Consider the expression of the form ah 2 + W + C, where A, B, C - valid numbers, and varied from zero. This mathematical expression is known as a square triple.

Recall that Ah 2 is a senior member of this square three-melan, and its senior coefficient.

But not always at square three decrees there are all three terms. Take for example an expression 3x 2 + 2x, where a \u003d 3, B \u003d 2, C \u003d 0.

We turn to the quadratic function y \u003d ah 2 + Vx + C, where A, B, C - any arbitrary numbers. This feature is quadratic, as it contains a member of the second degree, that is, x in the square.

It is fairly easy to build a chart of a quadratic function, for example, you can use the allocation method of a complete square.

Consider an example of constructing a graph of the function of the value of -3x 2 - 6x + 1.

For this, the first thing is that we remember the discharge scheme of the full square in three-half -3x 2 - 6x + 1.

I summarize the first two components for brackets. We have -3 multiplied by the amount x square plus 2x and add 1. By adding and taking a unit in brackets, we obtain the sum of the sum of the amount that can be collapsed. We obtain -3 multiply in the amount (x + 1) in the square minus 1 add 1. Revealing brackets and leading similar terms, expression comes out: -3 multiplied by the sum of the sum (x + 1) add 4.

We construct a graph of the obtained function by turning to the auxiliary coordinate system with the beginning at a point with coordinates (-1; 4).

In the picture from the video, this system is indicated by dotted lines. Tear the function in equals -3x 2 to the constructed coordinate system. For convenience, take control points. For example, (0; 0), (1; -3), (-1; -3), (2; -12), (-2; -12). At the same time, we postpone them in the constructed coordinate system. The parabol obtained during the construction is the required schedule. In the picture it is a red parabola.

By applying the method of isolating a complete square, we have a quadratic function of the form: y \u003d a * (x + 1) 2 + m.

The chart of parabola y \u003d ah 2 + bx + c is easy to get from parabola y \u003d ah 2 parallel transfer. This is confirmed by the theorem that can be proved by highlighting a full square of bico. The expression AH 2 + BX + C after successive transformations turns into an expression of the form: A * (x + L) 2 + m. Hatch the schedule. Perform a parallel movement of parabola y \u003d ah 2, combining the vertex with a point with coordinates (-l; m). It is important that x \u003d -l, which means -B / 2a. It means that this direct is the axis of parabola ah 2 + BX + C, its vertex is at a point with an abscissa x zero equal to minus B, divided by 2a, and the ordinate is calculated according to the cumbersome formula 4As - B 2 /. But this formula is not necessary to memorize. Since, substituting the value of the abscissa to the function, we obtain the ordinate.

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

We take the function y \u003d -3x 2 - 6x + 1. By drawing up the equation of the parabol axis, we have that x \u003d -1. And this value is the coordinate of the whip of the parabola. It remains to find only ordinate. Substituting the value -1 to the function, we obtain 4. The pearabol vertex is at the point (-1; 4).

The graph of the function y \u003d -3x 2 - 6x + 1 was obtained with a parallel transfer of the function of the function y \u003d -3x 2, it means that it behaves similarly. The senior coefficient is negative, so the branches are directed down.

We see that for any function of the form y \u003d ah 2 + BX + C, the easiest is the last question, that is, the direction of the branches of the parabola. If the coefficient is positive, then the branches are up, and if negative, then down.

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

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

We construct a graph of the function y \u003d 2x 2 - x + 1.

We define immediately - the graph is a parabola, the branches are directed upward, since the senior coefficient is 2, and this is a positive number. According to the formula we find the abscissa x zero, it is 1.5. To find ordinates, we remember that the zero is equal to the function of 1.5, when calculating, we obtain -3.5.

Top - (1.5; -3.5). Axis x \u003d 1.5. Take the points x \u003d 0 and x \u003d 3. y \u003d 1. Note these points. For three famous points, we build a desired schedule.

To build a graph function Ah 2 + BX + C, you must:

Find the coordinates of the top of the parabola and mark them in the picture, then hold the parabol axis;

On the axis oh, take two symmetrical, relative to the axis, parabola point, find the value of the function at these points and mark them on the coordinate plane;

After three points to build a parabola, if necessary, you can take a few more points and build a schedule on them.

In the following example, we will learn to find the greatest and smallest values \u200b\u200bof the function -2x 2 + 8x - 5 on the segment.

According to the algorithm: a \u003d -2, B \u003d 8, it means x zero equal to 2, and at zero - 3, (2; 3) - the top of the parabola, and x \u003d 2 is the axis.

Take the values \u200b\u200bx \u003d 0 and x \u003d 4 and we find the ordents of these points. This is -5. We build a parabola and determine that the smallest value of the function is -5 at x \u003d 0, and the largest 3, at x \u003d 2.

Square Threechlen called a polynomial 2nd degree, that is, the expression aX. 2 + bX. + c. , Where a. ≠ 0, b., c. - (usually specified) valid numbers called its coefficients, x. - variable value.

Note: coefficient a. It can be any valid number except zero. Indeed, if a. \u003d 0, then aX. 2 + bX. + c. = 0 · X. 2 + bX. + c. = 0 + bX. + c. = bX. + c.. In this case, the expression does not remain square, so it cannot be considered square Three. However, such expressions are bicked as, for example, 3 x. 2 − 2x. or x. 2 + 5 can be considered as square triple, if you add them to missing universes with zero coefficients: 3x. 2 − 2x. = 3x. 2 − 2x. + 0 and x. 2 + 5 = x. 2 + 0x. + 5.

If the task is to determine the values \u200b\u200bof the variable h., in which the square trigger takes zero values, i.e. aX. 2 + bX. + c. = 0, That have quadratic equation.

If there are valid roots x. 1 I. x. 2 of some square equation, then the corresponding three can be decomposed on linear multipliers: aX. 2 + bX. + c. = a.(x. − x. 1)(x. − x. 2)

Comment: If the square triple is considered on the set of integrated numbers with, which, perhaps you have not yet studied, it can always be laid on linear multipliers.

When another task is to determine all the values \u200b\u200bthat the result of calculating the square trothes can take the result of the variable h.. determine y. From the expression y. = aX. 2 + bX. + c., then we are dealing with quadratic function.

Wherein roots square equation are zeros of a quadratic function .

Square truder can also be represented as

This presentation is convenient to use when building a graph and study the properties of the quadratic function of a valid variable.

Quadratic function called a function specified by the formula y. = f.(x.), Where f.(x.) - Square truder. Those. formula of type

y. = aX. 2 + bX. + c.,

Where a. ≠ 0, b., c. - Any valid numbers. Or transformed formula

.

The chart of the quadratic function is parabola, the vertex of which is at the point .

Note: It does not say that the graph of the quadratic function called the parabola. It says that Parabola is written here. This is because such a curve of mathematics was discovered and called the parabola earlier (from Greek. Παραβολή - comparison, comparison, similarity), to the stage of detailed study of the properties and graphics of the quadratic function.

Parabola - Line of the intersection of a direct circular cone with a plane that does not pass through the vertex of the cone and parallel one of the samples of this cone.

Parabola has another interesting feature, which is also used as its definition.

Parabola It is a plurality of plane points, the distance from which to a certain point of the plane, called the focus of the parabola, is equal to the distance to a certain direct, called parabola's director.

Build sketch graphics Quadratic function can be by characteristic points .
For example, for function y \u003d X. 2 We take a point

Connecting them from hand, we build the right half of the parabola. Left we obtain symmetrical reflection relative to the axis of the ordinate.

For building sketch of the graph of the quadratic function of the general form As characteristic points, it is convenient to take the coordinates of its vertices, zeros of functions (roots of the equation), if there are, the point of intersection with the ordinate axis (when x. = 0, y \u003d C.) and symmetrical to it with a parabol axis point (- b. / a.; c.).

But in any case, only the sketch of the graphics of the quadratic function can be built by points, i.e. Approximate schedule. To build parabola For sure, it is necessary to use its properties: focus and director.
Arm with paper, line, carbon, two buttons and strong thread. Attach one button in about the center of the paper sheet - at a point that will be a focus of parabola. The second button is attached to the top of the smaller corner of the square. On the bases of the buttons, fasten the thread so that its length between the buttons is equal to a large carbon cathelet. Draw a direct line that is impassable through the focus of the future parabola, - the director of the parabola. Attach a ruler to the director, and the square to the line as shown in the figure. Move the kit along the line, while pressing the pencil to the paper and to the kitchen. Make sure the thread is stretched.

Measure the distance between the focus and the director (I remind you - the distance between the point and the direct is determined by perpendicular). This is the focal parameter parabola p.. In the coordinate system represented on the right figure, the equation of our parabola is: y \u003d x 2/ 2p.. In my drawing scale, a function schedule turned out y. = 0,15x 2.

Comment: To build a given parabola on a given scale, you need to do anything else, but in a different way. You need to start with the coordinate axes. Then draw the director and determine the position of the focus of parabola. And only then construct a tool from the square and a ruler. For example, in order to build a parabola on the checkered paper, the equation of which w. = x. 2, you need to position the focus at a distance of 0.5 cells from the directories.

Properties function w. = x. 2

1. Function Definition Area - All Numerical Direct: D.(f.) = R. = (−∞; ∞).
2. The function of the functions of the function is a positive sonar: E.(f.) \u003d, and increasing the function is performed on the interval. The values \u200b\u200bof this function cover the entire positive part of the actual axis, it is zero at the point, and does not have the greatest value.

   The slide 15 describes the properties of the function y \u003d AX 2, if negative. It is noted that its schedule also passes through the origin of the coordinates, but all its points except lie in the bottom half-plane. The symmetry of the graph relative to the axis is noted, and equal values \u200b\u200bof the function correspond to the opposite values \u200b\u200bof the argument. The function in the interval increases, decreases. The values \u200b\u200bof this function lie in the gap, it is zero at the point, and the smallest value does not have.

   
   

   Summarizing the considered characteristics, the slide 16 shows that the parabola branches are directed down when, and up - with. Parabola is symmetrical about the axis, and the top of the parabolla is located at the point of its intersection with the axis. Parabola y \u003d AX 2 vertex - the origin of the coordinates.

   Also, an important conclusion about parabola transformations is displayed on a slide 17. It presents options for conversion of the graph of the quadratic function. It is noted that the graph of the function y \u003d AX 2 is converted by a symmetric display of the graph with respect to the axis. It is also possible to compress or stretch the graph with respect to the axis.

   On the last slide, we are generalizing conclusions about the conversion of the graph of the function. The conclusions are presented that the function graph is obtained by symmetric conversion relative to the axis. A function graph is obtained from compression or stretching source graph from the axis. At the same time, the stretching from the axis is once observed in the case when. Compression to the axis in 1 / a times the schedule is formed in the case.

   
   

   The presentation "The Y \u003d AX 2 function, its schedule and properties" can be used by a teacher as a visual manual in the classroom. Algebra. Also, this manual well discloses the topic, giving an in-depth understanding of the subject, therefore it can be proposed for independent studies by students. This material will also help the teacher to give an explanation during distance learning.

   Lesson: How to build a parabola or quadratic function?

   THEORETICAL PART

   Parabola is a graph of the function described by formula AX 2 + BX + C \u003d 0.
   To build a parabola need to follow a simple action algorithm:

   1) Parabola formula Y \u003d AX 2 + BX + C,
   if a a\u003e 0. then the branches of parabola are directed up,
   and the branches of parabola are directed down.
   Free dick c. This point crosses the parabola with the Oy axis;

   2), it is found according to the formula x \u003d (- b) / 2a, found x we \u200b\u200bsubstitute in the parabola equation and find y.;

   3) Zero function Or, on another point of intersection of the parabola with the OX axis, they are also called the roots of the equation. To find the roots we equate equate to 0 aX 2 + BX + C \u003d 0;

   Types of equations:

   a) The full square equation has the form AX 2 + BX + C \u003d 0and is solved by discriminant;
   b) incomplete square equation AX 2 + BX \u003d 0. To solve it, you need to make x for brackets, then each multiplier to equate to 0:
   AX 2 + BX \u003d 0,
   x (ax + b) \u003d 0,
   x \u003d 0 and ax + b \u003d 0;
   c) incomplete square equation AX 2 + C \u003d 0. To solve it, unknown to transfer one way, and known to another. x \u003d ± √ (C / A);

   4) Find a few additional points to build a function.

   Practical part

   And so now on the example we will analyze all the actions:
   Example number 1:
   y \u003d x 2 + 4x + 3
   C \u003d 3 means Parabola crosses OY at point x \u003d 0 y \u003d 3. Parabola branches look up as a \u003d 1 1\u003e 0.
   a \u003d 1 b \u003d 4 c \u003d 3 x \u003d (- b) / 2a \u003d (- 4) / (2 * 1) \u003d - 2 y \u003d (-2) 2 +4 * (- 2) + 3 \u003d 4- 8 + 3 \u003d -1 Top is at point (-2; -1)
   Find the roots of the equation x 2 + 4x + 3 \u003d 0
   On discriminative find roots
   a \u003d 1 b \u003d 4 c \u003d 3
   D \u003d b 2 -4ac \u003d 16-12 \u003d 4
   x \u003d (- b ± √ (d)) / 2a
   x 1 \u003d (- 4 + 2) / 2 \u003d -1
   x 2 \u003d (- 4-2) / 2 \u003d -3
   
   Take several arbitrary points that are near the top x \u003d -2

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

   We substitute instead of x in the equation y \u003d x 2 + 4x + 3 values
   Y \u003d (- 4) 2 +4 * (- 4) + 3 \u003d 16-16 + 3 \u003d 3
   Y \u003d (- 3) 2 +4 * (- 3) + 3 \u003d 9-12 + 3 \u003d 0
   y \u003d (- 1) 2 +4 * (- 1) + 3 \u003d 1-4 + 3 \u003d 0
   y \u003d (0) 2 + 4 * (0) + 3 \u003d 0-0 + 3 \u003d 3
   Seen by the values \u200b\u200bof the function that the parabol is symmetric with respect to direct x \u003d -2

   Example number 2:
   y \u003d -X 2 + 4X
   c \u003d 0 So Parabola crosses OY at point x \u003d 0 y \u003d 0. Parabola branches look down as a \u003d -1 -1 Find the roots of the equation -x 2 + 4x \u003d 0
   An incomplete square equation of AX 2 + BX \u003d 0. To decide it, you need to make x for brackets, then each multiplier to equate to 0.
   x (-x + 4) \u003d 0, x \u003d 0 and x \u003d 4.
   
   Take several arbitrary points that are near the top x \u003d 2
   x 0 1 3 4
   0 3 3 0
   We substitute instead of the equation y \u003d -x 2 + 4x values
   Y \u003d 0 2 + 4 * 0 \u003d 0
   y \u003d - (1) 2 + 4 * 1 \u003d -1 + 4 \u003d 3
   Y \u003d - (3) 2 + 4 * 3 \u003d -9 + 13 \u003d 3
   Y \u003d - (4) 2 + 4 * 4 \u003d -16 + 16 \u003d 0
   It can be seen by the values \u200b\u200bof the function that Parabola is symmetric with respect to direct x \u003d 2

   Example number 3.
   y \u003d x 2 -4
   C \u003d 4 So Parabola crosses OY at point x \u003d 0 y \u003d 4. Parabola branches look up as a \u003d 1 1\u003e 0.
   a \u003d 1 b \u003d 0 C \u003d -4 x \u003d (- b) / 2a \u003d 0 / (2 * (1)) \u003d 0 y \u003d (0) 2 -4 \u003d -4 vertex is at point (0; -4 )
   Find the roots of the equation x 2 -4 \u003d 0
   An incomplete square equation of the form AX 2 + C \u003d 0. To solve it, unknown to transfer one way, and known to another. x \u003d ± √ (C / A)
   x 2 \u003d 4
   x 1 \u003d 2
   x 2 \u003d -2

   Take a few arbitrary points that are near the top x \u003d 0
   x -2 -1 1 2
   0 -3 -3 0
   We substitute instead of x equation y \u003d x 2 -4 values
   Y \u003d (- 2) 2 -4 \u003d 4-4 \u003d 0
   Y \u003d (- 1) 2 -4 \u003d 1-4 \u003d -3
   Y \u003d 1 2 -4 \u003d 1-4 \u003d -3
   y \u003d 2 2 -4 \u003d 4-4 \u003d 0
   Seen by the values \u200b\u200bof the function that parabola is symmetrical with respect to direct x \u003d 0

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