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

The parabola is present in the world of mathematics, physics and other sciences. Parabolas move along the trajectory artificial satellites who seek to go beyond solar system, the ball when playing volleyball also describes its trajectory. You need to be able to build a parabola. And to make it easy, you need to know how to find the top of the parabola.

The graph of the function y \u003d ax 2 + bx + c, where a is the first coefficient, b is the second coefficient, c is the free term, is called a parabola. But note the fact that a ≠ 0.

Each point of the parabola has symmetrical to it, except for one point, and this point is called a vertex. In order to find a point that is a vertex, you need to decide what a point on the chart is. A point on a graph is a specific coordinate along the x-axis and along the y-axis. It is denoted as (x; y). Let's figure out how to find the treasured numbers.

First way

If you want to know how to correctly calculate the coordinates of a vertex, then you only need to learn the formula x0 = -b/2a. Substituting the resulting number into the function, we get y0.

For example, y \u003d x 2 -8 x +15;

find the first, second coefficients and free term;

• a=1, b=-8, c=15;

substitute the values ​​of a and b into the formula;

• x0=8/2=4;

calculate y values;

• y0 = 16–32+15 = -1;

So the vertex is at the point (4;-1).

The branches of a parabola are symmetrical about the axis of symmetry, which goes through the top of the parabola. Knowing the roots of the equation, you can easily calculate the abscissa of the vertex of the parabola without much difficulty. Suppose that k and n are the roots of a quadratic equation. Then the point x0 is equidistant from the points k and n, and it can be calculated by the formula: x0 = (k + n)/2.

Consider the example y \u003d x 2 -6x + 5

1) Equate to zero:

• x2 -6x+5=0.

2) Find the discriminant using the formula: D = b 2 -4 ac:

• D \u003d 36–20 \u003d 16.

3) Find the roots of the equation using the formula (-b±√ D)/2a:

• 1 - the first root;
• 5 is the second root.

4) We calculate:

• x0 =(5+1)/2=3

Second way

Complementing to a full square is a great way to find out where a vertex is. Using this method, you can calculate the points x and y at the same time, without having to substitute x into initial example. Consider this method using the function as an example: y=x 2 +8 x +10.

1. First you need to equate the expression with a variable to 0. Then move c to the right side with opposite sign, that is, we get the expression x 2 + 8x = -10.

2. Now on the left side you need to make a full square. To do this, calculate (b/2) 2 and increase both sides of the equation result. In this case, you need to substitute 8 for b.

We get 16. Now add this number to both sides of the equation:

x 2 + 8x +16= 6.

3. It can be seen that the resulting expression is a full square. It can be represented in the form: (x + 4) 2 = 6.

4. Use this expression to find the coordinates of the vertex of the parabola. To calculate x, you need to equate it to 0. We get x = -4. The y-coordinate is equal to what is on the right side, that is, y = 6. The vertex of the parabola of this equation is (-4, 6).

Third way

If you know what a derivative is, then there is another formula for you. Regardless of where the "horns" of the parabola look, its top is the extremum point. For this method, the following algorithm must be applied:

1. Finding the first derivative by the formula f "(x) \u003d (ax² + bx + c) '\u003d 2ax + b.

2. Equating the derivative to 0. As a result, you will get 0 = 2ax + b, from here you can find what interests us.

Let's consider this method in more detail.

Given a function y = 4x²+16x-17;

• We write the derivative and equate to zero.

f "(x) \u003d (4x² + 16x-17) ' \u003d 8x + 16 \u003d 0

The most difficult thing in the construction is to correctly find the points of the function. For a detailed construction, you need to calculate 5-7 points (this is enough for a school course). To do this, select any value x and substitute it into this function. The result of the calculations will be the number of points along the y-axis. After that, we put the points we obtained on the coordinate plane. As a result, we get a parabola.

Let us consider in more detail the question of finding the points to be marked. For example, let's take the function y \u003d -x 2 +11 x -24 with a vertex at the point (5.5; -6.25).

1) Building a table

Find the right odds.

Write intermediate calculations on paper. This will not only make it easier to find the top, but also help you find your mistakes.

Do everything step by step. Follow the algorithm.

Please note that:

• You need to check if your solution is correct.
• You need to calm down. Solving any problems in mathematics requires experience. You just need to work out this topic, and then you will certainly succeed.

Video

This video will help you learn how to find the vertex of a parabola

Didn't get an answer to your question? Suggest a topic to the authors.

Nagaeva Svetlana Nikolaevna, teacher of mathematics at MAOU "Lyceum No. 1" in the city of Berezniki.

Project algebra lesson in grade 9(humanitarian profile).

“The deepest imprint leaves what a person has discovered himself.” (D. Poya.)

Lesson topic:"Derivation of formulas for calculating the coordinates of the vertex of a parabola".

Lesson Objectives: cognitive :

Expected Result:

- awareness, acceptance and resolution of the problem by students;

Formation of ways to obtain new knowledge through comparison and comparison of facts, a way from the particular to the general;

They learn the formulas for finding the coordinates of the vertex and the axis of symmetry of the parabola for functions of the form y = ax 2 +bx+c.

Lesson type: staging lesson learning task. Teaching methods- visual-illustrative, verbal, learning in cooperation, problematic, elements of critical thinking technology.

Equipment: computer, multimedia projector, demonstration screen, presentation slides on the topic: "Formulas for finding the coordinates of the vertex of a parabola"; sheets of A3 format; colored markers.

Technology- system-activity approach.

Lesson steps:

Psychological attitude (motivation).

Update basic knowledge(creating a situation of success).

Formulation of the problem.

Formulation of the topic and purpose of the lesson.

Solution to the problem.

Analysis of the course of solving the problem.

Application of the results of solving the problem in subsequent activities.

Summing up the lesson (the result of the “eyes” of the student, the result of the “eyes” of the teacher.).

Homework.

During the classes:

Psychological mood.

Task: Learns to solve a common problem and work in a team (work in groups of 5 people).

Guys, throughout last four In the lessons we have been studying the quadratic function, but our knowledge is not yet completely complete, so we continue to study the quadratic function in order to learn something new about this function.

Motivating students to independently set the topic and purpose of the lesson.

The table is filled in during the lesson.

Updating the basic knowledge and skills of students.Warm up. 1. Bracket the senior coefficient: 5x 2 + 25x -5; ax2 + bx + c. 2. Select the double product: ab; ax; b/a. 3.Square: b/2; c2/a; 2a/3b. 4. Present in the form of an algebraic sum: a - c; x –(-b/2a).

Explain how, knowing the form of the function graphy =ƒ( x ) , build graphs of functions:

A ) y =ƒ(x - a) , - with the help of parallel translation by a units to the right along the axis X;

b) y =ƒ(x) + b, - by means of a parallel translation b units up along the axis y;

V) y =ƒ(x- a) +b, ↔ on A units, ↕ on b units;

d) How to graph a function y = (x - 2) 2 + 3 ? What is her schedule?

Name the vertex of the parabola.
The graph is a parabola y = x 2 with vertex at (2; 3 ).

What are the coordinates of the vertex of the parabola: y=x -4x+5( problem). Why is it impossible to determine the coordinates of the vertex of the parabola by the form of the function?(a quadratic function has a different form).

Student activities:

Build speech constructions using functional terminology.

Discussion of answers. They compare, compare with previously studied functions, select and write down on the board the knowledge and skills that they may need to solve the problem in the “KNOW” column:

2.

3.

4.

In the column "I want to know": the top, the axis of symmetry of the parabola
.

Students can write in the “KNOW” and “WANT TO KNOW” columns functions as in general view, as well as special cases. Statement of the educational problem: find the coordinates of the vertex of the parabola, if the quadratic function is given in a general form y = ax + bx + c. Students formulate and write down the topic and purpose of the lesson in a notebook.(Derivation of formulas for calculating the coordinates of the parabola vertex. Learn to find the coordinates of the parabola vertex in a new way - using formulas).

Solution to the problem.

Student activities: Comparing the "old" knowledge with the new knowledge, students propose to highlight the full square. On concrete examples
;
and receive accordingly
;
. Find the coordinates of the vertex and the equation of the axis of symmetry. led new feature to a familiar sight.

Students select a full square for the function
; , compare the result obtained, draw a conclusion on this function. Find the coordinates of the vertex and the axis of symmetry.

Can you name the vertex and axis of the parabola if the function is given in a general way
, without highlighting the full square? How will you proceed in this case? And how to apply your previous experience in finding the vertex and axis of the parabola?

Student activities:

Based on the already existing knowledge, experience, students begin to understand that they need to go further, from the particular to the general, and conduct evidence in a general form.

New difficulties are emerging. Solution appears in groups: . Analysis of the course of solving the problem. One representative from each group is heard.

Compare and analyze records
And
, one general solution of the task is written in the notebook - formulas for the coordinates of the parabola vertex
.

Students conclude: the coordinates of the vertex and the axis of the parabola for the function
can be found in a rational way.

Application of the results of solving the problem in subsequent activities.

Student activities:

Solving tasks from textbook No. 121; 123. Find the coordinates of the vertex of the parabola in a new rational way. Write down the equation of the straight line, which is the axis of symmetry of the parabola.

Summing up (reflection learning activities at the lesson).

Let's go back to the table and fill in the "LEARNED" column.

The result of the lesson "through the eyes" of students:

The result "through the eyes of a teacher":

The goal of the lesson has been achieved.

The students recognized, accepted and resolved the problem.

In the process of solving a problem-solving task, students not only acquired new knowledge: the dependence of the coefficients of a square trinomial and the coordinates of the parabola vertex, the equation of the axis of symmetry, but the most important thing in the lesson is the formation of generalized ways of acquiring new knowledge, independent analysis of the problem and finding the unknown.

Homework: item 7 No. 122; 127 (b) ; 128.

P.S. The presented lesson was held on October 15, 2014 as part of the city seminar for teachers of mathematics on the topic "Formation of UUD in mathematics lessons."

At the stage "Application of the results ..." when solving tasks from the textbook, some students began to understand the value of their "discovery": more easy way finding the coordinates of the vertex and the equation of the axis of symmetry, while others did not hide their joy, because there is no need to "torment" with the selection of a full square. But most importantly, we did everything ourselves!

A parabola is the graph of a quadratic function. This line has significant physical significance. 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

In general representation, the quadratic function has the following form: y = ax 2 + bx + c. The graph of this 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 ​​are set to zero. quadratic equation of the following form: 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 in following expression: x 0 = 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 the general value of the function 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.

The graph of a quadratic function is called a parabola. This line has significant physical significance. Some celestial bodies move along parabolas. An antenna in the shape of a parabola focuses beams running parallel to the axis of symmetry of the parabola. Bodies thrown up at an angle reach the top point and fall down, also describing a parabola. Apparently, it is invariably useful to know the coordinates of the vertex of this movement.

Instruction

1. A quadratic function in general form is written by the equation: y = ax? + bx + c. The graph of this equation is a parabola, the branches of which are directed upwards (for a > 0) or downwards (for a< 0). Школьникам предлагается легко запомнить формулу вычисления координат вершины параболы. Вершина параболы лежит в точке x0 = -b/2a. Подставив это значение в квадратное уравнение, получите y0: y0 = a(-b/2a)? – b?/2a + c = – b?/4a + c.

2. It's easy for people who are friends with the representation of the derivative to find the top of the parabola. Regardless of the location of the branches of the parabola, its vertex is an extremum point (minimum if the branches are directed upwards, or maximum when the branches are directed downwards). In order to find the points of the supposed extremum of any function, it is necessary to calculate its first derivative and equate it to zero. In general, the derivative of a quadratic function is f "(x) \u003d (ax? + bx + c) '= 2ax + b. Equating to zero, you get 0 \u003d 2ax0 + b => x0 \u003d -b / 2a.

3. A parabola is a symmetrical line. The axis of symmetry passes through the top of the parabola. Knowing the points of intersection of the parabola with the coordinate axis X, it is easy to find the abscissa of the vertex x0. Let x1 and x2 be the roots of the parabola (the so-called points of intersection of the parabola with the abscissa axis, because these values ​​turn the quadratic equation ax? + bx + c to zero). Moreover, let |x2| > |x1|, then the vertex of the parabola lies in the middle between them and can be found from the following expression: x0 = ?(|x2| – |x1|).

A parabola is a graph of a quadratic function; in general terms, the equation of a parabola is written y=ax^2+bx+c, where a?0. This is a universal curve of the second order, which describes many phenomena in life, for example, the movement of a body being thrown and then a falling body, the shape of a rainbow, and therefore knowledge to detect parabola can be useful in life.

You will need

• is the formula of a quadratic equation;
• - a sheet of paper with a coordinate grid;
• - pencil, eraser;
• - Computer and Excel.

Instruction

1. First of all, find the top of the parabola. To find the abscissa of this point, take the value in front of x, divide it by twice the value in front of x^2 and multiply by -1 (formula x=-b/2a). Find the ordinate by substituting the resulting value into the equation or by the formula y \u003d (b ^ 2-4ac) / 4a. You have received the coordinates of the vertex point of the parabola.

2. The top of the parabola can also be found by another method. Because the vertex is the extremum of the function, then to calculate it, calculate the first derivative and equate it to zero. In general terms, you get the formula f(x)' = (ax? + bx + c)' = 2ax + b. And equating it to zero, you will come to the same formula - x \u003d -b / 2a.

3. Find out if the branches of the parabola point up or down. To do this, look at the exponent before x^2, that is, at a. If a>0, then the branches are directed upwards, if a

4. Construct the axis of symmetry of the parabola, it intersects the vertex of the parabola and is parallel to the y-axis. All points of the parabola will be equidistant from it, therefore it is allowed to build only one part, and then display it symmetrically with respect to the axis of the parabola.

5. Draw a parabola line. To do this, find several points by substituting various meanings x into equations and solving equality. It is convenient to detect the intersection with the axes, for this, substitute x=0 and y=0 into the equation. Raising one side, reflect it symmetrically about the axis.

6. It is allowed to erect parabola using the Excel program. To do this, open the newest document and select two columns in it, x and y \u003d f (x). In the first column, write down the x values ​​on the selected segment, and in the second column, write down the formula, say, =2B3*B3-4B3+1 or =2B3^2-4B3+1. In order not to write this formula every time, “stretch” it to each column by clicking on the small cross in the lower right corner and dragging it down.

7. Having received the table, click the menu "Insert" - "Chart". Select scatter plot, click next. In the window that appears, add a row by clicking the "Add" button. In order to prefer the required cells, click one by one on the buttons circled in red oval below, then select your columns with values. By clicking the "Finish" button, evaluate the result - the finished parabola .

Related videos

When looking for a quadratic function whose graph is a parabola, at one of the points you need to find coordinates peaks parabolas. How to do this analytically, using the equation given for the parabola?

Instruction

1. A quadratic function is a function of the form y=ax^2+bx+c, where a is the highest exponent (it must be strictly non-zero), b is the lowest exponent, c is a free term. This function gives its graph a parabola, the branches of which are directed either up (if a > 0) or down (if a<0). При a=0 квадратичная функция вырождается в линейную функцию.

2. Let's find the coordinate x0 peaks parabolas. It is found by the formula x0=-b/a.

3. y0=y(x0). To find the coordinate y0 peaks parabolas, you need to substitute the discovered value x0 into the function instead of x. Count what y0 is.

4. Coordinates peaks parabolas found. Write them down as coordinates of one point (x0,y0).

5. When constructing a parabola, remember that it is symmetrical about the axis of symmetry of the parabola, passing vertically through the vertex of the parabola, because the quadratic function is even. Consequently, it is enough to build only one branch of the parabola at the points, and complete the other symmetrically.

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For functions (or rather, their graphs), the representation of the largest value, including the local maximum, is used. The representation of the "top" is more likely associated with geometric shapes. The points of maxima of smooth functions (having a derivative) are easy to determine with the help of zeros of the first derivative.

Instruction

1. For points where the function is not differentiable, but constant, the largest value on the interval can look like a tip (for example, y=-|x|). At such points to the graph functions it is possible to draw as many tangents as desired, and the derivative for it does not easily exist. themselves functions of this type are usually given on segments. The points where the derivative functions is zero or does not exist, are called skeptical.

2. It turns out that to find the points of maxima functions y=f(x) one should: - detect skeptical points; - in order to prefer the maximum point, one should detect the sign of the derivative in the vicinity of the skeptical point. If during the passage of the point there is an alternation of the sign from "+" to "-", then there is a maximum.

3. Example. Detect highest values functions(see Fig.1).y=x+3 for x?-1 and y=((x^2)^(1/3)) –x for x>-1.

4. Rhine. y=x+3 for x?-1 and y=((x^2)^(1/3)) –x for x>-1. The function is set on segments intentionally, because in this case the goal is to display everything in one example. It is easy to check that for x=-1 the function remains constant. 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. y'>0 if x

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

Instruction

1. To find the coordinates of a vertex parabolas, use the following formula: x \u003d -b / 2a, where a is the indicator in front of x squared, and b is the indicator in front of x. Plug in your values ​​and calculate its value. After that, substitute the resulting value instead of x in the equation and calculate the ordinate of the vertex. Say, if you are given the equation y \u003d 2x ^ 2-4x + 5, then find the abscissa in the following way: x \u003d - (-4) / 2 * 2 \u003d 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).

2. Ordinate value parabolas it is allowed to detect without advance calculation of the abscissa. To do this, use the formula y \u003d -b ^ 2 / 4ac + c.

3. If you are familiar with derivative representation, discover peak parabolas with the help of derivatives, using a further property of any function: the first derivative of the function, equated to zero, points to the extremum points. Because the top parabolas, regardless of whether its branches are directed up or down, is an extreme point, calculate the derivative for your function. In its general form, it will look like f(x)=2ax+b. Equate it to zero and get the coordinates of the vertex parabolas, corresponding to your function.

4. Try to discover peak 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. Because the parabola is symmetric with respect to the directrix passing through peak, these points will be equidistant from the abscissa of the vertex. In order to detect it, we divide the distance between the points in half: x \u003d (Ix1-x2I) / 2.

5. If any of the exponents is zero (other than a), calculate the coordinates of the vertex parabolas with simplified formulas. Let's say if b=0, that is, the equation has the form y=ax^2+c, then the vertex will lie on the y-axis and its coordinates will be equal to (0;c). If not only the exponent b=0, but also c=0, then the vertex parabolas is located at the origin, the point (0;0).

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Starting from one point, the lines form an angle, where their common point is the vertex. In the section of theoretical algebra, there are often problems when you need to find the coordinates of this peaks, in order to then determine the equation of the straight line passing through the vertex.

Instruction

1. Before starting the process of finding coordinates peaks, decide on the initial data. Assume that the desired vertex belongs to the triangle ABC, in which the coordinates of the other 2 vertices are known, and also numerical values corners equal to "e" and "k" on side AB.

2. Align new system coordinates on one of the sides of the triangle AB in such a way that the preface of the coordinate system coincides with point A, the coordinates of which you know. The second vertex B will lie on the OX axis, and you also know its coordinates. Determine along the OX axis the value of the length of the side AB according to the coordinates and take it equal to "m".

3. Lower the perpendicular from the unfamiliar peaks C to the axis OX and to the side of the triangle AB, respectively. The resulting height is "y" and determines the value of one of the coordinates peaks C along the OY axis. Assume that the height "y" divides side AB into two segments equal to "x" and "m - x".

4. From the fact that you know the meaning of all corners triangles, which means that the values ​​of their tangents are also famous. Accept the tangents for corners adjacent to the side of the triangle AB, equal to tan(e) and tan(k).

5. Enter the equations for 2 lines passing along sides AC and BC respectively: y = tan(e) * x and y = tan(k) * (m – x). Then find the intersection of these lines by applying the transformed line equations: tan(e) = y/x and tan(k) = y/(m – x).

6. Assuming that tan(e)/tan(k) equals (y/x) /(y/ (m - x)) or later than "y" - (m - x) / x , you end up with the desired values coordinates equal to x = m / (tan(e)/tan(k) + e) ​​and y = x * tan(e).

7. Plug in the values corners(e) and (k), as well as the detected side value AB = m in the equations x = m / (tan(e)/tan(k) + e) ​​and y = x * tan(e).

8. Convert the new coordinate system to the initial coordinate system, since there is a one-to-one correspondence between them, and get the desired coordinates peaks triangle ABC.

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There is a whole cycle of identities in mathematics, among which significant place occupy quadratic equations. 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 required 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.

It is worth paying special attention to 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 a more detailed graph, you 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.