DEGREES OF COMPARISON OF ADJECTIVES In English, as in Russian, ...
Summary of the lesson in algebra for the 8th grade of the secondary school
Lesson topic: Function
The purpose of the lesson:
Educational: define the concept of a quadratic function of the form (compare the graphs of functions and ), show the formula for finding the coordinates of the parabola vertex (teach how to apply this formula in practice); to form the ability to determine the properties of a quadratic function from a graph (finding the axis of symmetry, the coordinates of the parabola vertex, the coordinates of the points of intersection of the graph with the coordinate axes).
Developing: the development of mathematical speech, the ability to correctly, consistently and rationally express one's thoughts; development of the skill of correct writing of a mathematical text using symbols and notations; development of analytical thinking; development of cognitive activity of students through the ability to analyze, systematize and generalize the material.
Educational: education of independence, the ability to listen to others, the formation of accuracy and attention in written mathematical speech.
Type of lesson: learning new material.
Teaching methods:
generalized-reproductive, inductive-heuristic.
Requirements for the knowledge and skills of students
know what a quadratic function of the form is, the formula for finding the coordinates of the vertex of a parabola; be able to find the coordinates of the parabola vertex, the coordinates of the points of intersection of the function graph with the coordinate axes, determine the properties of a quadratic function from the function graph.
Equipment:
Lesson Plan
Organizational moment (1-2 min)
Knowledge update (10 min)
Presentation of new material (15 min)
Consolidation of new material (12 min)
Summing up (3 min)
Homework (2 min)
During the classes
Organizing time
Greeting, checking absentees, collecting notebooks.
Knowledge update
Teacher: In today's lesson we will study a new topic: "Function". But first, let's review what we've learned so far.
Front poll:
What is a quadratic function? (A function where the given real numbers, , a real variable, is called a quadratic function.)
What is the graph of a quadratic function? (The graph of a quadratic function is a parabola.)
What are the zeros of a quadratic function? (The zeros of a quadratic function are the values at which it vanishes.)
List the properties of a function. (The values of the function are positive at and equal to zero at ; the graph of the function is symmetrical with respect to the ordinate axes; at the function increases, at - decreases.)
List the properties of a function. (If , then the function takes positive values for , if , then the function takes negative values for , the value of the function is only 0; the parabola is symmetrical about the ordinate axis; if , then the function increases for and decreases for , if , then the function increases for , decreases - at .)
Presentation of new material
Teacher: Let's start learning new material. Open your notebooks, write down the date and the topic of the lesson. Pay attention to the board.
Write on the board: Number.
Function .
Teacher: On the blackboard you see two graphs of functions. The first graph and the second . Let's try to compare them.
You know the properties of the function. Based on them, and comparing our graphs, we can highlight the properties of the function.
So, what do you think, what will determine the direction of the branches of the parabola?
Pupils: The direction of the branches of both parabolas will depend on the coefficient.
Teacher: Absolutely right. You can also notice that both parabolas have an axis of symmetry. For the first function graph, what is the axis of symmetry?
Pupils: For a parabola of the form, the axis of symmetry is the y-axis.
Teacher: Right. What is the axis of symmetry of a parabola?
Pupils: The axis of symmetry of the parabola is the line that passes through the top of the parabola, parallel to the y-axis.
Teacher: Right. So, we will call the axis of symmetry of the function graph a straight line passing through the vertex of the parabola, parallel to the y-axis.
And the top of the parabola is a point with coordinates . They are determined by the formula:
Write the formula in your notebook and circle it in a box.
Writing on the board and in notebooks
Parabola vertex coordinates.
Teacher: Now, to make it more clear, let's look at an example.
Example 1: Find the coordinates of the vertex of the parabola 
.
Solution: According to the formula
Teacher: As we have already noted, the axis of symmetry passes through the top of the parabola. Look at the desk. Draw this picture in your notebook.
Writing on the board and in notebooks:
Teacher: In the drawing: - the equation of the axis of symmetry of the parabola with the vertex at the point where is the abscissa of the vertex of the parabola.
Consider an example.
Example 2: From the graph of the function, determine the equation for the axis of symmetry of the parabola.
The equation of the axis of symmetry has the form: , hence, the equation of the axis of symmetry of the given parabola.
Answer: - the equation of the axis of symmetry.
Fixing new material
Teacher: There are tasks on the board that need to be solved in class.
Blackboard writing: No. 609(3), 612(1), 613(3)
Teacher: But first, let's solve an example not from a textbook. We will decide at the blackboard.
Example 1: Find the coordinates of the vertex of a parabola
Solution: According to the formula
Answer: the coordinates of the vertex of the parabola.
Example 2: Find the coordinates of the parabola intersection points 
with coordinate axes.
Solution: 1) With axis:
Those. 
According to Vieta's theorem:
Points of intersection with the abscissa axis (1;0) and (2;0).
Summary of the lesson in algebra for the 8th grade of the secondary school
Lesson topic: Function
The purpose of the lesson:
· Educational: define the concept of a quadratic function of the form (compare the graphs of functions and), show the formula for finding the coordinates of the parabola vertex (teach how to apply this formula in practice); to form the ability to determine the properties of a quadratic function from a graph (finding the axis of symmetry, the coordinates of the parabola vertex, the coordinates of the points of intersection of the graph with the coordinate axes).
· Educational: the development of mathematical speech, the ability to correctly, consistently and rationally express one's thoughts; development of the skill of correct writing of a mathematical text using symbols and notation; development of analytical thinking; development of cognitive activity of students through the ability to analyze, systematize and generalize the material.
· Educational: education of independence, the ability to listen to others, the formation of accuracy and attention in written mathematical speech.
Lesson type: learning new material.
Teaching methods:
generalized-reproductive, inductive-heuristic.
Requirements for the knowledge and skills of students
know what a quadratic function of the form is, the formula for finding the coordinates of the vertex of a parabola; be able to find the coordinates of the parabola vertex, the coordinates of the points of intersection of the function graph with the coordinate axes, determine the properties of a quadratic function from the function graph.
Equipment:
Lesson Plan
I. Organizational moment (1-2 minutes)
II. Knowledge update (10 min)
III. Presentation of new material (15 min)
IV. Consolidation of new material (12 min)
V. Debriefing (3 min)
VI. Homework (2 min)
During the classes
I. Organizational moment
Greeting, checking absentees, collecting notebooks.
II. Knowledge update
Teacher: In today's lesson we will learn a new topic: "Function". But first, let's review what we've learned so far.
Front poll:
1) What is called a quadratic function? (A function where the given real numbers are a real variable is called a quadratic function.)
2) What is the graph of a quadratic function? (The graph of a quadratic function is a parabola.)
3) What are the zeros of a quadratic function? (The zeros of a quadratic function are the values at which it vanishes.)
4) List the properties of the function. (The values of the function are positive at and equal to zero at; the graph of the function is symmetrical with respect to the ordinate axes; at the function increases, at - decreases.)
5) List the properties of the function. (If, then the function takes positive values at, if, then the function takes negative values at, the value of the function is only 0; the parabola is symmetrical about the y-axis; if, then the function increases at and decreases at, if, then the function increases at, decreases - at.)
III. Presentation of new material
Teacher: Let's start learning new material. Open your notebooks, write down the date and the topic of the lesson. Pay attention to the board.
whiteboard writing: Number.
Function.
Teacher: On the board you see two graphs of functions. The first graph, and the second. Let's try to compare them.
You know the properties of the function. Based on them, and comparing our graphs, we can highlight the properties of the function.
So, what do you think, what will determine the direction of the branches of the parabola?
Students: The direction of the branches of both parabolas will depend on the coefficient.
Teacher: Quite right. You can also notice that both parabolas have an axis of symmetry. For the first function graph, what is the axis of symmetry?
Students: For a parabola of the form, the axis of symmetry is the y-axis.
Teacher: Right. What is the axis of symmetry of a parabola?
Students: The axis of symmetry of a parabola is a line that passes through the vertex of the parabola, parallel to the y-axis.
Teacher: Correctly. So, we will call the axis of symmetry of the function graph a straight line passing through the vertex of the parabola, parallel to the y-axis.
And the top of the parabola is a point with coordinates. They are determined by the formula:
Write the formula in your notebook and circle it in a box.
Writing on the board and in notebooks
Parabola vertex coordinates.
Teacher: Now, to make it more clear, let's look at an example.
Example 1: Find the coordinates of the vertex of the parabola.
Solution: According to the formula
we have:
Teacher: As we have already noted, the axis of symmetry passes through the top of the parabola. Look at the desk. Draw this picture in your notebook.
Writing on the board and in notebooks:
Teacher: In the drawing: - the equation of the axis of symmetry of the parabola with the top at the point where the abscissa of the top of the parabola.
Consider an example.
Example 2: From the graph of the function, determine the equation for the axis of symmetry of the parabola.
The equation of the axis of symmetry has the form: , hence, the equation of the axis of symmetry of the given parabola.
Answer: - the equation of the axis of symmetry.
IV. Fixing new material
Teacher: There are tasks on the board that need to be solved in class.
whiteboard writing: № 609(3), 612(1), 613(3)
Teacher: But first, let's solve a non-textbook example. We will decide at the blackboard.
Example 1: Find the coordinates of the vertex of a parabola
Solution: According to the formula
we have:
Answer: the coordinates of the vertex of the parabola.
Example 2: Find the coordinates of the points of intersection of the parabola with the coordinate axes.
Solution: 1) With axis:
According to Vieta's theorem:
Points of intersection with the abscissa axis (1;0) and (2;0).
2) With axle:
Point of intersection with the y-axis (0;2).
Answer: (1;0), (2;0), (0;2) are the coordinates of the points of intersection with the coordinate axes.
No. 609(3). Find the coordinates of the vertex of the parabola
Solution: Abscissa of the top of the parabola:
Parabola vertex ordinate:
Answer: - the coordinates of the top of the parabola.
No. 612(1). Does the axis of symmetry of the parabola pass through the point (5;10)?
Solution: Symmetry axis equation: .
Find the abscissa of the top of the parabola: . So, the equation of the axis of symmetry looks like. Schematically draw this parabola:
Therefore, the axis of symmetry passes through the point (5;10).
No. 613(3). Find the coordinates of the points of intersection of the parabola with the coordinate axes.
Solution: 1) With axis:
We are looking for a discriminant:
This means that there are no points of intersection with the abscissa axis.
Point of intersection with the y-axis (0;12).
Answer: (0;12) - the coordinates of the point of intersection with the ordinate axis, the parabola does not intersect with the abscissa axis.
V. Debriefing
Teacher: In today's lesson, we studied a new topic: "Function", learned how to find the coordinates of the vertex of the parabola, the coordinates of the points of intersection of the parabola with the coordinate axes. In the next lesson, we will continue to solve problems on this topic.
VI. Homework
Teacher: Homework is written on the board. Write it down in your diaries.
Writing on the board and in the diaries: §38, no. 609(2), 612(2), 613(2).
Literature
1. Alimov Sh.A. Algebra Grade 8
2. Sarantsev G.I. Methods of teaching mathematics in high school
3. Mishin V.I. Private methodology for teaching mathematics in high school
Consider an expression of the form ax 2 + in + c, where a, b, c are real numbers, and is different from zero. This mathematical expression is known as the square trinomial.
Recall that ax 2 is the leading term of this square trinomial, and is its leading coefficient.
But the square trinomial does not always have all three terms. Take for example the expression 3x 2 + 2x, where a=3, b=2, c=0.
Let's move on to the quadratic function y \u003d ax 2 + in + c, where a, b, c are any arbitrary numbers. This function is quadratic because it contains a term of the second degree, that is, x squared.
It is quite easy to plot a quadratic function, for example, you can use the full square method.
Consider an example of plotting a function y equals -3x 2 - 6x + 1.
To do this, the first thing to remember is the scheme for highlighting the full square in the trinomial -3x 2 - 6x + 1.
We take out -3 from the first two terms in brackets. We have -3 times the sum of x plus 2x and add 1. Adding and subtracting the unit in brackets, we get the formula for the square of the sum, which can be collapsed. We get -3 times the sum (x + 1) squared minus 1, add 1. Opening the brackets and adding like terms, the expression comes out: -3 times the square of the sum (x + 1) add 4.
Let's build a graph of the resulting function by going to the auxiliary coordinate system with the origin at the point with coordinates (-1; 4).
In the figure from the video, this system is indicated by dotted lines. We bind the function y equals -3x 2 to the constructed coordinate system. For convenience, we take control points. For example, (0;0), (1;-3), (-1;-3), (2;-12), (-2;-12). At the same time, we set aside them in the constructed coordinate system. The parabola obtained during the construction is the graph we need. In the figure, this is a red parabola.
Applying the full square selection method, we have a quadratic function of the form: y = a * (x + 1) 2 + m.
The graph of the parabola y \u003d ax 2 + bx + c is easy to obtain from the parabola y \u003d ax 2 by parallel translation. This is confirmed by a theorem that can be proved by taking the full square of the binomial. The expression ax 2 + bx + c after successive transformations turns into an expression of the form: a * (x + l) 2 + m. Let's draw a graph. Let's perform a parallel movement of the parabola y \u003d ax 2, combining the vertex with the point with coordinates (-l; m). The important thing is that x = -l, which means -b / 2a. So this line is the axis of the parabola ax 2 + bx + c, its vertex is at the point with the abscissa x, zero is equal to minus b divided by 2a, and the ordinate is calculated by the cumbersome formula 4ac - b 2 /. But this formula is not necessary to memorize. Since, by substituting the value of the abscissa into the function, we get the ordinate.
To determine the axis equation, the direction of its branches and the coordinates of the parabola vertex, consider the following example.
Let's take the function y \u003d -3x 2 - 6x + 1. Having drawn up the equation for the axis of the parabola, we have that x \u003d -1. And this value is the x-coordinate of the top of the parabola. It remains to find only the ordinate. Substituting the value -1 into the function, we get 4. The top of the parabola is at the point (-1; 4).
The graph of the function y \u003d -3x 2 - 6x + 1 was obtained by parallel transfer of the graph of the function y \u003d -3x 2, which means that it behaves similarly. The leading coefficient is negative, so the branches are directed downwards.
We see that for any function of the form y = ax 2 + bx + c, the easiest question is the last question, that is, the direction of the branches of the parabola. If the coefficient a is positive, then the branches are up, and if negative, then they are down.
The next most difficult question is the first question, because it requires additional calculations.
And the second one is the most difficult, because, in addition to calculations, knowledge of the formulas by which x is zero and y is zero is also needed.
Let's plot the function y \u003d 2x 2 - x + 1.
We determine immediately - the graph is a parabola, the branches are directed upwards, since the leading coefficient is 2, and this is a positive number. According to the formula, we find the abscissa x is zero, it is equal to 1.5. To find the ordinate, remember that zero is equal to a function of 1.5, when calculating we get -3.5.
Top - (1.5; -3.5). Axis - x=1.5. Take the points x=0 and x=3. y=1. Note these points. Based on three known points, we build the required graph.
To plot the function ax 2 + bx + c, you need:
Find the coordinates of the vertex of the parabola and mark them in the figure, then draw the axis of the parabola;
On the x-axis, take two points that are symmetrical about the axis of the parabola, find the value of the function at these points and mark them on the coordinate plane;
Through three points, construct a parabola, if necessary, you can take a few more points and build a graph based on them.
In the following example, we will learn how to find the largest and smallest values of the function -2x 2 + 8x - 5 on the segment.
According to the algorithm: a \u003d -2, b \u003d 8, then x zero is 2, and zero y is 3, (2; 3) is the top of the parabola, and x \u003d 2 is the axis.
Let's take the values x=0 and x=4 and find the ordinates of these points. This is -5. We build a parabola and determine that the smallest value of the function is -5 at x=0, and the largest is 3 at x=2.
Tasks on the properties and graphs of a quadratic function cause, as practice shows, serious difficulties. This is rather strange, because the quadratic function is passed in the 8th grade, and then the entire first quarter of the 9th grade is "extorted" by the properties of the parabola and its graphs are built for various parameters.
This is due to the fact that forcing students to build parabolas, they practically do not devote time to "reading" graphs, that is, they do not practice comprehending the information received from the picture. Apparently, it is assumed that, having built two dozen graphs, a smart student himself will discover and formulate the relationship between the coefficients in the formula and the appearance of the graph. In practice, this does not work. For such a generalization, serious experience in mathematical mini-research is required, which, of course, most ninth-graders do not have. Meanwhile, in the GIA they propose to determine the signs of the coefficients precisely according to the schedule.
We will not demand the impossible from schoolchildren and simply offer one of the algorithms for solving such problems.
So, a function of the form y=ax2+bx+c is called quadratic, its graph is a parabola. As the name suggests, the main component is ax 2. That is a should not be equal to zero, the remaining coefficients ( b and With) can be equal to zero.
Let's see how the signs of its coefficients affect the appearance of the parabola.
The simplest dependence for the coefficient a. Most schoolchildren confidently answer: "if a> 0, then the branches of the parabola are directed upwards, and if a < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой a > 0.
y = 0.5x2 - 3x + 1
In this case a = 0,5
And now for a < 0:
y = - 0.5x2 - 3x + 1
In this case a = - 0,5
Influence of coefficient With also easy enough to follow. Imagine that we want to find the value of a function at a point X= 0. Substitute zero into the formula:
y = a 0 2 + b 0 + c = c. It turns out that y = c. That is With is the ordinate of the point of intersection of the parabola with the y-axis. As a rule, this point is easy to find on the graph. And determine whether it lies above zero or below. That is With> 0 or With < 0.
With > 0:
y=x2+4x+3
With < 0
y = x 2 + 4x - 3
Accordingly, if With= 0, then the parabola will necessarily pass through the origin:
y=x2+4x
More difficult with the parameter b. The point by which we will find it depends not only on b but also from a. This is the top of the parabola. Its abscissa (axis coordinate X) is found by the formula x in \u003d - b / (2a). In this way, b = - 2ax in. That is, we act as follows: on the graph we find the top of the parabola, determine the sign of its abscissa, that is, we look to the right of zero ( x in> 0) or to the left ( x in < 0) она лежит.
However, this is not all. We must also pay attention to the sign of the coefficient a. That is, to see where the branches of the parabola are directed. And only after that, according to the formula b = - 2ax in determine sign b.
Consider an example:
Branches pointing upwards a> 0, the parabola crosses the axis at below zero means With < 0, вершина параболы лежит правее нуля. Следовательно, x in> 0. So b = - 2ax in = -++ = -. b < 0. Окончательно имеем: a > 0, b < 0, With < 0.
