Quadratic function graph and shifts. Quadratic function and its graph

A quadratic function is a function of the form:
y=a*(x^2)+b*x+c,
where a is the coefficient for the highest degree of unknown x,
b - coefficient for unknown x,
and c is a free member.
The graph of a quadratic function is a curve called a parabola. The general view of the parabola is shown in the figure below.

Fig.1 General view of the parabola.

There are a few in various ways plotting a quadratic function. We will look at the main and most general of them.

Algorithm for plotting a quadratic function y=a*(x^2)+b*x+c

1. Construct a coordinate system, mark a unit segment and label coordinate axes.

2. Determine the direction of the parabola branches (up or down).
To do this, you need to look at the sign of the coefficient a. If there is a plus, then the branches are directed upward, if there is a minus, then the branches are directed downward.

3. Determine the x coordinate of the vertex of the parabola.
To do this, you need to use the formula Xvertex = -b/2*a.

4. Determine the coordinate at the vertex of the parabola.
To do this, substitute into the equation Uvershiny = a*(x^2)+b*x+c instead of x, the value of Xverhiny found in the previous step.

5. Plot the resulting point on the graph and draw an axis of symmetry through it, parallel to the Oy coordinate axis.

6. Find the points of intersection of the graph with the Ox axis.
To do this, you need to solve the quadratic equation a*(x^2)+b*x+c = 0 using one of known methods. If the equation does not have real roots, then the graph of the function does not intersect the Ox axis.

7. Find the coordinates of the point of intersection of the graph with the Oy axis.
To do this, we substitute the value x=0 into the equation and calculate the value of y. We mark this and a point symmetrical to it on the graph.

8. Find the coordinates of an arbitrary point A(x,y)
To do this, choose an arbitrary value for the x coordinate and substitute it into our equation. We get the value of y at this point. Plot the point on the graph. And also mark a point on the graph that is symmetrical to point A(x,y).

9. Connect the obtained points on the graph with a smooth line and continue the graph beyond extreme points, to the end of the coordinate axis. Label the graph either on the leader or, if space allows, along the graph itself.

Example of plotting

As an example, let's plot a quadratic function given by the equation y=x^2+4*x-1
1. Draw coordinate axes, label them and mark a unit segment.
2. Coefficient values ​​a=1, b=4, c= -1. Since a=1, which is greater than zero, the branches of the parabola are directed upward.
3. Determine the X coordinate of the vertex of the parabola Xvertices = -b/2*a = -4/2*1 = -2.
4. Determine the coordinate Y of the vertex of the parabola
Vertices = a*(x^2)+b*x+c = 1*((-2)^2) + 4*(-2) - 1 = -5.
5. Mark the vertex and draw the axis of symmetry.
6. Find the intersection points of the graph of the quadratic function with the Ox axis. We solve the quadratic equation x^2+4*x-1=0.
x1=-2-√3 x2 = -2+√3. We mark the obtained values ​​on the graph.
7. Find the points of intersection of the graph with the Oy axis.
x=0; y=-1
8. Choose an arbitrary point B. Let it have coordinate x=1.
Then y=(1)^2 + 4*(1)-1= 4.
9. Connect the resulting points and sign the graph.

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QUADRATIC FUNCTION, a mathematical FUNCTION whose value depends on the square of the independent variable, x, and is given, respectively, by a quadratic POLYNOMIAL, for example: f(x) = 4x2 + 17 or f(x) = x2 + 3x + 2. see also SQUARE THE EQUATION … Scientific and technical encyclopedic dictionary

Quadratic function - Quadratic function is a function of the form y= ax2 + bx + c (a ≠ 0). Graph K.f. - a parabola, the vertex of which has coordinates [ b/ 2a, (b2 4ac) / 4a], for a> 0 the branches of the parabola are directed upward, for a< 0 –вниз… …

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An affine quadratic function on an affine space S is any function Q: S→K that has the form Q(x)=q(x)+l(x)+c in vectorized form, where q is a quadratic function, l is a linear function, c is a constant. Contents 1 Shifting the reference point 2 ... ... Wikipedia

An affine quadratic function on an affine space is any function that has the form in vectorized form, where is a symmetric matrix, a linear function, a constant. Contents... Wikipedia

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objective function- - [Ya.N.Luginsky, M.S.Fezi Zhilinskaya, Yu.S.Kabirov. English-Russian dictionary of electrical engineering and power engineering, Moscow, 1999] objective function In extremal problems, a function whose minimum or maximum is required to be found. This… … Technical Translator's Guide

Objective function- in extremal problems, a function whose minimum or maximum needs to be found. This is a key concept in optimal programming. Having found the extremum of C.f. and, therefore, having determined the values ​​of the controlled variables that go to it... ... Economic-mathematical dictionary

Books

• Set of tables. Mathematics. Graphs of functions (10 tables), . Educational album of 10 sheets. Linear function. Graphical and analytical assignment of functions. Quadratic function. Transforming the graph of a quadratic function. Function y=sinx. Function y=cosx.…
• The most important function of school mathematics is quadratic - in problems and solutions, Petrov N.N.. The quadratic function is the main function school course mathematics. No wonder. On the one hand, the simplicity of this function, and on the other hand, deep meaning. Many tasks of school...

A function of the form where is called quadratic function.

Graph of a quadratic function – parabola.

Let's consider the cases:

I CASE, CLASSICAL PARABOLA

That is , ,

To construct, fill out the table by substituting the x values ​​into the formula:

Mark the points (0;0); (1;1); (-1;1), etc. on the coordinate plane (the smaller the step we take the x values ​​(in in this case step 1), and the more x values ​​we take, the smoother the curve will be), we get a parabola:

It is easy to see that if we take the case , , , that is, then we get a parabola that is symmetrical about the axis (oh). It’s easy to verify this by filling out a similar table:

II CASE, “a” IS DIFFERENT FROM UNIT

What will happen if we take , , ? How will the behavior of the parabola change? With title="Rendered by QuickLaTeX.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 visible that the points from the table for the parabola (1;1), (-1;1) were transformed into points (1;4), (1;-4), that is, with the same values, the ordinate of each point is multiplied by 4. This will happen to all key points of the original table. We reason similarly in the cases of pictures 2 and 3.

And when the parabola “becomes wider” than the parabola:

Let's summarize:

1)The sign of the coefficient determines the direction of the branches. With title="Rendered by QuickLaTeX.com" height="14" width="47" style="vertical-align: 0px;"> ветви направлены вверх, при - вниз. !}

2) Absolute value coefficient (modulus) is responsible for the “expansion” and “compression” of the parabola. The larger , the narrower the parabola; the smaller |a|, the wider the parabola.

III CASE, “C” APPEARS

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

IV CASE, “b” APPEARS

When will the parabola “break away” from the axis and finally “walk” along the entire coordinate plane? When will it cease to be equal?

Here to construct a parabola we need formula for calculating the vertex: , .

So at this point (as at point (0;0) new system coordinates) we will build a parabola, which we can already do. If we are dealing with the case, then from the vertex we put one unit segment to the right, one up, - the resulting point is ours (similarly, a step to the left, a step up is our point); if we are dealing with, for example, then from the vertex we put one unit segment to the right, two - upward, etc.

For example, the vertex of a parabola:

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

When constructing a parabola after finding the coordinates of the vertex veryIt is convenient to consider the following points:

1) parabola will definitely pass through the point . Indeed, substituting x=0 into the formula, we obtain that . That is, the ordinate of the point of intersection of the parabola with the axis (oy) is . In our example (above), the parabola intersects the ordinate at point , since .

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

3) Equating to , we find out the points of intersection of the parabola with the axis (oh). To do this, we solve the equation. Depending on the discriminant, we will get one (, ), two ( title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">, ) или нИсколько () точек пересечения с осью (ох) !} . In the previous example, our root of the discriminant is not an integer; when constructing, it doesn’t make much sense for us to find the roots, but we clearly see that we will have two points of intersection with the axis (oh) (since title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">), хотя, в общем, это видно и без дискриминанта.!}

So let's work it out

Algorithm for constructing a parabola if it is given in the form

1) determine the direction of the branches (a>0 – up, a<0 – вниз)

2) we find the coordinates of the vertex of the parabola using the formula , .

3) we find the point of intersection of the parabola with the axis (oy) using the free term, construct a point symmetrical to this point relative to the symmetry axis of the parabola (it should be noted that it happens that it is unprofitable to mark this point, for example, because the value is large... we skip this point...)

4) At the found point - the vertex of the parabola (as at the point (0;0) of the new coordinate system) we construct a parabola. If title="Rendered by QuickLaTeX.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 points of intersection of the parabola with the axis (oy) (if they have not yet “surfaced”) by solving the equation

Example 1

Example 2

Note 1. If the parabola is initially given to us in the form , where are some numbers (for example, ), then it will be even easier to construct it, because we have already been given the coordinates of the vertex . Why?

Let's take a quadratic trinomial and isolate the complete square in it: Look, we got that , . You and I previously called the vertex of a parabola, that is, now, .

For example, . We mark the vertex of the parabola on the plane, we understand that the branches are directed downward, the parabola is expanded (relative to ). That is, we carry out points 1; 3; 4; 5 from the algorithm for constructing a parabola (see above).

Note 2. If the parabola is given in a form similar to this (that is, presented as a product of two linear factors), then we can immediately see the points of intersection of the parabola with the axis (ox). In this case – (0;0) and (4;0). For the rest, we act according to the algorithm, opening the brackets.