Lesson: How to construct 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:

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

3)Function zeros or, in other words, 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 solved by the discriminant;
b) Incomplete quadratic equation of the form ax 2 +bx=0. To solve it, you need to take x out of 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 unknowns to one side, and the knowns to the other. x =±√(c/a);

4) Find several additional points to construct the function.

PRACTICAL PART

And so now, using an example, we will analyze everything step by step:
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 vertex is at point (-2;-1)
Let's find the roots of the equation x 2 +4x+3=0
Using the discriminant we find the roots
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

Let's take several arbitrary points that are located near the vertex x = -2

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

Substitute instead of 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 function values ​​that the parabola is symmetrical 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 since a=-1 -1 Let's 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 brackets, then equate each factor to 0.
x(-x+4)=0, x=0 and x=4.

Let's take several arbitrary points that are located near the vertex x=2
x 0 1 3 4
y 0 3 3 0
Substitute instead of 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 function values ​​that the parabola is symmetrical about 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 up 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 point (0;-4 )
Let's 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 unknowns to one side, and the knowns to the other. x =±√(c/a)
x 2 =4
x 1 =2
x 2 =-2

Let's take several arbitrary points that are located near the vertex x=0
x -2 -1 1 2
y 0 -3 -3 0
Substitute instead of 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 function values ​​that the parabola is symmetrical about the straight line x = 0

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As practice shows, tasks on the properties and graphs of a quadratic function cause serious difficulties. This is quite strange, because they study the quadratic function in the 8th grade, and then throughout the first quarter of the 9th grade they “torment” the properties of the parabola and build its graphs for various parameters.

This is due to the fact that when forcing students to construct parabolas, they practically do not devote time to “reading” the graphs, that is, they do not practice comprehending the information received from the picture. Apparently, it is assumed that, after constructing a dozen or two 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 most ninth-graders, of course, do not possess. Meanwhile, the State Inspectorate proposes to determine the signs of the coefficients using 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 called quadratic, its graph is a parabola. As the name suggests, the main term is ax 2. That is A should not be equal to zero, the remaining coefficients ( b And With) can equal zero.

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

The simplest dependence 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

Impact of the coefficient With It's also pretty easy to follow. Let's 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. Typically, 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 = 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 top of the parabola. Its abscissa (axis coordinate X) is found by the formula x in = - b/(2a). Thus, b = - 2ax in. That is, we proceed as follows: we find the vertex of the parabola on the graph, 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, that's not all. We also need to pay attention to the sign of the coefficient A. That is, look at where the branches of the parabola are directed. And only after that, according to the formula b = - 2ax in determine the sign b.

Let's look at an example:

The branches are directed upwards, which means A> 0, the parabola intersects the axis at below zero, that is 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.

Algebra lesson notes for 8th grade secondary school

Lesson topic: Function

The purpose of the lesson:

· Educational: define the concept of a quadratic function of the form (compare 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 develop the ability to determine the properties of a quadratic function from a graph (finding the axis of symmetry, the coordinates of the vertex of a parabola, the coordinates of the points of intersection of the graph with the coordinate axes).

· Developmental: development of mathematical speech, the ability to correctly, consistently and rationally express one’s thoughts; developing the skill of correctly writing 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: fostering independence, the ability to listen to others, developing accuracy and attention in written mathematical speech.

Lesson type: learning new material.

Teaching methods:

generalized reproductive, inductive heuristic.

Requirements for students' knowledge and skills

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 vertex of a parabola, the coordinates of the points of intersection of the graph of a function with the coordinate axes, and use the graph of a function to determine the properties of a quadratic function.

Equipment:

Lesson Plan

I. Organizational moment (1-2 min)

II. Updating knowledge (10 min)

III. Presentation of new material (15 min)

IV. Consolidating new material (12 min)

V. Summing up (3 min)

VI. Homework assignment (2 min)

During the classes

I. Organizational moment

Greeting, checking absentees, collecting notebooks.

II. Updating knowledge

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

Frontal survey:

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

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 ordinate 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 topic of the lesson. Pay attention to the board.

Writing on the board: 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 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: Absolutely right. You can also notice that both parabolas have an axis of symmetry. In the first graph of the function, what is the axis of symmetry?

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

Teacher: Right. What is the axis of symmetry of a parabola?

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

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

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

Write the formula in your notebook and circle it in a frame.

Writing on the board and in notebooks

Coordinates of the vertex of the parabola.

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 vertex of the parabola. Look at the blackboard. Draw this picture in your notebook.

Write on the board and in notebooks:

Teacher: In the drawing: - equation of the axis of symmetry of a parabola with the vertex at the point where the abscissa is the vertex of the parabola.

Let's look at an example.

Example 2: Using the graph of the function, determine the equation for the axis of symmetry of the parabola.

The equation for the axis of symmetry has the form: , which means the equation for the axis of symmetry of this parabola is .

Answer: - equation of the axis of symmetry.

IV. Consolidation of new material

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

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

Teacher: But first, let's solve an example not from the textbook. We will decide at the board.

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

Solution: According to the formula

Answer: coordinates of the vertex of the parabola.

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

Solution: 1) With axis:

Those.

According to Vieta's theorem:

The points of intersection with the x-axis are (1;0) and (2;0).

2) With axle:

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

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

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

Determining the values ​​of the coefficients of a quadratic function from a graph.

Methodological development by Sagnaeva A.M.

MBOU secondary school No. 44, Surgut, Khanty-Mansi Autonomous Okrug-Yugra .

Ι. Finding the coefficient A

• Using the graph of a parabola, we determine the coordinates of the vertex (m,n)

2. Using the graph of a parabola, we determine the coordinates of any point A (X 1 ;y 1 )

3. We substitute these values ​​into the formula of a quadratic function specified in a different form:

y=a(x-m)2+n

4. solve the resulting equation.

Oh 1 ;y 1 )

parabola

ΙΙ. Finding the coefficient b

1. First we find the value of the coefficient a

2. In the formula for the abscissa of a parabola m= -b/2a substitute the values m And a

3. Calculate the value of the coefficient b .

Oh 1 ;y 1 )

parabola

ΙΙΙ. Finding the coefficient c

1. We find the ordinate of the point of intersection of the parabola graph with the Oy axis, this value is equal to the coefficient With, i.e. dot (0;s)-the point of intersection of the parabola graph with the Oy axis.

2. If it is impossible to find the point of intersection of the parabola with the Oy axis from the graph, then we find the coefficients a,b

(see steps Ι, ΙΙ)

3. Substitute the found values a, b ,A(x 1; at 1 ) into the equation

y=ax 2 +bx+c and we find With.

Oh 1 ;y 1 )

parabola

Tasks

clue

Ιx 2 Ι, and x 1 0, because a The ordinate of the point of intersection of the parabola with the OY axis is the coefficient c Answer: 5 c x 1 x 2 "width="640"

• The branches of the parabola are directed downwards,

• The roots have different signs, Ι x 1 ΙΙх 2 Ι, and x 1 0, because a
• The ordinate of the point of intersection of the parabola with the OY axis is the coefficient With

X 1

X 2

P Clue

0 x 1 +x 2 = - b/a 0. a 0. Answer: 5 "width="640"

1.The branches of the parabola are directed downwards, which means a

• x 1 +x 2 = - b/a 0. a 0.

0 because the branches of the parabola are directed upward; 2. c=y(0)3. The vertex of the parabola has a positive abscissa: in this case a is 0, therefore b4. D0, because the parabola intersects the OX axis at two different points. "width="640"

The figure shows a graph of the function y=ax 2 +bx+c. Indicate the signs of the coefficients a, b, c and the discriminant D.

Solution:

1. a0, because the branches of the parabola are directed upward;

3. The vertex of the parabola has a positive abscissa:

in this case a 0, therefore b

4. D0, because the parabola intersects the OX axis at two different points.

The picture shows a parabola

Specify values k And t .

Find the coordinates of the vertex of the parabola and write the function whose graph is shown in the figure.

Find where are the abscissas of the intersection points

parabolas and horizontal straight lines (see figure).

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, and the practical application of the methods used for solving problems in physics.

Providing a high degree of clarity, this material will help the teacher to increase the effectiveness of teaching and provide an opportunity to more rationally distribute time in the lesson. With the help of animation effects, highlighting concepts and important points in color, students' attention is focused on the subject being studied, and better memorization of definitions and the course of reasoning when solving problems is achieved.

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 asked 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, with a≠0. Separately, on slide 4 it is noted for remembering that the domain of definition of this function is the entire axis of real values. Conventionally, this statement is denoted by D(x)=R.

An example of a quadratic function is its important application in physics - the formula for the dependence of the path during uniformly accelerated motion on time. At the same time, in physics lessons, students study formulas for various types of motion, so they will need the ability to solve such problems. On slide 5, students are reminded that when a body moves with acceleration and at the beginning of the time count the distance traveled and the speed of movement are known, then the functional dependence representing such movement will be expressed by the formula S = (at 2)/2+v 0 t+S 0 . Below is an example of turning this formula into a given quadratic function if the values ​​of acceleration = 8, initial speed = 3 and initial 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 plotting the function y=3x 2 . First, the table indicates the correspondence between the function values ​​and the argument values. It is noted that the difference between the constructed graph of the function y=3x 2 and the graph of the function y=x 2 is that each value will be three times greater than the corresponding one. This difference is well tracked in the table view. Nearby, in the graphical representation, the difference in the narrowing of the parabola is also clearly visible.

The next slide looks at plotting the quadratic function y=1/3 x 2. To construct a graph, you need to indicate in the table the values ​​of the function at a number of its points. 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 in the graph. Its parabola is more expanded relative to the ordinate axis than the parabola of the function y=x 2.

Examples help to understand the general rule, according to which you can then more simply and quickly construct the corresponding graphs. On slide 9, a separate rule is highlighted that the graph of the quadratic function y=ax 2 can be constructed depending on the value of the coefficient by stretching or narrowing the graph. If a>1, then the graph stretches from the x-axis by a factor. 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. Next, 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, stating that such a graph will also be called a parabola.

Slide 14 discusses the properties of the quadratic function y=ax 2 when positive. It is noted that its graph passes through the origin, and all points except lie in the upper half-plane. The symmetry of the graph relative to the ordinate axis is noted, specifying that opposite values ​​of the argument correspond to the same function values. It is indicated that the interval of decrease of this function is (-∞;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 has no 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 its points, except, lie in the lower half-plane. The graph is symmetrical about the axis, and opposite values ​​of the argument correspond to equal values ​​of the function. The function increases on the interval and decreases on. The values ​​of this function lie in the interval, it is equal to zero at a point, and has no minimum value.

Summarizing the characteristics considered, on slide 16 it is concluded that the branches of the parabola are directed downwards at, and upwards at. The parabola is symmetrical about the axis, and the vertex of the parabola is located at the point of its intersection with the axis. The vertex of the parabola y=ax 2 is the origin.

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

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

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