A parabola is a graph quadratic function. This line has significant physical significance. To make it easier to find the vertex of the parabola, you need to draw it. Then you can easily see its top on the chart. But to construct 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 the parabola
IN general idea the quadratic function has the following form: y = ax 2 + bx + c. Schedule given equation is a parabola. When the value is a › 0, its branches are directed upward, and when the value is a ‹ 0, they are directed downward. To construct a parabola on a graph, you need to know three points if it runs along the ordinate axis. Otherwise, four construction points must be known.
When finding the abscissa (x), you need to take the coefficient of (x) from the given polynomial formula, and then divide by the double coefficient of (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.
Next, by 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.
Let's consider next example, which will help us understand how to find the vertex of a parabola.
x 2 -9=0. IN in this case 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 coordinate (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 |, which means the vertex of the parabola is located in the middle between them. Thus, it can be found using the following expression: x 0 = ½(|x 2 | - |x 1 |).
Finding the area of the figure
To find the area of a figure on 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 image it in a Cartesian coordinate system.
First, using the method described above, the coordinate of the vertex of the (x) axis is determined, then the axis (y), after which the vertex of the parabola is found. Now we need 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 you should 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 the antiderivative value of the function in general form. To do this, you should 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 = x 2 +1 and x + y = 3.
The abscissas of the intersection points are found: x 1 = -2 and x 2 = 1.
We assume that y 2 = 3, and y 1 = x 2 + 1, substitute the values in 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 the figure that it limits.
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, 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, 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).
Many technical, economic and social issues are predicted using curves. The most used type among them is the parabola, or more precisely, 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.
Start the search
Before moving on to finding the coordinates of the vertex of a parabola, let’s get acquainted with the definition itself and its properties. In the classical sense, a parabola is such an arrangement of points that removed at the same distance from a specific point(focus, point F), as well as from a line that does not pass through point F. Consider this definition more detailed in Figure 1.
Figure 1. Classic view of a parabola
The picture shows the classic form. The focus is point F. The directrix in this case will be considered the straight line of the Y axis (highlighted in red). From the definition, you can make sure that absolutely any point on the curve, not counting the focus, has a similar one on the other side, located 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 director. 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 entry in the form of a formula:
In the indicated formula, the letter “s” denotes the parameter of the parabola, 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:
This formula is used when solving problems in 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 entry.
This is interesting!: proof
Calculation of coefficients and main points of a parabola
The main parameters usually 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 equation of a parabola 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 parameter s(half the distance between the directrix and the focus). If the function is presented in the form (2), then x zero is calculated using the formula:
That is, looking at this formula, we can say 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 defined by the following equation:
Vertex value on the ordinate axis
The numerical value of the location of the vertex for formula (2) on the ordinate 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 as mentioned earlier.
If the classical form of notation is given, then it will 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 form of notation (2), the axis of symmetry of the parabola, in the classical representation, will coincide with the ordinate axis.
Important! When solving problems using the parabola equation, first of all, identify the main values that are already known. Moreover, it will be useful if the missing parameters are determined. This approach will provide more “room for maneuver” in advance and a more rational decision. In practice, try to use notation (2). It is easier to understand (you don’t have to “reverse Descartes’ coordinates”), and the vast majority of tasks are adapted specifically to this form of notation.
Constructing a parabolic curve
Using a common form of notation, before constructing a parabola, you need to find its vertex. Simply put, you need to perform the following algorithm:
- Find the coordinate of the 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 values of Y and construct a curve.
Those. The algorithm is not complicated, the main emphasis is on how to find the vertex of a parabola. The further construction process can be considered mechanical.
Provided that three points are given, the coordinates of which are known, first of all it is necessary to create an equation for 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), those points that are known are used, respectively (for example, A (, B (, C (). In this way we find the equation of a parabola using 3 points. From the practical side, this approach is not the most “ pleasant”, but it gives a clear result, on the basis of which the curve itself is subsequently constructed.
When constructing a parabola, always there must be an axis of symmetry. The formula for the axis of symmetry to write (2) will look like this:
Those. Finding the axis of symmetry to which all points of the curve are symmetrical is not difficult. More precisely, it is equal to the first coordinate of the vertex.
Illustrative examples
Example 1. Let's say we have the equation of a parabola:
You need to find the coordinates of the vertex of the parabola, and also check whether point D (10; 5) belongs to the given curve.
Solution: First of all, let’s check that the mentioned point belongs to the curve itself
From which we conclude that the specified point does not belong to the given curve. Let's 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 as follows (-1.25; -7.625). This suggests that our the parabola originates in the 3rd quarter of the Cartesian system coordinates
Example 2. Find the vertex of a 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 specified function will look like this (Figure 2):
Figure 2. Graph of a parabola passing through 3 points
Those. The graph of a parabola that passes through three given points will have a vertex in the 1st quarter. However, the branches of this curve are directed downwards, i.e. there is a displacement of the parabola from the origin. This construction could have been predicted by paying attention to the coefficients a, b, c.
In particular, if a<0, то ветки» будут направлены вниз. При a>1 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 ordinate axis. If c>0, then the parabola “creeps” upward, otherwise – down. Regarding the coefficient b, the degree of influence can be determined only by changing the form of writing the equation, bringing it to the following form:
If the coefficient b>0, then the coordinates of the vertex of the parabola will be shifted to the right by b units, if less, then by b units to the left.
Important! Using techniques for determining the displacement of a parabola on the coordinate plane sometimes helps to save time when solving problems or to find out about the possible intersection of a parabola with another curve before construction. Usually they look only at coefficient a, since it is this that gives a clear answer to the question posed.
Useful video: how to find the vertex of a parabola
Useful video: how to easily create a parabola equation from a graph
Conclusion
An algebraic process such as determining the vertices of a parabola is not complicated, but it is quite labor-intensive. In practice, they try to use the second form of notation in order to facilitate understanding of the graphical solution and the solution as a whole. Therefore, we strongly recommend using exactly this approach, and if you don’t remember the formula for vertex coordinates, then at least have a cheat sheet.
The graph of a quadratic function is called a parabola. This line has significant physical significance. Some move along parabolas celestial bodies. 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 detect 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 branches of the parabola are directed up or down. 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 using Excel. 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 menu “Insert” – “Chart”. 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 is used highest value, including the local maximum. The idea of “top” 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 for it does not easily exist. 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) 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, 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).
Video on the topic
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 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 we 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.
Video on the topic
Video on the topic
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|).
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 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).
Video on the topic
For functions (more precisely, their graphs), the concept of the largest value, including a local maximum, is used. The concept of “vertex” is rather associated with geometric figures. 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