What is called zeros of a quadratic function. How to build a parabola? What is Parabola? How are square equations solve? III case appears "C"

The function of the species where is called quadratic function.

Schedule of a quadratic function - parabola.

Consider cases:

I Classical Parabola

I.e , ,

For constructing, fill in the table, substituting the X values \u200b\u200bin the formula:

We note the points (0; 0); (1; 1); (-1; 1), etc. on the coordinate plane (how with a smaller step we take x (in this case Step 1), and the more we take the values \u200b\u200bx, the smaller it will be a curve), we get a parabola:

It is easy to see that if we take the case ,,, that is, we will get a parabola, symmetric about the axis (oh). Ensure this is easy by filling out a similar table:

II Case, "A" is excellent from one

What will happen if we take,,? How will the behavior of the parabola change? With Title \u003d "(! Lang: Rendered by QuickTex.com" height="20" width="55" style="vertical-align: -5px;"> парабола изменит форму, она “похудеет” по сравнению с параболой (не верите – заполните соответствующую таблицу – и убедитесь сами):!}

In the first picture (see above) it is clearly seen that points from the table for parabola (1; 1), (-1; 1) were transformed into points (1; 4), (1; -4), that is, with the same The values \u200b\u200bof the ordinate of each point multiplied on 4. This will happen to all key points of the source table. Similarly, we argue in cases of pictures 2 and 3.

And at Parabola "will become wider" parabola:

Let's submit:

1) The coefficient sign is responsible for the direction of the branches. With Title \u003d "(! Lang: Rendered by QuickTex.com" height="14" width="47" style="vertical-align: 0px;"> ветви направлены вверх, при - вниз. !}

2) Absolute value The coefficient (module) is responsible for "expansion", "compression" parabola. The larger, the more parabol, the less | A |, the wider Parabola.

III case appears "C"

Now let's enter into the game (that is, we consider the case when), we will consider parabollas of the species. It is not difficult to guess (you can always refer to the table), which will displace parabola along the axis up or down depending on the sign:

IV Case appears "B"

When will Parabola "breaks off" from the axis and will finally "walk" throughout the coordinate plane? When will cease to be equal.

Here for the construction of parabola we will need formula for calculating the vertex: , .

So at this point (as at the point (0; 0) of the new coordinate system) we will build a parabola, which we can permanently. If we are dealing with the case, then from the tops are laying one single segment to the right, one up, - the resulting point is our (similar to the step left, step up is our point); If we are dealing with, for example, then from the tops are laying one single segment to the right, two - up, etc.

For example, the vertex parabola:

Now the main thing is to understand that in this top we will build a parabola on the parabola pattern, because in our case.

When building a parabolla after finding the coordinates of the vertex is very It is convenient to consider the following points:

1) parabola will definitely pass through the point . Indeed, substituting in the formula x \u003d 0, we get that. That is, the ordinate of the point of intersection of the parabola with the axis (OU) is. In our example (above), Parabola crosses the ordinate axis at the point, since.

2) axis of symmetry parabola is straight, so all points of parabola will be symmetrical about it. In our example, we immediately take the point (0; -2) and we build a parabola symmetry with a symmetry relative to the axis, we get a point (4; -2) through which Parabola will pass.

3) Equating to, we learn the points of intersection of the parabola with the axis (oh). To do this, solve the equation. Depending on the discriminant, we will receive one (,), two (title \u003d "(! Lang: Rendered by QuickTex.com" height="14" width="54" style="vertical-align: 0px;">, ) или нИсколько () точек пересечения с осью (ох) !} . In the previous example, we have a root from the discriminant - not a whole number, when building it, it does not make sense to find the roots, but we see clearly that two points of intersection with the axis (oh) will have (since Title \u003d "(! Lang: Rendered By QuickLatex.com." height="14" width="54" style="vertical-align: 0px;">), хотя, в общем, это видно и без дискриминанта.!}

So let's work out

Algorithm for building a parabola if it is asked in the form

1) we determine the direction of the branches (A\u003e 0 - up, a<0 – вниз)

2) We find the coordinates of the vertex parabola by the formula.

3) We find the point of intersection of the parabola with the axis (OU) on a free member, we build a point, symmetrical about the axis of symmetry of the parabola (it should be noted, it happens that this point is unprofitable to note, for example, because the value is great ... I miss this item ...)

4) At the found point - the top of the parabola (as at the point (0; 0) of the new coordinate system) we build a parabola. If title \u003d "(! Lang: Rendered by QuickTextEx.com" height="20" width="55" style="vertical-align: -5px;">, то парабола становится у’же по сравнению с , если , то парабола расширяется по сравнению с !}

5) We find the point of intersection of the parabola with the axis (OU) (if they are still "not pop up"), solving equation

Example 1.

Example 2.

Note 1. If Parabola is initially set in the form, where there are some numbers (for example,), then it will be even easier to build it, because the coordinates of the vertices are already specified. Why?

Take square threechlen And we highlight the full square in it: look, so we got that. We have previously called the top of the parabola, that is, now.

For example, . We note on the plane the top of the parabola, we understand that the branches are directed down, Parabola is expanded (relatively). That is, we perform paragraphs 1; 3; four; 5 of the parabola construction algorithm (see above).

Note 2. If Parabola is set in a form, similar to this (that is, it is presented in the form of a work of two linear multipliers), then we are immediately visible to the point of intersection of the parabola with the axis (oh). In this case - (0; 0) and (4; 0). Otherwise, we act according to the algorithm, the opening of the bracket.

Many tasks need to calculate the maximum or minimum quadratic function. Maximum or minimum can be found if the source function is recorded in a standard form: or through the coordinates of the vertex parabola: f (x) \u003d a (x - h) 2 + k (\\ displaystyle f (x) \u003d a (x - h) ^ (2) + K). Moreover, the maximum or minimum of any quadratic function can be calculated using mathematical operations.

Steps

The quadratic function is recorded in standard form

Write down the function in standard form. The quadratic function is a function whose equation includes a variable x 2 (\\ displaystyle x ^ (2)). The equation may include or not include a variable. X (\\ DisplayStyle X). If the equation includes a variable with an indicator of more than 2, it does not describe the quadratic function. If necessary, bring similar members and rear them to record the function in standard form.

A chart of a quadratic function is a parabola. Parabola branches are directed up or down. If the coefficient A (\\ DisplayStyle A) With a variable x 2 (\\ displaystyle x ^ (2)) A (\\ DisplayStyle A)

Calculate -B / 2a. Value - B 2 A (\\ DisplayStyle - (\\ FRAC (B) (2a))) - This is coordinates X (\\ DisplayStyle X) The peaks of parabola. If the quadratic function is recorded in a standard form a x 2 + b x + c (\\ displaystyle ax ^ (2) + bx + c), use the coefficients when X (\\ DisplayStyle X) and x 2 (\\ displaystyle x ^ (2)) in the following way:

• In the function of the coefficients a \u003d 1 (\\ displaystyle a \u003d 1) and B \u003d 10 (\\ DisplayStyle B \u003d 10)
• As a second example, consider the function. Here a \u003d - 3 (\\ displaystyle a \u003d -3) and B \u003d 6 (\\ DisplayStyle B \u003d 6). Therefore, the "X" coordinate of the top of parabolas will calculate this:

1. Find the corresponding value F (X). Submold the found value "x" into the source function to find the corresponding value f (x). So you will find a minimum or maximum function.

   • In the first example f (x) \u003d x 2 + 10 x - 1 (\\ displaystyle f (x) \u003d x ^ (2) + 10x-1) You have calculated that the coordinate "X" of the pearable parabol is equal to x \u003d - 5 (\\ displayStyle x \u003d -5). In the original function instead X (\\ DisplayStyle X) Put - 5 (\\ DisplayStyle -5)
   • In the second example f (x) \u003d - 3 x 2 + 6 x - 4 (\\ displaystyle f (x) \u003d - 3x ^ (2) + 6x-4) You found that the coordinate "x" of the top of the parabola is equal to x \u003d 1 (\\ displaystyle x \u003d 1). In the original function instead X (\\ DisplayStyle X) Put 1 (\\ DISPLAYSTYLE 1)To find its maximum value:

2. Write down the answer. Re-read the condition of the task. If you need to find the coordinates of the top of the parabola, in response, write both values X (\\ DisplayStyle X) and Y (\\ DisplayStyle Y) (or f (x) (\\ DisplayStyle F (X))). If you need to calculate a maximum or minimum function, in response, write down only the value Y (\\ DisplayStyle Y) (or f (x) (\\ DisplayStyle F (X))). Once again, look at the sign of the coefficient A (\\ DisplayStyle A)To verify that you have calculated: a maximum or a minimum.

   The quadratic function is recorded through the coordinates of the vertex parabola

   1. Record the quadratic function through the coordinates of the parabola vertex. Such an equation is as follows:

      Determine the direction of parabola. To do this, look at the coefficient sign A (\\ DisplayStyle A). If the coefficient A (\\ DisplayStyle A) Positive, parabola is directed up. If the coefficient A (\\ DisplayStyle A) Negative, parabola is directed down. For example:

      Find the minimum or maximum function value. If the function is recorded through the coordinates of the pearabela vertex, minimum or maximum equal to the value of the coefficient K (\\ DisplayStyle K). In the examples above:

      Find the coordinates of the pearabela vertices. If the task is required to find the top of the parabola, its coordinates are equal (H, K) (\\ DisplayStyle (H, K)). Note when the quadratic function is recorded through the coordinates of the pearabela vertex, the subtraction operation must be enclosed in brackets. (X - H) (\\ DisplayStyle (X - H)), Therefore, the value H (\\ DisplayStyle H) Takes up with the opposite sign.

   How to calculate a minimum or maximum with the help of mathematical operations

   First consider the standard type of equation. Record the quadratic function in the standard form: f (x) \u003d a x 2 + b x + c (\\ displaystyle f (x) \u003d ax ^ (2) + bx + c). If necessary, bring similar members and rearrange them to obtain a standard equation.

   Find the first derivative. The first derivative of the quadratic function, which is recorded in a standard form, is equal to f '(x) \u003d 2 a x + b (\\ displaystyle f ^ (\\ prime) (x) \u003d 2AX + B).

   Derivative equate to zero. Recall that the derived function is equal to the angular coefficient of the function at a certain point. In a minimum or maximum, the angular coefficient is zero. Therefore, to find the minimum or maximum function value, the derivative must be equal to zero. In our example:

The quadratic function is called the function of the form:
y \u003d a * (x ^ 2) + b * x + c,
where a is the coefficient with a senior degree of unknown x,
b - the coefficient at the unknown x,
and with - a free member.
The graph of the quadratic function is a curve called parabola. General form Parabola is presented in the figure below.

Fig.1 General view of parabola.

There are several different ways to build a chart of a quadratic function. We will look at the main and most common one.

Algorithm for constructing a graph of a quadratic function y \u003d a * (x ^ 2) + b * x + c

1. Build a coordinate system, mark a single segment and sign coordinate axes.

2. Determine the direction of the branches of the parabola (up or down).
To do this, you need to look at the sign of the coefficient a. If plus - the branches are directed up, if the branches are sent down.

3. Determine the coordinate of the top of the parabola.
To do this, you need to use the Formula of Hvershins \u003d -b / 2 * a.

4. Determine the coordinate at the top of the parabola.
To do this, substitute the alien \u003d a * (x ^ 2) + b * x + c equation instead of x, found in the previous step the value of Hvershina.

5. Apply the resulting point on the chart and spend the axis of symmetry through it, parallel to the OU's coordinate axis.

6. Find the intersection points of the graph with the axis oh.
To do this, it is required to solve the square equation A * (x ^ 2) + b * x + c \u003d 0 by one of the known methods. If the equation does not have real roots, then the function graph does not cross the axis OH.

7. Find the coordinates of the point of intersection of the graph with the OU axis.
To do this, we substitute the value x \u003d 0 to equation and calculate the value of y. We celebrate this and symmetrical point on the chart.

8. We find the coordinates of arbitrary point A (X, Y)
To do this, select the arbitrary value of the coordinates x, and we substitute it in our equation. We get the value at this point. Apply a point on the chart. And also note the point on the chart, the symmetric point A (x, y).

9. Connect the received points on the smooth line chart and continue the schedule for extreme Pointsto the end of the coordinate axis. Sign the schedule either on the callout, or if the place is located along the graph.

An example of building a graphic

As an example, we construct a chart of a quadratic function given by the equation y \u003d x ^ 2 + 4 * x-1
1. We draw coordinate axes, we sign them and mark a single segment.
2. The values \u200b\u200bof the coefficients A \u003d 1, B \u003d 4, C \u003d -1. Since a \u003d 1, that more zero branch of parabola is directed up.
3. Determine the coordinate x of the top of the Hvershina parabola \u003d -b / 2 * a \u003d -4 / 2 * 1 \u003d -2.
4. Determine the coordinate at the top of the parabola
Spray \u003d A * (x ^ 2) + b * x + c \u003d 1 * ((- 2) ^ 2) + 4 * (- 2) - 1 \u003d -5.
5. We note the vertex and carry out the symmetry axis.
6. We find the point of intersection of the graph of the quadratic function with the axis oh. We solve the square equation x ^ 2 + 4 * x-1 \u003d 0.
x1 \u003d -2-√3 x2 \u003d -2 + √3. We mark the values \u200b\u200bobtained on the chart.
7. We find the point of intersection of the schedule with the OU axis.
x \u003d 0; y \u003d -1.
8. Select an arbitrary point B. Let it have a coordinate x \u003d 1.
Then y \u003d (1) ^ 2 + 4 * (1) -1 \u003d 4.
9. We connect the points received and subscribe a schedule.

If you want to participate in great life, then fill your head with mathematics while there is a possibility. She will give you a huge help in all your work.

M.I. Kalinin

One of the main functions school Mathematicsfor which the full theory is built and all properties are proved, is quadratic function. Pupils should clearly understand and know all these properties. In this case, the tasks for the quadratic function there is a great set - from very simple, which flow directly from the theory and formulas to the most difficult, the solution of which requires an analysis and a deep understanding of all properties of the function.

When solving tasks on a quadratic function practical value It has compliance between the algebraic description of the problem and its geometric interpretation - the image on the coordinate plane of the sketch of the function of the function. It is thanks to this feature that you always have the opportunity to verify the correctness and consistency of our theoretical reasoning.

Consider several tasks on the topic "Quadratic function" and focus on their detailed solution.

Task 1.

Find the amount of the total values \u200b\u200bof the number P, in which the vertex parabola y \u003d 1 / 3x 2 - 2px + 12p is located above the OX axis.

Decision.

Parabola branches are directed up (a \u003d 1/3\u003e 0). Since the top of the parabola lies above the OX axis, then Parabola does not intersect the abscissa axis (Fig. 1). So function

y \u003d 1 / 3x 2 - 2px + 12p does not have zeros,

a equation

1 / 3x 2 - 2px + 12p \u003d 0 does not have roots.

This is possible if the discriminant of the last equation turns out to be negative.

I calculate it:

D / 4 \u003d p 2 - 1/3 · 12p \u003d p 2 - 4p;

p 2 - 4P< 0;

p (P - 4)< 0;

p belongs to the interval (0; 4).

The sum of the integer values \u200b\u200bof the number P from the gap (0; 4): 1 + 2 + 3 \u003d 6.

Answer: 6.

Note that for the answer to the question of the task it was possible to solve inequality

y in\u003e 0 or (4ac - b 2) / 4a\u003e 0.

Task 2.

Find the number of integers of the number A, in which the abscissa and the ordinate of the vertex of parabola y \u003d (x - 9a) 2 + a 2 + 7a + 6 are negative.

Decision.

If the quadratic function is viewed

y \u003d a (x - n) 2 + m, the point with coordinates (m; n) is a pearabol vertex.

In our case

x \u003d 9a; y B \u003d a 2 + 7a + 6.

Since the abscissa, and the ordinate of the pearabol vertices must be negative, then the system of inequalities:

(9A.< 0,
(A 2 + 7A + 6< 0;

Resolved the resulting system:

(A.< 0,
((A + 1) (A + 6)< 0;

I will depict the solution of inequalities on the coordinate direct and give the final answer:

a belongs to the interval (-6; -1).

Integer values \u200b\u200bof the number A: -5; -four; -3; -2. Their number: 4.

Answer: 4.

Task 3.

Find the most integer value of the number M at which the quadratic function
y \u003d -2x 2 + 8x + 2m takes only negative values.

Decision.

Parabola branches are directed down (a \u003d -2< 0). Данная функция будет принимать только отрицательные значения, если ее график не будет иметь общих точек с осью абсцисс, т.е. уравнение -2x 2 + 8x + 2m = 0 не будет иметь корней. Это возможно, если дискриминант последнего уравнения будет отрицательным.

2x 2 + 8x + 2m \u003d 0.

We divide the coefficients of the equation on -2, we get:

x 2 - 4x - m \u003d 0;

D / 4 \u003d 2 2 - 1 · 1 · (-m) \u003d 4 + m;

The greatest integer value of the number M: -5.

Answer: -5.

To answer the question of the task, it was possible to solve the inequality y in< 0 или

(4AC - B 2) / 4A< 0.

Task 4.

Find the smallest value of the quadratic function Y \u003d AX 2 - (A + 6) X + 9, if it is known that the straight line X \u003d 2 is the axis of symmetry of its schedule.

Decision.

1) Since straight x \u003d 2 is the axis of symmetry of this graph, x B \u003d 2. We use the formula

x B \u003d -b / 2a, then x B \u003d (A + 6) / 2A. But x B \u003d 2.

Make an equation:

(A + 6) / 2A \u003d 2;

Then the function takes the view

y \u003d 2x 2 - (2 + 6) x + 9;

y \u003d 2x 2 - 8x + 9.

2) Parabola branches

The smallest meaning of this function is the ordinate of the top of the parabola (Fig. 2)which is easy to find using the formula

y B \u003d (4ac - b 2) / 4a.

y B \u003d (4 · 2 · 9 - 8 2) / 4 · 2 \u003d (72 - 64) / 8 \u003d 8/8 \u003d 1.

The smallest value of the function under consideration is 1.

Answer: 1.

Task 5.

Find the smallest whole value of the number A, in which a set of values \u200b\u200bof the function y \u003d x 2 - 2x + a and y \u003d -x 2 + 4x - a do not intersect.

Decision.

Find many values \u200b\u200bof each function.

I Method.

y 1 \u003d x 2 - 2x + a.

Apply formula

y B \u003d (4ac - b 2) / 4a.

y B \u003d (4 · 1 · a - 2 2) / 4 · 1 \u003d (4a - 4) / 4 \u003d 4 (A - 1) / 4 \u003d A - 1.

As the branches of parabola are directed up, then

E (y) \u003d.

E (y 2) \u003d (-∞; 4 - a].

Depict the obtained sets on the coordinate direct (Fig. 3).

The resulting sets will not intersect if the point with the coordinate 4 - a will be located to the left of the point with the coordinate A - 1, i.e.

4 - A.< a – 1;

The smallest whole value of the number A: 3.

Answer: 3.

Tasks for the location of the roots of the quadratic function, tasks with parameters and tasks that are reduced to quadratic functions are very popular on the exam. Therefore, when preparing for exams, you should pay close attention to them.

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The function of the form y \u003d a * x ^ 2 + b * x + c, where a, b, c is some real numbers, and it is different from zero, and x, y - variables, called a quadratic function. The graph of the quadratic function y \u003d a * x ^ 2 + b * x + c is a line called in mathematics parabola. General view of parabola Presented in the figure below.

It is worth noting that if the function is the coefficient a\u003e 0, then the parabol is directed upwards, and if the agrafic of the quadratic function is symmetrical relative to the symmetry axis. The axis of symmetry of the parabola is a direct spent through the point x \u003d (- b) / (2 * a), parallel to the OU axis.

The coordinates of the vertex parabola are determined by the following formulas:

x0 \u003d (- b) / (2 * a) y0 \u003d y (x0) \u003d (4 * A * C-B ^ 2) / 4 * a.

Figure below shows a graph of an arbitrary quadratic function. Construction of a graph of a quadratic function. Also in the picture is marked the top of the parabola and the axis of symmetry.

Depending on the value of the coefficient A, the top of the parabola will be the minimum or maximum value of the quadratic function. With a\u003e 0, the vertex is the minimum value of the quadratic function, while the maximum value does not exist. In case of symmetry passes through the top of the parabola. The area of \u200b\u200bdefinition of the quadratic function is all many real numbers R.

The quadratic function y \u003d a * x ^ 2 + b * x + c is always possible to convert to the form y \u003d a * (x + k) ^ 2 + p, where k \u003d b / (2 * a), p \u003d (4 * A * CB ^ 2) / (4 * a). To do this, it is necessary to highlight a full square.

Please note that the point with coordinates (-k; p) will be a pearabol vertex. The graph of the quadratic function y \u003d a * (x + k) ^ 2 + P can be obtained from the graph of the function y \u003d a * x ^ 2 using parallel transfer.

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