as the value of k decreases, y increases at x

Graph of the function for k

Graphs of functions like y = kx + m differ from graphs y = km only by a parallel shift.

A linear function is a function of the form

x-argument (independent variable),

y- function (dependent variable),

k and b are some constant numbers

The graph of the linear function is straight.

enough to plot the graph. two points, because through two points you can draw a straight line, and moreover, only one.

If k˃0, then the graph is located in the 1st and 3rd coordinate quarters. If k˂0, then the graph is located in the 2nd and 4th coordinate quarters.

The number k is called the slope of the direct graph of the function y(x)=kx+b. If k˃0, then the angle of inclination of the straight line y(x)= kx+b to the positive direction Ox is sharp; if k˂0, then this angle is obtuse.

The coefficient b shows the intersection point of the graph with the y-axis (0; b).

y(x)=k∙x-- a special case of a typical function is called direct proportionality. The graph is a straight line passing through the origin, so one point is enough to build this graph.

Linear function graph

Where coefficient k = 3, hence

The graph of the function will increase and have an acute angle with the Ox axis. coefficient k has a plus sign.

OOF of a linear function

FRF of a linear function

Except the case where

Also a linear function of the form

It is a general function.

B) If k=0; b≠0,

In this case, the graph is a straight line parallel to the Ox axis and passing through the point (0;b).

C) If k≠0; b≠0, then the linear function has the form y(x)=k∙x+b.

Example 1 . Plot the function y(x)= -2x+5

Example 2 . Find the zeros of the function y=3x+1, y=0;

are zeros of the function.

Answer: or (;0)

Example 3 . Determine function value y=-x+3 for x=1 and x=-1

y(-1)=-(-1)+3=1+3=4

Answer: y_1=2; y_2=4.

Example 4 . Determine the coordinates of their intersection point or prove that the graphs do not intersect. Let the functions y 1 =10∙x-8 and y 2 =-3∙x+5 be given.

If the graphs of functions intersect, then the value of the functions at this point is equal to

Substitute x=1, then y 1 (1)=10∙1-8=2.

Comment. You can also substitute the obtained value of the argument into the function y 2 =-3∙x+5, then we will get the same answer y 2 (1)=-3∙1+5=2.

y=2 - ordinate of the intersection point.

(1;2) - the point of intersection of the graphs of the functions y \u003d 10x-8 and y \u003d -3x + 5.

Answer: (1;2)

Example 5 .

Construct graphs of functions y 1 (x)= x+3 and y 2 (x)= x-1.

It can be seen that the coefficient k=1 for both functions.

It follows from the above that if the coefficients of a linear function are equal, then their graphs in the coordinate system are parallel.

Example 6 .

Let's build two graphs of the function.

The first graph has the formula

The second graph has the formula

In this case, we have a graph of two straight lines intersecting at the point (0; 4). This means that the coefficient b, which is responsible for the height of the rise of the graph above the x-axis, if x=0. So we can assume that the coefficient b of both graphs is 4.

Editors: Ageeva Lyubov Alexandrovna, Gavrilina Anna Viktorovna

Class: 8

Presentation for the lesson

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Lesson type: lesson in discovering new knowledge.

Basic goals:

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

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: On fig. graphs of functions y \u003d kx are shown 2 .

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

4) a sample for self-examination of independent work.

Handout:

1) card:

1, 2 group:

Plot Functions y= 2X 2 , y = 4X

3, 4 group:

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

2) reflection card:

DURING THE CLASSES

1. Motivation for learning activities

Goals:

• organize the actualization of the requirements for the student from the side of educational activities;
• organize the activities of students to set the thematic framework: we continue to work with functions;
• create conditions for the emergence of the student's internal need for inclusion in educational activities.

Organization of the educational process at stage 1:

- Hello! What interesting things have you learned 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 features.
- How will you work today? (With nice mood).
- I wish you success!

2. Actualization of knowledge and fixation of difficulties in individual activity

Goals:

• to update the educational content, necessary and sufficient for the perception of new material.
• fix updated methods of action in speech and in signs;
• organize a synthesis of updated methods of action;
• motivate to complete an individual task;
• to organize independent fulfillment of an individual task for new knowledge;
• organize the fixation of individual difficulties in the performance by students of an individual task or in its justification.

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

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

- In what coordinate quarters does the graph of the function lie? y = x 2 , what is the domain of definition and range of this function, the intervals of increase and decrease? (Graph of the function y = x 2 lies in 1 and 2 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 \u003d x 2 takes non-negative values; increases at x > 0, decreasing at 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 is 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) According to what plan do you usually build function graphs? Slide 7.

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

- What did you repeat?

- And now, using everything that you have just repeated and learned, I suggest that you complete the following task:
Plot Functions y= 2X 2 , y = - 4X 2 and determine in which coordinate quarters the graphs of these functions are located. Make a conclusion about how the graph is located depending on the coefficient k.

Students work on graph paper.

Who doesn't get results?
What couldn't you do? (I could not__________________)
- Show the results, who completed the construction.
How can you prove that you did the right thing? (I should___________)
What will you use for proof? (____________.)
What couldn't you do?
What rule did you use when building?
- That you can not do?

3. Identifying the causes of the difficulty

Goals:

• organize the correlation of their actions with the standards used (algorithm, concept, etc.);
• on this basis, organize the identification and fixation in external speech of the cause of the difficulty - those specific knowledge and skills that are not enough 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 problem arise?
- What is the cause of the difficulty? (We have no way to determine how the graph of the function y \u003d kx2 is located depending on the coefficient k.)

4. Problematic explanation of new knowledge

Goals:

• organize the goal setting of the lesson;
• organize the clarification and coordination of the topic of the lesson;
• organize a leading or encouraging dialogue on the problematic introduction of new knowledge;
• organize the use of subject actions with models, schemes, properties, etc.;
• organize the fixation of a new mode of action in speech;
• organize the fixation of a new mode of action in signs;
• correlation of new knowledge with the rule in a textbook, reference book, dictionary, etc.
• organize the fixation of overcoming the difficulty.

Organization of the educational process at stage 4:

- State the purpose of your activity. (Find a way to determine how the graph of the function y \u003d 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.

1, 2 group:

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

3, 4 group:

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

Each group is given a card. (Students may use a textbook or reference book if they have difficulty.)

- Present your version of the algorithm.

Each of the groups presents its own version, the rest complement, clarify. After agreement, the rule is posted on the board:

The teacher adds:

- Each of the lines you draw is called a parabola. In this case, the point (0;0) is called the vertex of the parabola, and the axis at is the axis of symmetry of the parabola.
The “speed of aspiration” of the branches of the parabola up (down), the “degree of steepness” of the parabola depends on the value of the coefficient k.
What have you discovered now?
- What should you do now?

5. Primary consolidation in external speech

Target: organize the assimilation by children of a new mode of action with their pronunciation in external speech.

Organization of the educational process at stage 5:

- In what coordinate quarters are the graphs of functions 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 blackboard.

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

Goals:

• to organize independent performance by students of standard tasks for a new mode of action;
• based on the results of independent work, organize the identification and correction of errors;
• 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 on a card is offered. slide 9.

On fig. graphs of functions are shown at = kx 2 .

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

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

What rules did you use to complete the task?
- Who has difficulty - how to determine the sign of the coefficient k?
- Who had difficulty in determining the value of the coefficient k?
Who did the job right?

7. Inclusion in the knowledge system and repetition

Goals:

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

Organization of the educational process at stage 7:

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

- Identify the term that was repeated many times today in the lesson. (Graph)

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

3. Find the range of the function y \u003d - 5x2

a) at = –15X 2 b) at = – 9X 2 in) at = – X 2 G) at = – 5X 2

c uh f and

5. Indicate the intervals of increasing the function y \u003d - 5x 2

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

h about and t

6. Specify the smallest value of the function y \u003d - 5x 2

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

s to d in.

Physics tasks: slide 17.

The path traveled by the body in the first t seconds of free fall is calculated by the formula: H = gt 2 /2, where g\u003d 9.8 m / s 2. Find from the graph the dependence of H on t:

A) the distance that a falling stone will fly in the first 6 seconds;
B) the time during which the stone will fly the first 250 m?

8. Reflection of activities in the lesson

Goals:

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

Organization of the educational process at stage 8:

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

Students work with reflection cards (P).

Homework: slide 18.

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

Bibliography:

1. A.G. Mordkovich. Algebra, class 8. In two parts. Textbook for students of educational institutions. M.: Mnemozina. 2011.
2. Internet resources.

2). Then we build a graph of a linear function y \u003d -3x + 6 y x y \u003d -3x + 6

Functions whose graphs are parallel to the x-axis 2nd case: K=0 In this case, the function takes the form y=b y Y=2 Y=-3 Y=0 x

If k is greater than zero, then the lines are located in the first and third quadrants. The larger the coefficient, the closer the straight line is pressed to the Oy axis, and the smaller the coefficient, the closer the straight line is to the Ox axis. That is, the larger the slope, the greater the angle between the straight line and the x-axis.

5 Y \u003d 2x +6 Y \u003d 2x - 5 x y Two lines are parallel if they have the same angle of inclination, and it depends on the slope k 0 Two lines are parallel if they have the same slope.
Conclusions 1. A function of the form y = kx + b, where k and b are some numbers, is called a linear function. Line graph is a straight line. 2. A function of the form y= kx is called direct proportionality, and its graph passes through the origin. 3. The graph of the function y \u003d b is parallel to the x-axis and passes through the point with coordinates (0; b). 4. The coefficient k is called the slope. It determines the angle of inclination of the straight line to the x-axis. 5. If two different lines have equal slope coefficients, then the graphs of these functions will be parallel, if their slope coefficients are not equal, then the graphs will intersect.