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How to find the vertex of the graph of a quadratic function. Quadratic function

In mathematics there is a whole cycle of identities, among which quadratic equations occupy a significant place. Such equalities can be solved both separately and to construct graphs on the coordinate axis. equations are the points of intersection of the parabola and the straight line oh.

General view

IN general view has the following structure:

Both individual variables and entire expressions can be considered as “X”. For example:

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

In the case when the role of x is an expression, 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)=a, then the equation takes the form a 2 +3a+2=0.

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

and 1 =(-3-1)/2*1=-2;

and 2 =(-3+1)/2*1=-1.

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

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

The roots are the x-coordinate value of the point where the parabola intersects the x-axis. In principle, their value is not so important if the task is only to find the vertex of the parabola. But for plotting a graph, the roots play an important role.

Let's return to the initial 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 x-coordinate value of the desired point.

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

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

x VP =-b/2a=-3/2*1;

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

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

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

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

Worth paying special attention to the odds quadratic equation.

Coefficient a affects the direction of the parabola. In the case where he has negative value, the branches will be directed downwards, and with a positive sign - upwards.

The coefficient b indicates how wide the parabola arm will be. The higher its value, the wider it will be.

Coefficient c indicates the displacement of the parabola along the OS 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 necessary to find the roots of the equation.

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

x 2 =(-b-V - D)/2a

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

After this, we mark the resulting values ​​at the top of the parabola. For a more detailed graph, you need to find a few more points. To do this, choose any value of x 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 op-amp axis.

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

Content:

The vertex of a parabola is its highest or lowest point. To find the vertex of a parabola, you can use a special formula or the square addition method. Below is how to do this.

Steps

1 Formula for finding the vertex

  1. 1 Find the values ​​of a, b, and c. In a quadratic equation, the coefficient at x 2 = a, at x= b, constant (coefficient without variable) = c. For example, take the equation: y = x 2 + 9x + 18. Here a = 1, b= 9, and c = 18.
  2. 2 Use the formula to calculate the x coordinate value of a vertex. The vertex is also the point of symmetry of the parabola. Formula for finding the x coordinate of a parabola: x = -b/2a. Substitute the appropriate values ​​into it to calculate x.
    • x=-b/2a
    • x=-(9)/(2)(1)
    • x=-9/2
  3. 3 Substitute the found x value into the original equation to calculate the y value. Now that you know the value of x, simply plug it into the original equation to find y. Thus, the formula for finding the vertex of a parabola can be written as a function: (x, y) = [(-b/2a), f(-b/2a)]. This means that to find y, you must first find x using the formula, and then substitute the value of x into the original equation. Here's how it's done:
    • y = x 2 + 9x + 18
    • y = (-9/2) 2 + 9(-9/2) +18
    • y = 81/4 -81/2 + 18
    • y = 81/4 -162/4 + 72/4
    • y = (81 - 162 + 72)/4
    • y = -9/4
  4. 4 Write the x and y values ​​as a pair of coordinates. Now that you know that x = -9/2 and y = -9/4, write them down as coordinates in the form: (-9/2, -9/4). The vertex of the parabola is located at coordinates (-9/2, -9/4). If you need to draw this parabola, then its vertex lies at the bottom point, since the coefficient of x 2 is positive.

2 Complement to a perfect square

  1. 1 Write down the equation. Completing a perfect square is another way to find the vertex of a parabola. By using this method, you will find the x and y coordinates right away, without having to substitute x into the original equation. For example, given the equation: x 2 + 4x + 1 = 0.
  2. 2 Divide each coefficient by the coefficient of x 2 . In our case, the coefficient of x 2 is 1, so we can skip this step. Dividing by 1 won't change anything.
  3. 3 Move the constant to the right side of the equation. Constant is a coefficient without a variable. Here it is "1". Move 1 to the right by subtracting 1 from both sides of the equation. Here's how to do it:
    • x 2 + 4x + 1 = 0
    • x 2 + 4x + 1 -1 = 0 - 1
    • x 2 + 4x = - 1
  4. 4 Complete the left side of the equation to make it a complete square. To do this, just find (b/2) 2 and add the result to both sides of the equation. Substitute "4" for b, since "4x" is the coefficient b of our equation.
    • (4/2) 2 = 2 2 = 4. Now add 4 to both sides of the equation and you get:
      • x 2 + 4x + 4 = -1 + 4
      • x 2 + 4x + 4 = 3
  5. 5 Let's simplify the left side of the equation. We see that x 2 + 4x + 4 – perfect square. It can be written as: (x + 2) 2 = 3
  6. 6 Use it to find the x and y coordinates. You can find x by simply equating (x + 2) 2 to 0. Now that (x + 2) 2 = 0, we calculate x: x = -2. The y coordinate is a constant on the right side of a perfect square. So y = 3. The vertex of the parabola of the equation is x 2 + 4x + 1 = (-2, 3)
  • Identify a, b, and c correctly.
  • Record preliminary calculations. This will not only help during the work process, but will also allow you to see where mistakes were made.
  • Do not disturb the order of calculations.

Warnings

  • Check your answer!
  • Make sure you know how to determine the coefficients a, b, and c. If you don't know, the answer will be wrong.
  • No – solving such problems requires practice.

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 coordinate plane(the smaller the step we take the values ​​of x (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 stop being 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 with respect 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 quadratic trinomial and select a 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 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.

Perhaps everyone knows what a parabola is. But we’ll look at how to use it correctly and competently when solving various practical problems below.

First, let us outline the basic concepts that algebra and geometry give to this term. Let's consider everything possible types this chart.

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

Let's find out: how to correctly construct the desired curve using the equation, what you need to pay attention to. Let's see the basics practical application 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 branches of the function drawing along the abscissa axis.

The canonical equation is:

y 2 = 2 * p * x,

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

In algebra it will be 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 abscissa axis.

The range of values ​​of the function – (-∞, M) or (M, +∞) depends on the direction of the branches of the curve. 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 a curve of this type from an expression, you need to determine the sign before the first parameter algebraic expression. If a ˃ 0, then they are directed upward. If it's the other way around, 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 determine it? There is a special formula. When b is not equal to 0, we need to look for the coordinates of this point.

Formulas for finding the vertex:

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

Example.

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

For a line like this:

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

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

Parabola displacement

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

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

Example.

We have: b = 2, c = 3.

This means that the classic form 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 using given parameters.

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

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

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

D = (b 2 - 4 * a * c).

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

The presence of roots of a parabola depends on the result:

  • D ˃ 0, then x 1, 2 = (-b ± D 0.5) / (2 * a);
  • D = 0, then x 1, 2 = -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 ordinate axis;
  • find the intersection with the x-axis.

Example 1.

Given the function y = x 2 - 5 * x + 4. It is necessary to construct a parabola. We follow the algorithm:

  1. a = 1, therefore, the branches are directed upward;
  2. extremum coordinates: x = - (-5) / 2 = 5/2; y = (5/2) 2 - 5 * (5/2) + 4 = -15/4;
  3. intersects with the ordinate axis at the value y = 4;
  4. let's find the discriminant: D = 25 - 16 = 9;
  5. looking for roots:
  • X 1 = (5 + 3) / 2 = 4; (4, 0);
  • X 2 = (5 - 3) / 2 = 1; (1, 0).

Example 2.

For the function y = 3 * x 2 - 2 * x - 1 you need to construct a parabola. We act according to the given algorithm:

  1. a = 3, therefore, the branches are directed upward;
  2. extremum coordinates: x = - (-2) / 2 * 3 = 1/3; y = 3 * (1/3) 2 - 2 * (1/3) - 1 = -4/3;
  3. will intersect with the y-axis at the value y = -1;
  4. let's find the discriminant: D = 4 + 12 = 16. So the roots are:
  • X 1 = (2 + 4) / 6 = 1; (1;0);
  • X 2 = (2 - 4) / 6 = -1/3; (-1/3; 0).

Using the obtained points, you can construct a parabola.

Directrix, eccentricity, focus of a parabola

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

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

Eccentricity (constant) = 1.

Conclusion

We looked at a topic that schoolchildren 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 a displacement along the axes, and, having a construction algorithm, you can draw its graph.

Instructions

Quadratic function in general form it is written by the equation: y = ax² + bx + c. The graph of this equation is , the branches of which are directed upward (for a > 0) or downward (for a< 0). Школьникам предлагается просто запомнить формулу вычисления координат вершины . Вершина параболы в точке x0 = -b/2a. Подставив это значение в квадратное , получите y0: y0 = a(-b/2a)² - b²/2a + c = - b²/4a + c.

For people familiar with the concept of derivative, it is easy to find the vertex of a parabola. Regardless of the position of the branches of a parabola, its vertex is a point (minimum if the branches are directed upward, or when the branches are directed downward). To find the supposed extremum points of any , you need to calculate its first derivative and equate it to zero. In general, the derivative is equal to f"(x) = (ax² + bx + c)" = 2ax + b. Equating to zero, you get 0 = 2ax0 + b => x0 = -b/2a.

A parabola is a symmetrical line. The axis passes through the vertex of the parabola. Knowing the points of the parabola with the X coordinate axis, you can easily find the abscissa of the vertex x0. Let x1 and x2 be the roots of the parabola (the so-called intersection points of the parabola with the x-axis, since these values ​​make the quadratic equation ax² + bx + c vanish). Moreover, let |x2| > |x1|, then the vertex of the parabola lies halfway between them and can be found from the following expression: x0 = ½(|x2| - |x1|).

Video on the topic

Sources:

  • Quadratic function
  • formula for finding the vertex of a parabola

A parabola is a graph of a quadratic function; in general, the equation of a parabola is written y=ax^2+bx+c, where a≠0. This is a universal second-order curve that describes many phenomena in life, for example, the movement of a tossed and then falling body, the shape of a rainbow, so the ability to find parabola can be very useful in life.

You will need

  • - quadratic equation formula;
  • - a sheet of paper with a coordinate grid;
  • - pencil, eraser;
  • - computer and Excel program.

Instructions

First of all, find the vertex of the parabola. To find the abscissa of this point, take the coefficient of x, divide it by twice the coefficient of x^2 and multiply by -1 ( x = -b/2a). Find the ordinate by substituting the resulting value into the equation or using the formula y=(b^2-4ac)/4a. You have obtained the coordinates of the vertex point of the parabola.

The vertex of a parabola can be found in another way. Since it is the extremum of the function, to calculate it, calculate the first derivative and equate it to zero. In general, you will get the formula f(x)" = (ax? + bx + c)" = 2ax + b. And by equating it to zero, you will come to the same formula - x=-b/2a.

Find out whether the parabola's branches are directed upward or downward. To do this, look at the coefficient in front of x^2, that is, a. If a>0, then the branches are directed upward, if a

Coordinates peaks parabolas have been found. Write them down as the coordinates of a single point (x0,y0).

Video on the topic

For functions (more precisely, their graphs) the concept is used highest value, including the local maximum. The concept of “peak” is more likely associated with geometric shapes. The maximum points of smooth functions (having a derivative) are easy to determine using the zeros of the first derivative.

Instructions

For points at which the function is not differentiable but continuous, the largest value on the interval can have the form of a tip (at y=-|x|). At such points functions You can draw as many tangents as you like; the tangents simply do not exist for it. Sami functions This type is usually specified on segments. Points at which the derivative functions equal to zero or does not exist are called critical.

Rheaning. y=x+3 for x≤-1 and y=((x^2)^(1/3)) –x for x>-1. The function is specified on segments deliberately, since in this case the goal is to display everything in one example. It is easy that for x=-1 the function remains continuous.y'=1 for x≤-1 and y'=(2/3)(x^(-1/3))-1=(2-3(x^ (1/3))/(x^(1/3)) for x>-1. y'=0 for x=8/27. y' does not exist for x=-1 and x=0. In this case, y '>0 if x

Video on the topic

A parabola is one of the second-order curves; its points are constructed in accordance with a quadratic equation. The main thing in constructing this curve is to find top parabolas. This can be done in several ways.

Instructions

To find the coordinates of a vertex parabolas, use the following formula: x=-b/2a, where a is the coefficient before x in, and b is the coefficient before x. Plug in your values ​​and calculate it. Then substitute the resulting value for x into the equation and calculate the ordinate of the vertex. For example, if you are given the equation y=2x^2-4x+5, then find the abscissa as follows: x=-(-4)/2*2=1. Substituting x=1 into the equation, calculate the y-value for the vertex parabolas: y=2*1^2-4*1+5=3. So the top parabolas has coordinates (1;3).

The value of the ordinate parabolas can be found without first calculating the abscissa. To do this, use the formula y=-b^2/4ac+c.

If you are familiar with the concept of derivative, find top parabolas using derivatives, using the following property of any: the first derivative of a function, equal to zero, points to. Since the top parabolas, regardless of whether its branches are directed up or down, point , calculate the derivative for your function. In general, it will look like f(x)=2ax+b. Equate it to zero and get the coordinates of the vertex parabolas, corresponding to your function.

Try to find top parabolas, taking advantage of its property such as symmetry. To do this, find the intersection points parabolas with the x axis, equating the function to zero (substituting y = 0). By solving the quadratic equation, you will find x1 and x2. Since the parabola is symmetrical with respect to the directrix passing through top, these points will be equidistant from the abscissa of the vertex. To find it, we divide