Parabola is a chart of a quadratic function. This line has a significant physical value. In order to make it easier to find the top of the parabola, you need to draw it. Then the graph can be easily seen on the chart. But to build a parabola, you need to know how to find Parabola's points and how to find the coordinates of Parabola.
We find the point and the top of the parabola
In the general presentation, the quadratic function has the following form: Y \u003d AX 2 + BX + C. The graph of this equation is Parabola. With the value of A\u003e 0, its branches are directed upwards, and with the value of \u003c0 - down. To build a parabola on the chart, you need to know three points if it passes along the ordinate axis. Otherwise, four construction points should be known.
When the abscissa (x), it is necessary to take the coefficient at (x) from the given formula of the polynomial, and then divided into a double coefficient at (x 2), after which multiply by number - 1.
In order to find the ordinate, it is necessary to find the discriminant, then multiply it to 1, after which it is divided into the coefficient at (x 2), by first multiplying it to 4.
Further, substituting the numerical values, the top of the parabola is calculated. For all the calculations, it is advisable to use the engineering calculator, and when drawing graphs and parabola use a ruler and a luminary, this will significantly increase the accuracy of your calculations.
Consider the following example, which will help us understand how to find the top of the parabola.
x 2 -9 \u003d 0. In this case, the coordinates of the vertices are calculated as follows: Point 1 (-0 / (2 * 1); point 2 - (0 ^ 2-4 * 1 * (- 9)) / (4 * 1)). Thus, the coordinates of the vertices are values \u200b\u200b(0; 9).
We find the abscissue of the vertices
After you learned how to find a parabola, and you can calculate the points of its intersection with the axis of coordinates (x), you can easily calculate the abscissa of the vertices.
Let (x 1) and (x 2) are the roots of parabola. Parabola roots are points of its intersection with the abscissa axis. These values \u200b\u200brefer to zero square equation of the following form: AX 2 + BX + C.
At the same time | x 2 | \u003e | x 1 |, So the top of the parabola is located in the middle between them. Thus, it can be found at the following expression: x 0 \u003d ½ (| x 2 | - | x 1 |).
Find the area of \u200b\u200bFigure
To find the area of \u200b\u200bthe shape on the coordinate plane you need to know the integral. And in order to apply it, it is enough to know certain algorithms. In order to find the area limited by parabolami, it is necessary to produce its image in the Cartesian coordinate system.
At first, according to the method described above, the coordinate of the top of the axis (x) is determined, then the axis (y) is determined, after which the top of the parabola is located. Now you should define the integration limits. As a rule, they are indicated in the condition of the problem using variables (a) and (b). These values \u200b\u200bshould be placed in the upper and lower parts of the integral, respectively. Next should beo in general The value of the function and multiply it to (DX). In the case of parabola: (x 2) DX.
Then you need to calculate in the general form a primary function value. To do this, use the special table of values. Substituting the integration limits there, there is a difference. This difference will be an area.
As an example, consider the system of equations: y \u003d x 2 +1 and x + y \u003d 3.
There are abscisses of intersection points: x 1 \u003d -2 and x 2 \u003d 1.
We assume that in 2 \u003d 3, and in 1 \u003d x 2 + 1, we substitute the values \u200b\u200binto the above formula and we obtain a value of 4.5.
Now we learned how to find a parabola, as well as based on this data, calculate the area of \u200b\u200bthe figure, which it limits.
The function of the species where is called quadratic function.
Schedule of a quadratic function - parabola.
Consider cases:
I Classical Parabola
I.e , ,
For constructing, fill in the table, substituting the X values \u200b\u200bin the formula:
We note the points (0; 0); (1; 1); (-1; 1), etc. On the coordinate plane (with a smaller step, we take the value of x (in this case, step 1), and the more we take the values \u200b\u200bof x, the smaker will be the curve), we get a parabola:
It is easy to see that if we take the case ,,, that is, we will get a parabola, symmetric about the axis (oh). Ensure this is easy by filling out a similar table:
II Case, "A" is excellent from one
What will happen if we take,,? How will the behavior of the parabola change? With Title \u003d "(! Lang: Rendered by QuickTex.com" height="20" width="55" style="vertical-align: -5px;"> парабола изменит форму, она “похудеет” по сравнению с параболой (не верите – заполните соответствующую таблицу – и убедитесь сами):!}
In the first picture (see above) it is clearly seen that points from the table for parabola (1; 1), (-1; 1) were transformed into points (1; 4), (1; -4), that is, with the same The values \u200b\u200bof the ordinate of each point multiplied on 4. This will happen to all key points of the source table. Similarly, we argue in cases of pictures 2 and 3.
And at Parabola "will become wider" parabola:
Let's summarize:
1) The coefficient sign is responsible for the direction of the branches. With Title \u003d "(! Lang: Rendered by QuickTex.com" height="14" width="47" style="vertical-align: 0px;"> ветви направлены вверх, при - вниз. !}
2) Absolute value The coefficient (module) is responsible for "expansion", "compression" parabola. The larger, the more parabol, the less | A |, the wider Parabola.
III case appears "C"
Now let's enter into the game (that is, we consider the case when), we will consider parabollas of the species. It is not difficult to guess (you can always refer to the table), which will displace parabola along the axis up or down depending on the sign:
IV Case appears "B"
When will Parabola "breaks off" from the axis and will finally "walk" throughout the coordinate plane? When will cease to be equal.
Here for the construction of parabola we will need formula for calculating the vertex: , .
So at this point (as at the point (0; 0) of the new coordinate system) we will build a parabola, which we can permanently. If we are dealing with the case, then from the tops are laying one single segment to the right, one up, - the resulting point is our (similar to the step left, step up is our point); If we are dealing with, for example, then from the tops are laying one single segment to the right, two - up, etc.
For example, the vertex parabola:
Now the main thing is to understand that in this top we will build a parabola on the parabola pattern, because in our case.
When building a parabolla after finding the coordinates of the vertex is very It is convenient to consider the following points:
1) parabola will definitely pass through the point . Indeed, substituting in the formula x \u003d 0, we get that. That is, the ordinate of the point of intersection of the parabola with the axis (OU) is. In our example (above), Parabola crosses the ordinate axis at the point, since.
2) axis of symmetry parabola is straight, so all points of parabola will be symmetrical about it. In our example, we immediately take the point (0; -2) and we build a parabola symmetry with a symmetry relative to the axis, we get a point (4; -2) through which Parabola will pass.
3) Equating to, we learn the points of intersection of the parabola with the axis (oh). To do this, solve the equation. Depending on the discriminant, we will receive one (,), two (title \u003d "(! Lang: Rendered by QuickTex.com" height="14" width="54" style="vertical-align: 0px;">, ) или нИсколько () точек пересечения с осью (ох) !} . In the previous example, we have a root from the discriminant - not a whole number, when building it, it does not make sense to find the roots, but we see clearly that two points of intersection with the axis (oh) will have (since Title \u003d "(! Lang: Rendered By QuickLatex.com." height="14" width="54" style="vertical-align: 0px;">), хотя, в общем, это видно и без дискриминанта.!}
So let's work out
Algorithm for building a parabola if it is asked in the form
1) we determine the direction of the branches (A\u003e 0 - up, a<0 – вниз)
2) We find the coordinates of the vertex parabola by the formula.
3) We find the point of intersection of the parabola with the axis (OU) on a free member, we build a point, symmetrical about the axis of symmetry of the parabola (it should be noted, it happens that this point is unprofitable to note, for example, because the value is great ... I miss this item ...)
4) At the found point - the top of the parabola (as at the point (0; 0) of the new coordinate system) we build a parabola. If title \u003d "(! Lang: Rendered by QuickTextEx.com" height="20" width="55" style="vertical-align: -5px;">, то парабола становится у’же по сравнению с , если , то парабола расширяется по сравнению с !}
5) We find the point of intersection of the parabola with the axis (OU) (if they are still "not pop up"), solving equation
Example 1.
Example 2.
Note 1. If Parabola is initially set in the form, where there are some numbers (for example,), then it will be even easier to build it, because the coordinates of the vertices are already specified. Why?
Take the square three stale and highlight the full square in it: look, so we got that. We have previously called the top of the parabola, that is, now.
For example, . We note on the plane the top of the parabola, we understand that the branches are directed down, Parabola is expanded (relatively). That is, we perform paragraphs 1; 3; four; 5 of the parabola construction algorithm (see above).
Note 2. If Parabola is set in a form, similar to this (that is, it is presented in the form of a work of two linear multipliers), then we are immediately visible to the point of intersection of the parabola with the axis (oh). In this case - (0; 0) and (4; 0). Otherwise, we act according to the algorithm, the opening of the bracket.
Instruction
The quadratic function is generally written by the equation: Y \u003d Ax² + BX + C. The graph of this equation is, the branches of which are directed up (at a\u003e 0) or down (when 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.
People familiar with the concept of a derivative, easy to find the top of the parabola. Regardless of the position of the branches of the parabola, its top is a point (minimum if the branches are directed up, or when the branches are directed down). To find the points of the alleged extremum of any, it is necessary to calculate its first derivative and equate it to zero. In general, the derivative is equal to F "(x) \u003d (Ax² + BX + C)" \u003d 2AX + B. Equating to zero, you will get 0 \u003d 2AX0 + B \u003d\u003e x0 \u003d -b / 2a.
Parabola is a symmetric line. The axis passes through the top of the parabola. Knowing the parabola points with the x coordinate axis, you can easily find the abscissa of the vertices x0. Let X1 and X2 be the rooram roots (so called the points of intersection of the parabola with the abscissa axis, since these values \u200b\u200bare treated with a square equation AX² + BX + C at zero). At the same time let | x2 | \u003e | x1 |, then the top of the parabola lies in the middle between them and can be found from the following expression: x0 \u003d ½ (| x2 | - | x1 |).
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Sources:
- Quadratic function
- the formula for finding the vertex of parabola
Parabola is a chart of a quadratic function, in a general form, the parabola equation is written y \u003d Ah ^ 2 + BX + C, where a ≠ 0. This is a second-order universal curve, which describes many phenomena in life, for example, the movement of the thrown-up and then falling body, the form of the rainbow, so the ability to find parabola May be very useful in life.
You will need
- - Formula quadratic equation;
- - sheet of paper with a coordinate grid;
- - pencil, eraser;
- - Computer and Excel program.
Instruction
First of all, find the pearabol vertex. To find the abscissa of this point, take the coefficient before x, divide it to the double coefficient before X ^ 2 and multiply to -1 (x \u003d -b / 2a). Find ordinates, substituting the value obtained to the equation or by the formula y \u003d (b ^ 2-4ac) / 4a. You got the coordinates of the point of the top of the parabola.
The peakin of parabola can be found in another way. Since it is an extremum of the function, then to calculate it, calculate the first derivative and equate it to zero. In general, do you get the formula f (x) "\u003d (AX? + BX + C)" \u003d 2AX + B. And equating it to zero, you will come to the same formula itself - x \u003d -b / 2a.
Find out whether the branches of parabola are directed up or down. To do this, look at the coefficient before X ^ 2, that is, at a. If a\u003e 0, then the branches are directed up, if a
Coordinates vershins Parabolas found. Record them in the form of coordinates of one point (x0, y0).
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For functions (more precisely their graphs) is used the greatest value, including local maximum. The concept of "Top" is rather connected with geometric figures. The points of the maxima of smooth functions (having a derivative) are easy to determine with zeros of the first derivative.
Instruction
For points in which the function is not differentiable, but is continuous, the view is the highest at the gap may be the view of the isge (on y \u003d - | x |). At such points to functions You can spend arbitrarily tangent for her simply does not exist. Ourselves functions This type is usually set on segments. Points in which the derivative functions equal to zero or does not exist, called critical.
Decision. y \u003d x + 3 at x≤-1 and y \u003d ((x ^ 2) ^ (1/3)), with x\u003e -1. The function is defined in the segments intentionally, since in this case the target is pursued to display everything in one example. It is easy that at x \u003d -1 function remains continuous .Y '\u003d 1 at x≤-1 and y' \u003d (2/3) (x ^ (- 1/3)) - 1 \u003d (2-3 (x ^ (1/3)) / (x ^ (1/3)) at x\u003e -1. Y '\u003d 0 at x \u003d 8 / 27. y' does not exist at x \u003d -1 and x \u003d 0. In this y '\u003e 0 if x
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Parabola is one of the second order curves, its points are constructed in accordance with the square equation. The main thing in building this curve is to find top parabola. This can be done in several ways.
Instruction
To find the coordinates of the vertices parabola, use the following formula: x \u003d -b / 2a, where a is the coefficient before x in, and b is the coefficient before x. Substitute your values \u200b\u200band calculate it. Then substitute the obtained value instead of the equation and calculate the ordinate of the vertex. For example, if you are given an equation y \u003d 2x ^ 2-4x + 5, then find the abscissa as follows: x \u003d - (- 4) / 2 * 2 \u003d 1. Substituting x \u003d 1 to the equation, calculate the value of the vertex parabola: y \u003d 2 * 1 ^ 2-4 * 1 + 5 \u003d 3. Thus, the vertex parabola It has coordinates (1; 3).
The value of ordinate parabola Can be found without prior calculation of the abscissa. To do this, use the formula y \u003d -b ^ 2 / 4as + s.
If you are familiar with the concept of the derivative, find top parabola With the help of derivatives, using the following property of any: The first derivative function equal to zero indicates. Since the top parabolaRegardless of whether its branches are directed up or down, point, calculate the derivative for your function. In general, it will have the form f (x) \u003d 2ach + b. Eclay it to zero and get the coordinates of the vertices parabolacorresponding to your function.
Try to find top parabola, Taking advantage of its property as symmetry. To do this, find the intersection points parabola With the axis oh, equating the function to zero (substituting y \u003d 0). Deciding the square equation, you will find x1 and x2. Since Parabola is symmetrical about the directress passing through topThese points will be equidistant of the abscissa of the vertex. To find it, we divide
The graph of the quadratic function is called parabola. This line has a weighty physical value. On parabolam, some celestial bodies are moving. The antenna in the form of a parabola focuses rays, walking parallel to the axis of the parabola symmetry. Bodies flown up at an angle, reach the top point and fall down, also describing the parabola. Apparently that the coordinates of the vertices of this movement are invariably suitable.
Instruction
1. The quadratic function in general is written by the equation: Y \u003d AX? + BX + C. The graph of this equation is parabola, whose branches are directed up (at a\u003e 0) or down (with 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, a friend with a derivative representation, it is easy to detect a pearabol vertex. Alone on the location of the branches of the parabola, its vertex is an extremum point (minimum, if the branches are directed up, or the maximum when the branches are directed down). In order to detect the points of believable extremum, any feature, it is necessary to calculate its first derivative and equate it to zero. In the universal form, the derivative of the quadratic function is F "(x) \u003d (AX? + BX + C) '\u003d 2AX + B. Equating to zero, you get 0 \u003d 2AX0 + B \u003d\u003e x0 \u003d -b / 2a.
3. Parabola is a symmetric line. The axis of symmetry passes through the top of the parabola. Knowing the point of intersection of the parabola with the axis of the x coordinate, is allowed to easily detect the abscissa of the vertices x0. Let X1 and X2 be the rooram roots (so call the point of intersection of the parabola with the abscissa axis, from the fact that these values \u200b\u200bdo the square equation AX? + BX + C to zero). At the same time let me | x2 | \u003e | x1 |, then the top of the parabola lies in the middle between them and can be detected from further expression: x0 \u003d? (| x2 | - | x1 |).
Parabola is a chart of a quadratic function, in the universal form of the parabola equation is written y \u003d Ah ^ 2 + BX + C, where or? 0. This is a second-order universal curve, which describes many phenomena in life, say, move the movement of the thrown and after this falling body, the form of rainbow, the following knowledge to detect parabola Maybe a bit of life in life.
You will need
- - formula of a quadratic equation;
- - sheet of paper with a coordinate grid;
- - pencil, eraser;
- - Computer and Excel program.
Instruction
1. First, detect the pearabol vertex. In order to detect the abscissa of this point, take the indicator before x, divide it to the double rate before X ^ 2 and multiply to -1 (formula x \u003d -b / 2a). Detected the ordinate by substituting the value obtained to the equation either by the formula y \u003d (b ^ 2-4ac) / 4a. You got the coordinates of the point of the top of the parabola.
2. The vertex of parabola is permitted to detect and otherwise. Because the vertex is an extremum of the function, then to calculate it, calculate the first derivative and equate it to zero. In general, you get the formula f (x) '\u003d (AX? + BX + C)' \u003d 2AX + B. And equating it to zero, you will come to the same formula itself - x \u003d -b / 2a.
3. Learn whether the branches of parabola are directed up either down. To do this, look at the indicator before X ^ 2, that is, at a. If a\u003e 0, then the branches are directed up, if a
4. Build the axis of the symmetry of Parabola, it crosses the top of the parabola and parallel to the OU axis. All points of parabola will be equidistant of it, it is decided to decline only one part, and then it is symmetrically to display it regarding the axis of parabola.
5. Build a parabola line. To do this, detect several points, substituting the different values \u200b\u200bof x in the equation and solve equality. It is comfortable to detect the intersection with the axes, for this, substitute in the equality x \u003d 0 and y \u003d 0. Erecting one side, reflect it symmetrically regarding the axis.
6. Allowed to raise parabola Using Excel. 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 values \u200b\u200bof x on the selected section, and in the second column, write down the formula, say, \u003d 2V3 * B3-4B3 + 1 or \u003d 2B3 ^ 2-4V3 + 1. In order not to write this formula every time, "stretch" her for each column, pressing the mouse to a small cross in the lower right corner and pulling down.
7. Having received a table, press the "Insert" menu - "diagram". Select a point diagram, click "Next". In the window that appears, add a number by clicking the Add button. In order to prefer the necessary cells, click alternately by the buttons circled with a red oval below, then select your columns with values. By clicking the "Finish" button, rate the result - Ready parabola .
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When finding a quadratic function, the schedule of which is parabola, in one of the items you need to detect coordinates vershins Parabola. How to do it analytically, applying the equation specified for parabolava?
Instruction
1. The quadratic function is the function of the form y \u003d AX ^ 2 + BX + C, where a is a senior indicator (it must be nonzero), B is the junior indicator, C is a free member. This function gives its parabola chart, whose branches are directed or up (if a\u003e 0), or down (if a<0). При a=0 квадратичная функция вырождается в линейную функцию.
2. Detect x0 coordinate vershins Parabola. It is by formulax0 \u003d -b / a.
3. y0 \u003d y (x0). Details to detect the y0 coordinate vershins Parabolas, it is necessary to substitute the detected X0 value into return to X0. Consider what is equal to y0.
4. Coordinates vershins Parabolas are detected. Record them in the form of coordinates of one point (x0, y0).
5. When building a parabola, remember that it is symmetrical about the axis of symmetry of the parabola, passing vertically through the top of the parabola, because The quadratic function is even. Incidentally, it is quite possible to build only one branch of parabola, and it is different to complete symmetrically.
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For functions (rather, their graphs) is used the representation of the greatest value, including the local maximum. The presentation of the "Top" is rather due to geometric shapes. The points of the maxima of smooth functions (having a derivative) are easy to determine with the submool zeros of the first derivative.
Instruction
1. For points in which the function is not differentiable, but is constant, the view is the highest at the interval may be a view of the isge (for example y \u003d - | x |). At such points to the schedule functions It is allowed to spend how preferably a lot of tangents and the derivative for it easily does not exist. Ourselves functions This type is usually set on segments. Points in which the derivative functions equal to zero or does not exist, referred to as skeptical.
2. It turns out to find high points functions Y \u003d F (x) It follows: - detect skeptic points; - In order to prefer the maximum point, the sign of the derivative in the neighborhood of the skeptic point should be detected. If, when passing a point, an alternation of a sign with "+" on "-" occurs, then there is a maximum.
3. Example. Detect the greatest values functions (see Fig. 1) .Y \u003d x + 3 with x? -1 and y \u003d (((x ^ 2) ^ (1/3))) at x\u003e -1.
4. Decision. y \u003d x + 3 with x? -1 and y \u003d ((x ^ 2) ^ (1/3)) with x\u003e -1. The function is specified in the segments intentionally, because in this case the target is pursued to display everything in one example. It is easy to check that at x \u003d -1 the function remains constant .Y '\u003d 1 with x? -1 and y' \u003d (2/3) (x ^ (- 1/3)) - 1 \u003d (2-3 (x ^ (1/3)) / (x ^ (1/3)) at x\u003e -1. Y '\u003d 0 at x \u003d 8 / 27. y' does not exist at x \u003d -1 and x \u003d 0. y '\u003e 0 if x
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Parabola is one of the second order curves, its points are erected in accordance with the square equation. The main thing in the construction of this oblique is to detect top parabola . This is allowed to make several methods.
Instruction
1. In order to detect the coordinates of the vertices parabola , Take advantage of the further formula: x \u003d -b / 2a, where A is an indicator before x in a square, and B is an indicator before x. Substitute your values \u200b\u200band calculate its value. After that, substitute the value obtained instead of x into the equation and calculate the ordinate of the vertex. Let's say if you are given equation y \u003d 2x ^ 2-4x + 5, then detect the abscissus in a further effect: x \u003d - (- 4) / 2 * 2 \u003d 1. Substituting x \u003d 1 to the equation, calculate the value of the vertex parabola : y \u003d 2 * 1 ^ 2-4 * 1 + 5 \u003d 3. Thus, the vertex parabola It has coordinates (1; 3).
2. The value of ordinate parabola It is permitted to detect and without the previously calculated abscissa. To do this, use the formula y \u003d -b ^ 2 / 4as + s.
3. If you are familiar with the presentation of the derivative, detect top parabola With the help of derivatives, using a further feature of any function: the first derivative function equal to zero indicates extremum points. Because the top parabola Alone on whether its branches are directed up either down, is an extremum point, calculate the derivative for your function. In general, it will have the form f (x) \u003d 2ach + b. Eclay it to zero and get the coordinates of the vertices parabola corresponding to your function.
4. Try to detect top parabola , Taking advantage of its property as symmetry. To do this, detect the intersection points parabola With the axis oh, equating the function to zero (substituting y \u003d 0). Deciding the square equation, you will find x1 and x2. Because Parabola is symmetrical about the director passing through top These points will be equidistant of the abscissa of the vertex. In order to detect it, we divide the distance between the points of pressure: x \u003d (IX1-x2i) / 2.
5. If any of the indicators is zero (in addition to a), calculate the coordinates of the vertices parabola On lightweight formulas. For example, if B \u003d 0, that is, the equation has the form y \u003d ah ^ 2 + C, then the peak will lie on the OU axis and its coordinates will be equal (0; c). If not only the indicator B \u003d 0, but also C \u003d 0, then the vertex parabola Located at the beginning of the coordinates, point (0; 0).
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Based on the same point, the straight line form an angle where the universal point for them is a vertex. In the section of the theoretical algebra often there are tasks when it is necessary to detect the coordinates of this vershins In order to determine the equation passing through the top of the line.
Instruction
1. Before starting the process of finding coordinates vershins Decide on the initial data. Please accept the desired peak belongs to the ABC triangle, in which the coordinates of the 2nd vertices, as well as numerical values corners equal to "E" and "K" on the side of AB.
2. Align the new coordinate system with one side of the AB triangle thus in order to preface the coordinate system coincided with the point A, the coordinates of which you are famous. The second vertex B will lie on the OX axis, and its coordinates are also famous. Determine on the axis oh, the value of the length of the side AB according to the coordinates and take it equal to "M".
3. Lower perpendicular from unfamiliar vershins C on the axis oh and on the side of the AB triangle, respectively. The resulting height "y" and determines the value of one of the coordinates vershins C along the Oy axis. Please accept the height "y" divides the side of AB by two segments equal to "x" and "m - x".
4. From what you behave the meanings of all corners The triangle, it means that they are famous and the meanings of their tangents. Take tangent values \u200b\u200bfor corners adjacent to the side of the AB triangle equal to Tan (E) and Tan (k).
5. Enter the equations for 2 straight lines passing around AC and BC, respectively: y \u003d tan (E) * x and y \u003d tan (k) * (m - x). After that, detect the intersection of these direct, using the transformed direct equations: tan (e) \u003d y / x and tan (k) \u003d y / (m - x).
6. If we assume that tan (e) / tan (k) is equal to (Y / X) / (Y / (M - X)) or later the reduction of "Y" - (M - X) / X, as a result, you will get the desired values Coordinates equal to x \u003d m / (tan (e) / tan (k) + e) \u200b\u200band y \u003d x * tan (E).
7. Substitute values corners (E) and (k), as well as the found value of the side AB \u003d M in the equation x \u003d m / (tan (e) / tan (k) + e) \u200b\u200band y \u003d x * tan (E).
8. Convert a new coordinate system to the initial coordinate system, from the fact that a mutually unequivocal compliance is established between them, and get the desired coordinates vershins Triangle ABC.
Video on the topic
Video on the topic
Content:
The top of the parabola is the highest or lowest point. To find the top of the parabola, you can take advantage of the special formula or addition to the complete square. The following is described how to do it.
Steps
1 formula for the top of the vertex
- 1 Find the values \u200b\u200ba, b, and c. IN square equation The coefficient is x 2 = a for x. \u003d b, constant (coefficient without variable) \u003d c. For example, take the equation: y. = x 2 + 9x + 18. Here a. = 1, b. \u003d 9, and c. = 18.
- 2
Use the formula to calculate the x vertex coordinate value. The peak is also a point of symmetry parabola. Formula for finding coordinates X Parabola: x \u003d -b / 2a. Submold in it the corresponding values \u200b\u200bfor calculation x..
- x \u003d -b / 2a
- x \u003d - (9) / (2) (1)
- x \u003d -9 / 2
- 3
Submold the value X to the original equation to calculate the value y. Now that you know the value of X, simply substitute it into the original equation to find Y. Thus, the formula for finding a parabola vertex can be written as a function: (x, y) \u003d [(-b / 2a), f (-b / 2a)]. This means that in order to find Y, it is necessary to first find X by the formula, and then substitute the value x into the original equation. This is how it is done:
- y \u003d x 2 + 9x + 18
- y \u003d (-9/2) 2 + 9 (-9/2) +18
- y \u003d 81/4 -81/2 + 18
- y \u003d 81/4 -162/4 + 72/4
- y \u003d (81 - 162 + 72) / 4
- y \u003d -9/4.
- 4 Record the X and Y values \u200b\u200bin the form of a coordinate pair. Now that you know that x \u003d -9/2, and y \u003d -9/4, write them down as coordinates in the form: (-9/2, -9/4). The top of Parabola is located by coordinates (-9/2, -9/4). If you need to draw this parabola, then its vertex lies at the bottom point, since the coefficient at x 2 is positive.
2 addition to the full square
- 1 Write down the equation. Supplement to a complete square is another way to find the top of the parabola. Applying this method, you will find the x and y coordinates immediately, without the need to substitute X into the original equation. For example, an equation is given: x 2 + 4x + 1 \u003d 0.
- 2 Divide each coefficient to the coefficient at x 2. In our case, the coefficient at x 2 is 1, so we can skip this step. Decision on 1 will not change anything.
- 3
Transfer the equation permanent into the right-hand side. Permanent - coefficient without a variable. Here it is "1". Transfer 1 to the right by subtracting 1 from both parts of the equation. Here's how to do it:
- x 2 + 4x + 1 \u003d 0
- x 2 + 4x + 1 -1 \u003d 0 - 1
- x 2 + 4x \u003d - 1
- 4
Complete the left part of the equation to a complete square. To do this, just find (b / 2) 2 And add the result to both parts of the equation. Substitute "4" instead b.Since "4x" is the coefficient of our equation.
- (4/2) 2 \u003d 2 2 \u003d 4. Now add 4 to both parts of the equation and get:
- x 2 + 4x + 4 \u003d -1 + 4
- x 2 + 4x + 4 \u003d 3
- (4/2) 2 \u003d 2 2 \u003d 4. Now add 4 to both parts of the equation and get:
- 5 We simplify the left part of the equation. We see that x 2 + 4x + 4 - full square. It can be recorded in the form: (x + 2) 2 \u003d 3
- 6 Use it to find the X and Y coordinates. You can find X, simply equating (x + 2) 2 to 0. Now that (x + 2) 2 \u003d 0, calculate x: x \u003d -2. Y coordinate is constant in the right side of a full square. So, y \u003d 3. Top Parabola equation x 2 + 4x + 1 \u003d (-2, 3)
- Correctly define a, b, and c.
- Record preliminary calculations. This will not only help in the process of work, but also will allow you to see where the errors are made.
- Do not disturb the order of calculations.
Warnings
- Check your answer!
- Make sure you know how to determine the coefficient a, b, and c. If you do not know, the answer will be wrong.
- Not - the solution of such tasks requires practice.