How to find the vertex of a parabola of a function graph. How to find the vertex of a parabola: three formulas

11.10.2019

Everyone knows what a parabola is. But how to use it correctly, competently in solving various practical problems, we will understand below.

First, let us denote the basic concepts that algebra and geometry give to this term. Consider everything possible types this chart.

We learn all the main characteristics of this function. Let's understand the basics of constructing a curve (geometry). Let's learn how to find the top, other basic values of the graph of this type.

We will find out: how the required curve is correctly constructed according to the equation, what you need to pay attention to. Let's see the main practical use this unique value in human life.

What is a parabola and what does it look like

Algebra: This term refers to the graph of a quadratic function.

Geometry: This is a second-order curve that has a number of specific features:

Canonical parabola equation

The figure shows a rectangular coordinate system (XOY), an extremum, the direction of the function drawing branches along the abscissa axis.

The canonical equation is:

y 2 \u003d 2 * p * x,

where the coefficient p is the focal parameter of the parabola (AF).

In algebra, it is written differently:

y = a x 2 + b x + c (recognizable pattern: y = x 2).

Properties and Graph of a Quadratic Function

The function has an axis of symmetry and a center (extremum). The domain of definition is all values of the x-axis.

The range of values of the function - (-∞, M) or (M, +∞) depends on the direction of the curve branches. The parameter M here means the value of the function at the top of the line.

How to determine where the branches of a parabola are directed

To find the direction of this type of curve from an expression, you need to specify the sign in front of the first parameter of the algebraic expression. If a ˃ 0, then they are directed upwards. Otherwise, down.

How to find the vertex of a parabola using the formula

Finding the extremum is the main step in solving many practical problems. Of course, you can open special online calculators but it's better to be able to do it yourself.

How to define it? There is a special formula. When b is not equal to 0, we must look for the coordinates of this point.

Formulas for finding the top:

x 0 \u003d -b / (2 * a);
y 0 = y (x 0).

Example.

There is a function y \u003d 4 * x 2 + 16 * x - 25. Let's find the vertices of this function.

For such a line:

x \u003d -16 / (2 * 4) \u003d -2;
y = 4 * 4 - 16 * 2 - 25 = 16 - 32 - 25 = -41.

We get the coordinates of the vertex (-2, -41).

Parabola offset

The classic case is when in a quadratic function y = a x 2 + b x + c, the second and third parameters are 0, and = 1 - the vertex is at the point (0; 0).

Movement along the abscissa or ordinate axes is due to a change in the parameters b and c, respectively. The shift of the line on the plane will be carried out exactly by the number of units, which is equal to the value of the parameter.

Example.

We have: b = 2, c = 3.

This means that the classic view of the curve will shift by 2 unit segments along the abscissa axis and by 3 along the ordinate axis.

How to build a parabola using a quadratic equation

It is important for schoolchildren to learn how to correctly draw a parabola according to the given parameters.

By analyzing expressions and equations, you can see the following:

The point of intersection of the desired line with the ordinate vector will have a value equal to c.
All points of the graph (along the x-axis) will be symmetrical with respect to the main extremum of the function.

In addition, the intersections with OX can be found by knowing the discriminant (D) of such a function:

D \u003d (b 2 - 4 * a * c).

To do this, you need to equate the expression to zero.

The presence of parabola roots depends on the result:

D ˃ 0, then x 1, 2 = (-b ± D 0.5) / (2 * a);
D \u003d 0, then x 1, 2 \u003d -b / (2 * a);
D ˂ 0, then there are no points of intersection with the vector OX.

We get the algorithm for constructing a parabola:

determine the direction of the branches;
find the coordinates of the vertex;
find the intersection with the y-axis;
find the intersection with the x-axis.

Example 1

Given a function y \u003d x 2 - 5 * x + 4. It is necessary to build a parabola. We act according to the algorithm:

a \u003d 1, therefore, the branches are directed upwards;
extremum coordinates: x = - (-5) / 2 = 5/2; y = (5/2) 2 - 5 * (5/2) + 4 = -15/4;
intersects with the y-axis at the value y = 4;
find the discriminant: D = 25 - 16 = 9;
looking for roots

X 1 \u003d (5 + 3) / 2 \u003d 4; (4, 0);
X 2 \u003d (5 - 3) / 2 \u003d 1; (ten).

Example 2

For the function y \u003d 3 * x 2 - 2 * x - 1, you need to build a parabola. We act according to the above algorithm:

a \u003d 3, therefore, the branches are directed upwards;
extremum coordinates: x = - (-2) / 2 * 3 = 1/3; y = 3 * (1/3) 2 - 2 * (1/3) - 1 = -4/3;
with the y-axis will intersect at the value y \u003d -1;
find the discriminant: D \u003d 4 + 12 \u003d 16. So the roots:

X 1 \u003d (2 + 4) / 6 \u003d 1; (1;0);
X 2 \u003d (2 - 4) / 6 \u003d -1/3; (-1/3; 0).

From the obtained points, you can build a parabola.

Directrix, eccentricity, focus of a parabola

Based canonical equation, the focus F has coordinates (p/2, 0).

Straight line AB is a directrix (a kind of parabola chord of a certain length). Her equation is x = -p/2.

Eccentricity (constant) = 1.

Conclusion

We considered the topic that students study in high school. Now you know, looking at the quadratic function of a parabola, how to find its vertex, in which direction the branches will be directed, whether there is an offset along the axes, and, having a construction algorithm, you can draw its graph.

Function of the form , where is called quadratic function.

Graph of quadratic function − parabola.

Consider the cases:

CASE I, CLASSICAL PARABOLA

That is , ,

To build, fill in the table by substituting x values into the formula:

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

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

II CASE, "a" DIFFERENT FROM ONE

What will happen if we take , , ? How will the behavior of the parabola change? With title="(!LANG: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):!}

The first picture (see above) clearly shows 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 argue similarly in the cases of pictures 2 and 3.

And when the parabola "becomes wider" parabola:

Let's recap:

1)The sign of the coefficient is responsible for the direction of the branches. With title="(!LANG: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”, “compression” of the parabola. The larger , the narrower the parabola, the smaller |a|, the wider the parabola.

CASE III, "C" APPEARS

Now let's put into play (that is, we consider the case when ), we will consider parabolas of the form . It is easy to guess (you can always refer to the table) that the parabola will move up or down along the axis, depending on the sign:

IV CASE, "b" APPEARS

When will the parabola “tear off” from the axis and will finally “walk” along the entire coordinate plane? When it ceases to be equal.

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

So at this point (as at the point (0; 0) new system coordinates) we will build a parabola, which is already within our power. If we are dealing with the case, then from the top we set aside one single 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 top we set aside one single segment to the right, two - up, 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 template, because in our case.

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

1) parabola must pass through the point . Indeed, substituting x=0 into the formula, we get that . That is, the ordinate of the point of intersection of the parabola with the axis (oy), this is. In our example (above), the parabola intersects the y-axis at , 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 a parabola symmetrical about the axis of symmetry, 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 (ox). To do this, we solve the equation. Depending on the discriminant, we will get one (, ), two ( title="(!LANG:Rendered by QuickLaTeX.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 an integer, when building it, it makes little sense for us to find the roots, but we can clearly see that we will have two points of intersection with the (oh) axis (since title = "(!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 constructing a parabola if it is given in the form

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

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

3) we find the point of intersection of the parabola with the axis (oy) by the free term, we build a point symmetrical to the given one with respect to the axis of symmetry 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 top of the parabola (as at the point (0; 0) of the new coordinate system), we build a parabola. If title="(!LANG: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 themselves have not yet “surfaced”), solving the equation

Example 1

Example 2

Remark 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 build it, because we have already been given the coordinates of the vertex . Why?

Let's take a square trinomial and select a full square in it: Look, here we got that , . We previously called the top of the parabola, that is, now,.

For example, . We mark the top of the parabola on the plane, we understand that the branches are directed downwards, the parabola is expanded (relatively). That is, we perform steps 1; 3; four; 5 from the algorithm for constructing a parabola (see above).

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

A parabola is the graph of a quadratic function. This line has a significant physical value. In order to make it easier to find the top of the parabola, you need to draw it. Then you can easily see its top on the chart. But to build a parabola, you need to know how to find the points of the parabola and how to find the coordinates of the parabola.

Finding the Points and Vertex of a Parabola

AT general idea the quadratic function has the following form: y = ax 2 + bx + c. schedule given equation is a parabola. When the value a > 0, its branches are directed upwards, and when the value a ‹ 0 - downwards. To build a parabola on a graph, you need to know three points if it runs along the y-axis. Otherwise, four construction points must be known.

When finding the abscissa (x), it is necessary to take the coefficient at (x) from the given polynomial formula, and then divide by twice the coefficient at (x 2), and then multiply by the number - 1.

In order to find the ordinate, you need to find the discriminant, then multiply it by - 1, and then divide by the coefficient at (x 2), after multiplying it by 4.

Further, substituting numerical values, the vertex of the parabola is calculated. For all calculations, it is advisable to use an engineering calculator, and when drawing graphs and parabolas, use a ruler and a lumograph, this will significantly increase the accuracy of your calculations.

Consider the following example to help us understand how to find the vertex of a parabola.

x 2 -9=0. In this case, the vertex coordinates are calculated as follows: point 1 (-0/(2*1); point 2 -(0^2-4*1*(-9))/(4*1)). Thus, the coordinates of the vertex are the values (0; 9).

Finding the abscissa of the vertex

Once you know how to find a parabola and can calculate its intersection points with the x-axis, you can easily calculate the abscissa of the vertex.

Let (x 1) and (x 2) be the roots of the parabola. The roots of a parabola are the points of its intersection with the x-axis. These values nullify the following quadratic equation: ax 2 + bx + c.

Moreover, |x 2 | > |x 1 |, then the vertex of the parabola is located in the middle between them. Thus, it can be found by the following expression: x 0 \u003d ½ (|x 2 | - |x 1 |).

Finding the area of a figure

To find the area of a figure on the coordinate plane, you need to know the integral. And to apply it, it is enough to know certain algorithms. In order to find the area bounded by parabolas, it is necessary to produce its image in the Cartesian coordinate system.

First, according to the method described above, the coordinate of the top of the axis (x) is determined, then the axis (y), after which the top of the parabola is found. Now it is necessary to determine the limits of integration. As a rule, they are indicated in the problem statement using variables (a) and (b). These values should be placed in the upper and lower parts of the integral, respectively. Next, enter in general view function value and multiply it by (dx). In the case of a parabola: (x 2)dx.

Then you need to calculate in general terms the antiderivative value of the function. To do this, use a special table of values. Substituting the limits of integration there, the difference is found. This difference will be the area.

As an example, consider the system of equations: y \u003d x 2 +1 and x + y \u003d 3.

The abscissas of the intersection points are found: x 1 \u003d -2 and x 2 \u003d 1.

We believe that y 2 \u003d 3, and y 1 \u003d x 2 + 1, we substitute the values \u200b\u200bin the above formula and get a value equal to 4.5.

Now we have learned how to find a parabola, and also, based on this data, calculate the area of \u200b\u200bthe figure that it limits.

There is a whole cycle of identities in mathematics, among which quadratic equations occupy a significant place. Similar equalities can be solved both separately and for plotting graphs on the coordinate axis. equations are the intersection points of the parabola and the line ox.

General form

In general, it has the following structure:

In the role of "x" can be considered both individual variables and entire expressions. For example:

(x+7) 2 +3(x+7)+2=0.

In the case when an expression acts as x, it is necessary to represent it as a variable and find After that, equate the polynomial to them and find x.

So, if (x + 7) \u003d a, then the equation takes the form a 2 + 3a + 2 \u003d 0.

D=3 2 -4*1*2=1;

and 1 \u003d (-3-1) / 2 * 1 \u003d -2;

and 2 \u003d (-3 + 1) / 2 * 1 \u003d -1.

With roots equal to -2 and -1, we get the following:

x+7=-2 and x+7=-1;

Roots are the x-coordinate value of the point of intersection of the parabola with the x-axis. In principle, their value is not so important if the task is only to find the top of the parabola. But for plotting, the roots play an important role.

Let's go back to the original equation. To answer the question of how to find the vertex of a parabola, you need to know the following formula:

where x vp is the value of the x-coordinate of the desired point.

But how do you find the vertex of a parabola without a y-coordinate value? We substitute the obtained value of x into the equation and find the desired variable. For example, let's solve the following equation:

Find the value of the x-coordinate for the top of the parabola:

x VP \u003d -b / 2a \u003d -3 / 2 * 1;

Find the value of the y-coordinate for the top of the parabola:

y \u003d 2x 2 + 4x-3 \u003d (-1.5) 2 + 3 * (-1.5) -5;

As a result, we get that the top of the parabola is at the point with coordinates (-1.5; -7.25).

A parabola is a connection of points that has a vertical line. For this reason, its very construction is not difficult. The most difficult thing is to make the correct calculations of the coordinates of the points.

Worth paying Special attention on the coefficients of the quadratic equation.

The coefficient a affects the direction of the parabola. In the event that he has negative meaning, the branches will be directed downwards, and with a positive sign - upwards.

The coefficient b shows how wide the arm of the parabola will be. The larger its value, the wider it will be.

The coefficient c indicates the displacement of the parabola along the y-axis relative to the origin.

We have already learned how to find the vertex of a parabola, and to find the roots, we should be guided by the following formulas:

where D is the discriminant that is needed to find the roots of the equation.

x 1 \u003d (-b + V - D) / 2a

x 2 \u003d (-b-V - D) / 2a

The resulting x values will correspond to zero y values, because they are points of intersection with the x-axis.

After that, we mark the obtained values \u200b\u200bin the top of the parabola. For more detailed graphics need to find a few more points. To do this, we select any value of x that is allowed by the domain of definition, and substitute it into the equation of the function. The result of the calculations will be the coordinate of the point along the y-axis.

To simplify the plotting process, you can draw a vertical line through the top of the parabola and perpendicular to the x-axis. This will be with the help of which, having one point, you can designate a second one, equidistant from the drawn line.

Many technical, economic and social issues predicted using curves. The most used type among them is the parabola, or rather, half of it. An important component of any parabolic curve is its vertex, the determination of the exact coordinates of which sometimes plays a key role not only in the display of the process itself, but also for subsequent conclusions. How to find its exact coordinates will be discussed in this article.

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Before moving on to finding the coordinates of the parabola vertex, let's get acquainted with the definition itself and its properties. In the classical sense, a parabola is such an arrangement of points that distant at the same distance from a particular point(focus, point F), as well as from a straight line that does not pass through point F. Consider this definition more detailed in Figure 1.

Figure 1. Classic view of a parabola

The figure shows the classic form. The focus is point F. In this case, the directrix will be considered the straight line of the Y axis (highlighted in red). From the definition, one can make sure that absolutely any point of the curve, not counting the focus, has a similar one on the other side, removed at the same distance from the axis of symmetry as itself. Moreover, the distance from any of the points on the parabola equal to the distance to the directrix. Looking ahead, let's say that the center of the function does not have to be at the origin, and the branches can be directed in different directions.

A parabola, like any other function, has its own notation in the form of a formula:

In this formula, the letter "s" denotes the parabola parameter, which is equal to the distance from the focus to the directrix. There is also another form of recording, indicated by GMT, which has the form:

Such a formula is used in solving problems from the field of mathematical analysis and is used more often than the traditional one (due to convenience). In the future, we will focus on the second record.

It is interesting!: proof

Calculation of the coefficients and main points of the parabola

Among the main parameters, it is customary to include the location of the vertex on the abscissa axis, the coordinates of the vertex on the ordinate axis, and the directrix parameter.

Numerical value of the vertex coordinate on the x-axis

If the parabola equation is given in the classical form (1), then the value of the abscissa at the desired point will be equal to half the value of the parameter s(half the distance between the directrix and the focus). If the function is presented in the form (2), then zero x is calculated by the formula:

That is, looking at this formula, it can be argued that the vertex will be in the right half relative to the y axis if one of the parameters a or b is less than zero.

The directrix equation is given by the following equation:

Vertex value on the y-axis

The numerical value of the location of the vertex for formula (2) on the y-axis can be found using the following formula:

From this we can conclude that if a<0, то the vertex of the curve will be in the upper half-plane, otherwise, at the bottom. In this case, the points of the parabola will have the same properties that were mentioned earlier.

If the classical notation is given, then it would be more rational to calculate the value of the location of the vertex on the abscissa axis, and through it the subsequent value of the ordinate. Note that for the notation (2), the symmetry axis of the parabola, in the classical representation, will coincide with the y-axis.

Important! When solving tasks using the parabola equation, first of all, highlight the main values \u200b\u200bthat are already known. Moreover, it would be useful if the missing parameters are determined. This approach will give more "room for maneuver" in advance and a more rational solution. In practice, try to use notation (2). It is easier to understand (you don't have to "flip the coordinates of Descartes"), moreover, the vast majority of tasks are adapted specifically for this form of notation.

Construction of a parabolic type curve

Using a common notation, before constructing a parabola, it is required to find its vertex. Simply put, you need to perform the following algorithm:

Find the coordinate of a vertex on the x-axis.
Find the coordinate of the vertex location on the Y axis.
Substituting different values of the dependent variable X, find the corresponding Y values and plot the curve.

Those. the algorithm is nothing complicated, the main focus is on how to find the vertex of the parabola. The further process of construction can be considered mechanical.

Provided that three points are given, the coordinates of which are known, first of all it is necessary to formulate the equation of the parabola itself, and then repeat the procedure that was described earlier. Because in equation (2) there are 3 coefficients, then, using the coordinates of the points, we calculate each of them:

(5.1).

(5.2).

(5.3).

In formulas (5.1), (5.2), (5.3), respectively, those points that are known are used (for example, A (, B (, C (. In this way we find the equation of a parabola at 3 points. From a practical point of view, this approach is not the most “ pleasant”, but it gives a clear result, on the basis of which the curve itself is subsequently built.

When constructing a parabola, always there must be an axis of symmetry. The symmetry axis formula for writing (2) will look like this:

Those. it is not difficult to find the axis of symmetry to which all points of the curve are symmetrical. More precisely, it is equal to the first coordinate of the vertex.

illustrative examples

Example 1. Let's say we have a parabola equation:

It is required to find the coordinates of the vertex of the parabola, and also to check whether the point D (10; 5) belongs to the given curve.

Solution: First of all, we check whether the mentioned point belongs to the curve itself

From where we conclude that the specified point does not belong to the given curve. Find the coordinates of the vertex of the parabola. From formulas (4) and (5) we obtain the following sequence:

It turns out that the coordinates at the top, at point O, are the following (-1.25; -7.625). This means that our the parabola originates in the 3rd quadrant of the Cartesian system coordinates.

Example 2. Find the vertex of the parabola, knowing the three points that belong to it: A (2;3), B (3;5), C (6;2). Using formulas (5.1), (5.2), (5.3), we find the coefficients of the parabola equation. We get the following:

Using the obtained values, we obtain the following equation:

In the figure, the given function will look like this (Figure 2):

Figure 2. Plot of a parabola passing through 3 points

Those. a parabola graph that goes through three given points will have a vertex at the 1st quadrant. However, the branches of this curve are directed downwards; there is an offset of the parabola from the origin. Such a construction could have been foreseen by paying attention to the coefficients a, b, c.

In particular, if a<0, то ветки» будут направлены вниз. При a>1 the curve will be stretched, and if less than 1 it will be compressed.

The constant c is responsible for the "movement" of the curve along the y-axis. If c>0, then the parabola "creeps" up, otherwise down. Regarding the coefficient b, it is possible to determine the degree of influence only by changing the form of the equation, bringing it to the following form:

If the coefficient b>0, then the coordinates of the parabola vertex will be shifted to the right by b units, if less, then by b units to the left.

Important! The use of techniques for determining the displacement of a parabola on the coordinate plane sometimes helps to save time when solving problems or to learn about the possible intersection of a parabola with another curve even before construction. Usually they look only at the coefficient a, since it is he who gives a clear answer to the question posed.

Useful video: how to find the vertex of a parabola

Useful video: how to easily write a parabola equation from a graph