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 of x squared, and b is the coefficient of x. Plug in your values and calculate its value. 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, taking advantage of the following property of any function: the first derivative of a function, equal to zero, indicates extremum points. Since the top parabolas, regardless of whether its branches are directed up or down, is an extremum point, calculate the derivative for your function. IN general view 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). Having decided 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, divide the distance between the points in half: x=(Ix1-x2I)/2.
If any of the coefficients is zero (except a), calculate the coordinates of the vertex parabolas using simplified formulas. For example, if b=0, that is, the equation has the form y=ax^2+c, then the vertex will lie on the oy axis and its coordinates will be equal to (0-c). If not only the coefficient b=0, but also c=0, then the vertex parabolas is located at the origin, point (0-0).
Instructions
A quadratic function in general form 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|).
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Sources:
- Quadratic function
- formula for finding the vertex of a parabola
A parabola is a graph quadratic function, in general, the parabola equation is written y=aх^2+bх+с, 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 point 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).
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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 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
The graph of a quadratic function is called a parabola. This line has significant physical significance. Some celestial bodies move along parabolas. A parabola-shaped antenna focuses rays running parallel to the axis of symmetry of the parabola. Bodies thrown upward at an angle reach the top point and fall down, also describing a parabola. Apparently, it is always useful to know the coordinates of the vertex of this movement.
Instructions
1. The quadratic function in its 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 upward (for a > 0) or downward (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. People familiar with the derivative representation can easily spot the vertex of a parabola. Regardless of the location of the branches of the parabola, its top is the point of extremum (minimum if the branches are directed upward, or maximum when the branches are directed downward). In order to find the supposed extremum points of any function, you need to calculate its first derivative and equate it to zero. In general, the derivative of a quadratic function is equal to f"(x) = (ax? + bx + c)' = 2ax + b. Equating to zero, you get 0 = 2ax0 + b => x0 = -b/2a.
3. A parabola is a symmetrical line. The axis of symmetry passes through the vertex of the parabola. Knowing the points of intersection 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 abscissa axis, because these values turn the quadratic equation ax? + bx + c to zero). At the same time, let |x2| > |x1|, then the vertex of the parabola lies in the middle between them and can be found from the further expression: x0 = ?(|x2| – |x1|).
A parabola is a graph of a quadratic function; in general, the equation of a parabola is written y=aх^2+bх+с, where a?0. This is a universal second-order curve that describes many phenomena in life, say, the movement of a thrown and then falling body, the shape of a rainbow, and therefore the knowledge to detect parabola It might come in handy in real life.
You will need
- – quadratic equation formula;
- – a sheet of paper with a coordinate grid;
- – pencil, eraser;
- – computer and Excel program.
Instructions
1. First, locate the vertex of the parabola. To find the abscissa of this point, take the exponent before x, divide it by twice the exponent before x^2 and multiply by -1 (formula 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.
2. The vertex of a parabola can also be detected using another method. Because the vertex is the extremum of the function, to calculate it, calculate the first derivative and equate it to zero. In general form 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.
3. Find out whether the parabola's branches are directed upward or downward. To do this, look at the indicator in front of x^2, that is, a. If a>0, then the branches are directed upward, 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 possible to construct only one part, and then symmetrically display it relative to the axis of the parabola.
5. Draw a line of a parabola. To do this, find several points by substituting different meanings x into the equations and solving the equality. It is convenient to detect the intersection with the axes; to do this, substitute x=0 and y=0 into the equality. Having raised one side, reflect it symmetrically about the axis.
6. Allowed to build parabola with help Excel programs. To do this, open the new document and select two columns in it, x and y=f(x). In the first column, write down the values of x 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. Once you have the table, click the “Insert” – “Chart” menu. Select the scatter plot, click Next. In the window that appears, add a row by clicking the “Add” button. To select the required cells, click one by one on the buttons circled in red oval below, then select your columns with values. By clicking the “Done” button, evaluate the result – the finished parabola .
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When searching 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?
Instructions
1. A quadratic function is a function of the form y=ax^2+bx+c, where a is the leading exponent (it must strictly be 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 upward (if a>0) or downward (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 formulax0=-b/a.
3. y0=y(x0).To detect the coordinate y0 peaks parabolas, you need to substitute the detected value x0 into the function instead of x. Calculate what y0 is equal to.
4. Coordinates peaks parabolas have been discovered. Write them down as the coordinates of a single point (x0,y0).
5. When constructing a parabola, remember that it is symmetrical about the axis of symmetry of the parabola, which passes vertically through the vertex of the parabola, because the quadratic function is even. Consequently, it is enough to construct only one branch of the parabola from 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 idea of “vertex” 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
1. For points at which the function is not differentiable but constant, the largest value on the interval may have the form of a tip (for example, y=-|x|). At such points to the graph functions it is possible to draw as many tangents as desired, and a derivative does not easily exist for it. Sami functions of this type are usually specified on segments. Points at which the derivative functions equal to zero or does not exist are called skeptical.
2. It turns out that to find the maximum points functions y=f(x) it is necessary: - to detect skeptical points; - in order to prefer the maximum point, it is necessary to detect the sign of the derivative in the vicinity of the skeptical point. If, when passing a point, the sign alternates from “+” to “-”, then a maximum occurs.
3. Example. Find the largest values functions(see Fig. 1).y=x+3 for x?-1 and y=((x^2)^(1/3)) –x for x>-1.
4. Rheaning. y=x+3 for x?-1 and y=((x^2)^(1/3)) –x for x>-1. The function is specified on segments deliberately, because in this case the goal is to display everything in one example. It is easy to check that at x=-1 the function remains constant. y'=1 at 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 raised in accordance with a quadratic equation. The main thing in constructing this oblique is to detect top parabolas. This can be done in several ways.
Instructions
1. To find the coordinates of the vertex parabolas, use the following formula: x = -b/2a, where a is the indicator before x squared, and b is the indicator before x. Plug in your values and calculate its value. After this, substitute the resulting value for x in the equation and calculate the ordinate of the vertex. Let's say, if you are given the equation y=2x^2-4x+5, then find the abscissa in the following way: 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).
2. The value of the ordinate parabolas can be detected without calculating the abscissa in advance. To do this, use the formula y=-b^2/4ac+c.
3. If you are familiar with derivative representation, discover top parabolas using derivatives, taking advantage of the further property of every function: the first derivative of a function, equal to zero, indicates the extremum points. Because the top parabolas, regardless of whether its branches are directed up or down, is an extremum point, calculate the derivative for your function. In 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 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). When you solve a quadratic equation, you will find x1 and x2. Because the parabola is symmetrical about the directrix passing through top, 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 = (Ix1-x2I)/2.
5. If any of the exponents is zero (besides a), calculate the coordinates of the vertex parabolas using 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 oy 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, point (0;0).
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Starting from one point, straight 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 line passing through the vertex.
Instructions
1. Before you begin the process of finding coordinates peaks, decide on the initial data. Accept that the desired vertex belongs to triangle ABC, in which the coordinates of the other 2 vertices, as well as the numerical values, are known corners, equal to “e” and “k” on side AB.
2. Combine new system coordinates on one of the sides of triangle AB in such a way that the preface of the coordinate system coincides with point A, the coordinates of which are known to you. The second vertex B will lie on the OX axis, and its coordinates are also known to you. Determine the length of side AB along the OX axis according to the coordinates and take it equal to “m”.
3. Lower the perpendicular from the unfamiliar peaks C to the OX axis and to the side of the triangle AB, respectively. The resulting height “y” 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. Because you know the meanings of all corners triangle, which means that the values of their tangents are also known. Take the tangent values 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 the 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. If you assume that tan(e)/tan(k) equals (y/x) /(y/ (m – x)) or later abbreviate “y” – (m – x) / x, you will end up with the desired values coordinates equal to x = m / (tan(e)/tan(k) + e) and y = x * tan(e).
7. Substitute values corners(e) and (k), as well as the detected value of the side AB = m into the equations x = m / (tan(e)/tan(k) + e) and y = x * tan(e).
8. Convert the new coordinate system to initial system coordinates, from the fact that a one-to-one correspondence has been established between them, and you will get the desired coordinates peaks triangle ABC.
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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 form
In general it 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 coefficients of a quadratic equation.
Coefficient a affects the direction of the parabola. In the case where he has negative meaning, the branches will be directed downwards, and when positive sign- up.
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 op 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.
Probably 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 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 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:
- 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 relative 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:
- a = 1, therefore, the branches are directed upward;
- extremum coordinates: x = - (-5) / 2 = 5/2; y = (5/2) 2 - 5 * (5/2) + 4 = -15/4;
- intersects with the ordinate axis at the value y = 4;
- let's find the discriminant: D = 25 - 16 = 9;
- looking for roots:
- X 1 = (5 + 3) / 2 = 4; (4, 0);
- X 2 = (5 - 3) / 2 = 1; (10).
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:
- a = 3, therefore, the branches are directed upward;
- extremum coordinates: x = - (-2) / 2 * 3 = 1/3; y = 3 * (1/3) 2 - 2 * (1/3) - 1 = -4/3;
- will intersect with the y-axis at the value y = -1;
- 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 the 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.
