As practice shows, tasks for the properties and graphs of a quadratic function cause serious difficulties. This is rather strange, because the quadratic function is passed in the 8th grade, and then the whole first quarter of the 9th grade is "forced out" the properties of the parabola and its graphs are plotted 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 obtained from the picture. Apparently, it is assumed that, having built a 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, it doesn't work that way. For such a generalization, a serious experience of mathematical mini-research is required, which, of course, most ninth-graders do not have. Meanwhile, the GIA proposes to determine the signs of the coefficients precisely according to the schedule.

We will not demand the impossible from schoolchildren and will simply offer one of the algorithms for solving such problems.

So, a function of the form y = ax 2 + bx + c is called quadratic, its graph is a parabola. As the name suggests, the main term is ax 2... I.e but should not be zero, other coefficients ( b and with) can be equal to zero.

Let's see how the signs of its coefficients affect the appearance of a parabola.

The simplest relationship for the coefficient but... Most schoolchildren confidently answer: "if but> 0, then the branches of the parabola are directed upwards, and if but < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой but > 0.

y = 0.5x 2 - 3x + 1

In this case but = 0,5

And now for but < 0:

y = - 0.5x2 - 3x + 1

In this case but = - 0,5

Influence of the coefficient with is also easy enough to trace. Let's imagine that we want to find the value of the function at the point NS= 0. Substitute zero in the formula:

y = a 0 2 + b 0 + c = c... Turns out that y = c... I.e with is the ordinate of the point of intersection of the parabola with the y-axis. Typically, this point is easy to find on a chart. And determine whether it lies above zero or below. I.e with> 0 or with < 0.

with > 0:

y = x 2 + 4x + 3

with < 0

y = x 2 + 4x - 3

Accordingly, if with= 0, then the parabola will necessarily pass through the origin:

y = x 2 + 4x

More difficult with the parameter b... The point at which we will find it depends not only on b but also from but... This is the apex of the parabola. Its abscissa (coordinate along the axis NS) is found by the formula x in = - b / (2a)... Thus, b = - 2х в... That is, we act as follows: on the chart we find the vertex of the parabola, we 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 but... That is, to see where the branches of the parabola are directed. And only after that, according to the formula b = - 2х в identify the sign b.

Let's consider an example:

The branches are directed upwards, which means but> 0, the parabola crosses the axis at below zero means with < 0, вершина параболы лежит правее нуля. Следовательно, x in> 0. Hence b = - 2х в = -++ = -. b < 0. Окончательно имеем: but > 0, b < 0, with < 0.

Abstract of a lesson in algebra for the 8th grade of a 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 vertex of a parabola (teach how to apply this formula in practice); to form the ability to determine the properties of a quadratic function according to the graph (finding the axis of symmetry, coordinates of the vertex of a parabola, coordinates of the points of intersection of the graph with the coordinate axes).

· Developing: development of mathematical speech, the ability to correctly, consistently and rationally express your thoughts; developing the skill of correctly writing a mathematical text using symbols and notation; development of analytical thinking; development of students' cognitive activity through the ability to analyze, systematize and generalize 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 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; to be able to find the coordinates of the vertex of the parabola, the coordinates of the points of intersection of the graph of the function with the coordinate axes, to determine the properties of the quadratic function from the graph of the function.

Equipment:

Lesson plan

I. Organizational moment (1-2 min)

II. Knowledge update (10 min)

III. Presentation of new material (15 min)

IV. Securing new material (12 min)

V. Summing up (3 min)

Vi. Homework (2 min)

During the classes

I. Organizational moment

Greetings, checking for absentees, collecting notebooks.

II. Knowledge update

Teacher: In today's lesson we will explore a new topic: "Function". But first, let's repeat the previously studied material.

Frontal poll:

1) What is called a quadratic function? (A function where given real numbers,, 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 symmetric with respect to the axes of the ordinates; 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 symmetric about the ordinate; 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 number and topic of the lesson. Pay attention to the board.

Writing on the blackboard: Number.

Function.

Teacher: On the board, you see two function graphs. The first is the 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 the direction of the branches of the parabola will depend on?

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. The first graph of the function, what is the axis of symmetry?

Students: For a parabola of the form, the axis of symmetry is the ordinate axis.

Teacher: Right. And what is the axis of symmetry of the parabola

Students: The axis of symmetry of a parabola is a line that passes through the apex of the parabola, parallel to the ordinate axis.

Teacher: Right. So, the axis of symmetry of the graph of the function will be called the straight line passing through the vertex of the parabola, parallel to the ordinate axis.

And the vertex of the parabola is the point with coordinates. They are determined by the formula:

Write the formula down in a notebook and frame it.

Writing on the board and in notebooks

The coordinates of the vertex of the parabola.

Teacher: Now, to make it clearer, let's look at an example.

Example 1: Find the coordinates of the vertex of the parabola.

Solution: By formula

Teacher: As we have already noted, the axis of symmetry passes through the vertex of the parabola. Look at the desk. Draw this drawing 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 the abscissa of the vertex of the parabola.

Let's look at an example.

Example 2: From the graph of the function, determine the equation of 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. Securing new material

Teacher: Written on the board are tasks that need to be solved in class.

Writing on the blackboard: № 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: By formula

Answer: the coordinates of the vertex of the parabola.

Example 2: Find the coordinates of the intersection points of a parabola with coordinate axes.

Solution: 1) With axis:

Those.

By Vieta's theorem:

The points of intersection with the abscissa axis (1; 0) and (2; 0).

2) With axle:

The point of intersection with the y-axis (0; 2).

Answer: (1; 0), (2; 0), (0; 2) - coordinates of points of intersection with coordinate axes.

No. 609 (3). Find the coordinates of the vertex of a parabola

Consider an expression of the form ax 2 + bx + c, where a, b, c are real numbers, and are 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 us pass to the quadratic function y = ax 2 + bx + c, where a, b, c are any arbitrary numbers. This function is quadratic, since 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 selection method.

Consider an example of plotting a function y is -3x 2 - 6x + 1.

To do this, the first thing we remember is the scheme for allocating a complete square in the trinomial -3x 2 - 6x + 1.

Take -3 out of the brackets for the first two terms. We have -3 multiplied by the sum of x square plus 2x and add 1. Adding and subtracting one in parentheses, we get a formula for the square of the sum, which can be folded. We get -3 multiplied by the sum (x + 1) squared minus 1 add 1. Expanding the parentheses and giving similar terms, we get the expression: -3 multiplied by the square of the sum (x + 1) add 4.

Let's build a graph of the resulting function, passing to the auxiliary coordinate system with the origin at the point with coordinates (-1; 4).

In the picture from the video, this system is indicated by dotted lines. Let us bind the function y is equal to -3x 2 to the constructed coordinate system. Let's take control points for convenience. For example, (0; 0), (1; -3), (-1; -3), (2; -12), (-2; -12). At the same time, we will postpone them in the constructed coordinate system. The resulting parabola is the graph we need. In the picture, it is a red parabola.

Applying the method of allocating a complete square, we have a quadratic function of the form: y = a * (x + 1) 2 + m.

The graph of the parabola y = ax 2 + bx + c is easy to obtain from the parabola y = ax 2 by parallel translation. This is confirmed by a theorem that can be proved by selecting the complete 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 carry out a parallel movement of the parabola y = ax 2, aligning the vertex with a point with coordinates (-l; m). The important thing is that x = -l, which means -b / 2a. This means that this straight 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 using the cumbersome formula 4ac - b 2 /. But you don't have to memorize this formula. Since, substituting the value of the abscissa into the function, we get the ordinate.

To determine the equation of the axis, the direction of its branches and the coordinates of the vertex of the parabola, consider the following example.

Take the function y = -3x 2 - 6x + 1. Having compiled the equation for the axis of the parabola, we have that x = -1. And this value is the x-coordinate of the vertex of the parabola. It remains to find only the ordinate. Substituting the value -1 into the function, we get 4. The vertex of the parabola is at the point (-1; 4).

The graph of the function y = -3x 2 - 6x + 1 was obtained with a parallel transfer of the graph of the function y = -3x 2, which means that it behaves similarly. The senior coefficient is negative, so the branches are directed downward.

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 upward, and if negative, then downward.

The first question is next in complexity, because it requires additional calculations.

And the most difficult is the second, since, in addition to calculations, knowledge of the formulas by which x is zero and y is zero are also needed.

Let's build a graph of the function y = 2x 2 - x + 1.

We determine immediately - the graph is a parabola, the branches are directed upward, since the senior coefficient is 2, and this is a positive number. Using the formula, we find the abscissa x 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.

Vertex - (1.5; -3.5). Axis - x = 1.5. Take the points x = 0 and x = 3. y = 1. Let's mark these points. Using three known points, we build the desired graph.

To plot the function ax 2 + bx + c, you must:

Find the coordinates of the vertex of the parabola and mark them in the figure, then draw the axis of the parabola;

On the ox axis, take two symmetric, about the axis, parabola points, find the value of the function at these points and mark them on the coordinate plane;

Build a parabola through three points, if necessary, you can take a few more points and build a graph based on them.

In the next example, we will learn how to find the largest and smallest values ​​of the function -2x 2 + 8x - 5 on a segment.

According to the algorithm: a = -2, b = 8, so x zero is 2, and y zero is 3, (2; 3) is the vertex of the parabola, and x = 2 is the axis.

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.

The presentation "Function y = ax 2, its graph and properties" is a visual aid that was created to accompany the teacher's explanation on this topic. This presentation discusses in detail the quadratic function, its properties, features of plotting, the practical application of the used methods for solving problems in physics.

Providing a high degree of clarity, this material will help the teacher to increase the effectiveness of teaching, will make it possible to more rationally allocate time in the lesson. With the help of animation effects, highlighting concepts and important points in color, the attention of students is focused on the subject under study, a better memorization of definitions and the course of reasoning is achieved when solving problems.

The presentation begins with an introduction to the title of the presentation and the concept of a quadratic function. The importance of this topic is emphasized. Students are invited to remember the definition of a quadratic function as a functional dependence of the form y = ax 2 + bx + c, in which is an independent variable, and are numbers, while a ≠ 0. Separately, on slide 4, it is noted for remembering that the domain of this function is the entire axis of real values. This statement is conventionally denoted by D (x) = R.

An example of a quadratic function is its important application in physics - the formula for the dependence of a path for uniformly accelerated motion on time. At the same time, in physics lessons, students study the formulas of various types of movement, so the ability to solve such problems will be necessary for them. On slide 5, students are reminded that when the body moves with acceleration and at the beginning of the countdown, the distance traveled and the speed of movement are known, then the functional dependence representing such a movement will be expressed by the formula S = (at 2) / 2 + v 0 t + S 0 ... Below is an example of converting this formula into a given quadratic function if the values ​​of acceleration = 8, start speed = 3 and start path = 18. In this case, the function will take the form S = 4t 2 + 3t + 18.

Slide 6 examines the form of the quadratic function y = ax 2, in which it is represented at. If = 1, then the quadratic function has the form y = x 2. It is noted that the graph of this function will be a parabola.

The next part of the presentation is devoted to plotting a quadratic function. It is proposed to consider the construction of the graph of the function y = 3x 2. First, the table notes the correspondence of the values ​​of the function to the values ​​of the argument. It is noted that the difference between the plotted graph of the function y = 3x 2 and the graph of the function y = x 2 is that each value of it will be three times greater than the corresponding one. In a tabular view, this difference is well tracked. The difference in the narrowing of the parabola is also clearly visible in the graphical representation next to it.

The next slide looks at plotting a quadratic function y = 1/3 x 2. To build a graph, it is necessary to indicate the values ​​of the function in a number of its points in the table. It is noted that each value of the function y = 1/3 x 2 is 3 times less than the corresponding value of the function y = x 2. This difference, in addition to the table, is clearly visible on the graph. Its parabola is more extended with respect to the ordinate than the parabola of the function y = x 2.

The examples help to understand the general rule according to which you can then more easily and quickly produce the construction of the corresponding graphs. On slide 9, a separate rule is highlighted that the graph of the quadratic function y = ax 2 can be plotted depending on the value of the coefficient by stretching or narrowing the graph. If a> 1, then the graph is stretched from the x-axis in times. If 0

The conclusion about the symmetry of the graphs of the functions y = ax 2 and y = -ax2 (at ≠ 0) relative to the abscissa axis is separately highlighted on slide 12 for memorization and is clearly displayed on the corresponding graph. Further, the concept of the graph of a quadratic function y = x 2 is extended to the more general case of the function y = ax 2, arguing that such a graph will also be called a parabola.

Slide 14 examines the properties of the quadratic function y = ax 2 when positive. It is noted that its graph passes through the origin of coordinates, and all points, except, lie in the upper half-plane. The symmetry of the graph with respect to the ordinate axis is noted, specifying that the opposite values ​​of the argument correspond to the same values ​​of the function. It is indicated that the interval of decreasing of this function (-∞; 0], and the increase of the function is performed on the interval. The values ​​of this function cover the entire positive part of the real axis, it is equal to zero at the point, and does not have the greatest value.

Slide 15 describes the properties of the function y = ax 2 if negative. It is noted that its graph also passes through the origin, but all of its points, except, lie in the lower half-plane. The symmetry of the graph about the axis is noted, and equal values ​​of the function correspond to opposite values ​​of the argument. The function increases on the interval, decreases on. The values ​​of this function lie in the interval, it is equal to zero at the point, and does not have the least value.

Summarizing the considered characteristics, slide 16 shows that the branches of the parabola are directed downward at, and upward - at. The parabola is symmetrical about the axis, and the apex of the parabola is located at the point of its intersection with the axis. The parabola y = ax 2 has a vertex - the origin.

Also, an important conclusion about parabola transformations is displayed on slide 17. It shows the options for transforming the graph of a quadratic function. It is noted that the graph of the function y = ax 2 is transformed by symmetric display of the graph about the axis. It is also possible to compress or stretch the graph about the axis.

The last slide draws general conclusions about the transformations of the function graph. Conclusions are presented that the graph of the function is obtained by symmetric transformation about the axis. A function graph is obtained by compressing or stretching the original graph from the axis. In this case, stretching from the axis in times is observed in the case when. By shrinking to the axis 1 / a times, the graph is formed in the case.

The presentation "Function y = ax 2, its graph and properties" can be used by the teacher as a visual aid in an algebra lesson. Also, this manual covers the topic well, giving an in-depth understanding of the subject, therefore it can be offered for independent study by students. Also, this material will help the teacher to give an explanation during distance learning.

Lesson: how to build a parabola or quadratic function?

THEORETICAL PART

A parabola is a graph of a function described by the formula ax 2 + bx + c = 0.
To build a parabola, you need to follow a simple algorithm of actions:

1) Parabola formula y = ax 2 + bx + c,
if a> 0 then the branches of the parabola are directed up,
otherwise the branches of the parabola are directed way down.
Free member c this point intersects the parabola with the OY axis;

2), it is found by the formula x = (- b) / 2a, we substitute the found x into the parabola equation and find y;

3)Function zeros or otherwise the points of intersection of the parabola with the OX axis, they are also called the roots of the equation. To find the roots, we equate the equation to 0 ax 2 + bx + c = 0;

Types of equations:

a) The complete quadratic equation has the form ax 2 + bx + c = 0 and is decided by the discriminant;
b) Incomplete quadratic equation of the form ax 2 + bx = 0. To solve it, you need to put x outside the brackets, then equate each factor to 0:
ax 2 + bx = 0,
x (ax + b) = 0,
x = 0 and ax + b = 0;
c) Incomplete quadratic equation of the form ax 2 + c = 0. To solve it, you need to move the unknown in one direction, and the known in the other. x = ± √ (c / a);

4) Find some additional points to build the function.

PRACTICAL PART

And so now, using an example, we will analyze everything according to the actions:
Example # 1:
y = x 2 + 4x + 3
c = 3 means the parabola intersects OY at the point x = 0 y = 3. The branches of the parabola look upward since a = 1 1> 0.
a = 1 b = 4 c = 3 x = (- b) / 2a = (- 4) / (2 * 1) = - 2 y = (-2) 2 +4 * (- 2) + 3 = 4- 8 + 3 = -1 the vertex is at the point (-2; -1)
Find the roots of the equation x 2 + 4x + 3 = 0
Find the roots by the discriminant
a = 1 b = 4 c = 3
D = b 2 -4ac = 16-12 = 4
x = (- b ± √ (D)) / 2a
x 1 = (- 4 + 2) / 2 = -1
x 2 = (- 4-2) / 2 = -3

Take some arbitrary points that are near the vertex x = -2

x -4 -3 -1 0
y 3 0 0 3

Substitute x into the equation y = x 2 + 4x + 3 values
y = (- 4) 2 +4 * (- 4) + 3 = 16-16 + 3 = 3
y = (- 3) 2 +4 * (- 3) + 3 = 9-12 + 3 = 0
y = (- 1) 2 +4 * (- 1) + 3 = 1-4 + 3 = 0
y = (0) 2 + 4 * (0) + 3 = 0-0 + 3 = 3
It can be seen from the values ​​of the function that the parabola is symmetric with respect to the straight line x = -2

Example # 2:
y = -x 2 + 4x
c = 0 means the parabola intersects OY at the point x = 0 y = 0. The branches of the parabola look down as a = -1 -1 Find the roots of the equation -x 2 + 4x = 0
Incomplete quadratic equation of the form ax 2 + bx = 0. To solve it, you need to take x out of the brackets, then equate each factor to 0.
x (-x + 4) = 0, x = 0 and x = 4.

Take some arbitrary points that are near the vertex x = 2
x 0 1 3 4
y 0 3 3 0
Substitute the x into the equation y = -x 2 + 4x values
y = 0 2 + 4 * 0 = 0
y = - (1) 2 + 4 * 1 = -1 + 4 = 3
y = - (3) 2 + 4 * 3 = -9 + 13 = 3
y = - (4) 2 + 4 * 4 = -16 + 16 = 0
It can be seen from the values ​​of the function that the parabola is symmetric with respect to the straight line x = 2

Example No. 3
y = x 2 -4
c = 4 means the parabola intersects OY at the point x = 0 y = 4. The branches of the parabola look upward since a = 1 1> 0.
a = 1 b = 0 c = -4 x = (- b) / 2a = 0 / (2 * (1)) = 0 y = (0) 2 -4 = -4 the vertex is at the point (0; -4 )
Find the roots of the equation x 2 -4 = 0
Incomplete quadratic equation of the form ax 2 + c = 0. To solve it, you need to move the unknown in one direction, and the known in the other. x = ± √ (c / a)
x 2 = 4
x 1 = 2
x 2 = -2

Take some arbitrary points that are near the vertex x = 0
x -2 -1 1 2
y 0 -3 -3 0
Substitute the x into the equation y = x 2 -4 values
y = (- 2) 2 -4 = 4-4 = 0
y = (- 1) 2 -4 = 1-4 = -3
y = 1 2 -4 = 1-4 = -3
y = 2 2 -4 = 4-4 = 0
It can be seen from the values ​​of the function that the parabola is symmetric with respect to the straight line x = 0

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