How to build a parabola? What is a parabola? How are quadratic equations solved? Functions and graphs Properties of the function ax2 bx c.

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 on a graph (finding the axis of symmetry, coordinates of the vertex of a parabola, coordinates of points of intersection of the graph with the coordinate axes).

Developing: the 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 notations; 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, 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

Organizational moment (1-2 min)

Knowledge update (10 min)

Presentation of new material (15 min)

Securing new material (12 min)

Summing up (3 min)

Homework (2 min)

During the classes

Organizing time

Greetings, checking for absentees, collecting notebooks.

Knowledge update

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

Frontal poll:

What is called a quadratic function? (A function where given real numbers,, real variable, is called a quadratic function.)

What is a squared function graph? (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 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.)

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 .)

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.

Chalkboard writing: Number.

Function.

Teacher: On the board, you see two graphs of functions. 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?

Pupils: For a parabola of a view, the axis of symmetry is the ordinate.

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

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

Teacher: Right. So, the axis of symmetry of the graph of the function will be called the 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 a parabola .

Solution: By formula

Teacher: As we have already noted, the axis of symmetry passes through the apex 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 a parabola with apex 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 a parabola.

The equation of the axis of symmetry has the form:, therefore, the equation of the axis of symmetry of the given parabola.

Answer: - the equation of the axis of symmetry.

Securing new material

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

Writing on the board: No. 609 (3), 612 (1), 613 (3)

Teacher: But first, let's solve an example not from the textbook. 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).

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.

We 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 equal to -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 collapsed. 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 isolating 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 can be easily obtained 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 upwards, 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 smallest value functions -5 at x = 0, and the greatest 3, at x = 2.

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, this does not work. For such a generalization, 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... That is a 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 a... Most schoolchildren confidently answer: "if a> 0, then the branches of the parabola are directed upward, and if a < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой a > 0.

y = 0.5x 2 - 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 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 X= 0. Substitute zero in 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. Typically, this point is easy to find on a chart. And determine whether it lies above zero or below. That is 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 a... This is the apex of the parabola. Its abscissa (coordinate along the axis X) is found by the formula x in = - b / (2a)... In this way, b = - 2х в... That is, we act as follows: on the chart we find the top 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 a... 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 a> 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: a > 0, b < 0, With < 0.

Lesson: how to construct a parabola or a 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 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 is 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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