Zavdannya z joke points overflow be-some figures ideologically primitive. Difficult in them they are used only through arithmetic;

Instructions

1. Dana zavdannya virіshuє to be analytically, if it is allowed to call not a little graphic straight and parabols. Most often the price is a great plus for the virile butt, which in the factory can be given to such a function, which is simpler and not simple.

2. It is a good idea to use the parabola algebra as a guide to the function of the form f (x) = ax ^ 2 + bx + c, where a, b, c are chain numbers, before the exponent a is a good zero. The function g (x) = kx + h, where k, h are chain numbers, starting with a straight line on the square.

3. Speck overflow straightі parabols - the whole starting point of both curves, when in this function there is an identical value, so f (x) = g (x). Given the permission to write down the rіvnyannya: ax ^ 2 + bx + c = kx + h, to give the power to evoke without any dots overflow .

4. In the case of the equal ax ^ 2 + bx + c = kx + h, it is necessary to transfer all warehouses to the left part and to bring the details: ax ^ 2 + (b-k) x + c-h = 0. It is too much of a problem to lose virility from a square one.

5. All the "ixies" are not a result at all, because the point on the area is characterized by two valid numbers (x, y). For a new solution, it is necessary to calculate the type "IGRIC". For this, it is necessary to put "ixi" either in the function f (x), or in the function g (x), tea for the point overflow correct: y = f (x) = g (x). Find out all the points of the parabola and straight .

6. To fix the material, it is very important to look at the solution on the butt. Let the parabola be defined by the function f (x) = x ^ 2-3x + 3, and the straight line - g (x) = 2x-3. Store the equal f (x) = g (x), so x ^ 2-3x + 3 = 2x-3. Transferring all warehouses to the left part, and bringing additional ones, deduct: x ^ 2-5x + 6 = 0. The root of the square ryvnyannya: x1 = 2, x2 = 3. Infecting the results of "IGRIC": y1 = g ( 1, y2 = g (x2) = 3. With this rank, all specks overflow: (2.1) і (3.3).

speck overflow straight it is allowed to approximate the value for the graph. Since it is often necessary to accurately coordinate the center of the point, or the graph will not be needed, because it is possible to select a point overflow, Know only rivnyannya straight.

Instructions

1. Leave two straight lines given by straight lines: A1 * x + B1 * y + C1 = 0 і A2 * x + B2 * y + C2 = 0. Point overflow to establish і one straight, і іnshiy. Visible from the first straight line x, we can recognize: x = - (B1 * y + C1) / A1. Pidstavami recognized the value in another equal: -A2 * (B1 * y + C1) / A1 + B2 * y + C2 = 0. Abo -A2B1 * y - A2C1 + A1B2 * y + A1C2 = 0, stars y = (A2C1 - A1C2) / (A1B2 - A2B1). The values ​​for the first straight forward are shown: A1 * x + B1 (A2C1 - A1C2) / (A1B2 - A2B1) + C1 = 0.A1 (A1B2 - A2B1) * x + A2B1C1 - A1B1C2 + A1B2C1 - A2B1C1 - = 0 (A1B1C1 A2B1) * x - B1C2 + B2C1 = 0 Then x = (B1C2 - B2C1) / (A1B2 - A2B1).

2. In the school course in mathematics, it is often straightforward to ask the level of mathematics with a good indicator, the definitions of the names and types. Leave two straight lines given by this rank: y1 = k1 * x + b1 і y2 = k2 * x + b2. Mabut, scho in point overflow y1 = y2, todі k1 * x + b1 = k2 * x + b2. Otrimuєmo, sho ordinate of a point overflow x = (b2 - b1) / (k1 - k2). By way of x, whether it be straightforward or not, y = k1 (b2 - b1) / (k1 - k2) + b1 = (k1b2 - b1k2) / (k1 - k2).

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rivnyannya parabolasє quadratic function. Isnu kіlka options folding ts'go rіvnyannya. Everything is based on what parameters are presented in the mind of the factory.

Instructions

1. The parabola є is crooked, like an arc behind its shape and a graph of statical function. Necessarily because of the fact that the collage is a parabola, the function is paired. The pairwise function does not change the value of the argument from the field of value when the sign changes: f (-x) = f (x) Read from the most primitive function: y = x ^ 2. , there is a build-up yak for correct, and for negative values ​​of the argument x. The point at x = 0, and at the same time, y = 0 is a point of minimum function.

2. Below you will find all the main options to stimulate the whole function and family. In order to display the graph of this function, it is necessary to destroy the graph of the function f (x) by a single one. The application can serve the function y = x ^ 2 + 3, then along the y axis destroy the function uphill on two units. If a function is given with an opposite sign, say y = x ^ 2-3, then the graph will destroy downward along the y axis.

3. Another kind of function, which can be given a parabola - f (x) = (x + a) ^ 2. In these types of graphics, navpaks, are drawn along the abscis axis (x axis) on a single. For the butt, you can use the following functions: y = (x +4) ^ 2 and y = (x-4) ^ 2. At the first drop, if the function with a plus sign, the graph will destroy on the x axis in the left, and in the other - to the right ... All colors are shown on the baby.

4. There is also parabolic fallowness in the form y = x ^ 4. With such drops, x = const, and y is steeply moving. However, only pair functions are welcome. parabolas often present in physical tasks, for example, the policy describes the line, I go like a parabola. the same kind parabolas MAK podovzhny peretin reflector headlights, lіkhtar. On a sinusoidal basis, the Danish graph is not periodic and growing.

Rada 4: Yak vis-à-vis the point overflowing straight from the area

Dane zavdannya to get the point overflow straight with a square є classic in the course of engineering graphics and using the methods of drawing geometry and graphic design on the armchair.

Instructions

1. Visible point value overflow straight in the area of ​​a private roztashuvannya (Figure 1). point їх overflow K to lay out i straightі of the area, it means that the backward projection of K2 lies on? 2 і l2. Tobto, K2 = l2 ?? 2, and її is a horizontal projection of K1, it starts at l1 behind an additional line of projection connection. overflow K (K2K1) will be unimpeachable without the visor of the additional planes. Similarly, the points overflow straight with wide areas of private roasting.

2. Visible point value overflow straight with the area of ​​the zagalny roztashuvannya. For little 2 in the space, have you given enough space? i straight l. For the value of the point overflow straight In the area of ​​the zagalny roztashuvannya vikoristovu

3. Is an additional area drawn through the line l? .To lodge the motives of the project, the area will be.

5. Point K overflow straight l and prompted lines overflow MN. Won і є bazhany point overflow straight and area.

6. There is a very common rule for solving singing tasks on a complex drawing. Example. by point overflow straight l with the area of ​​the zagalny roztashuvannya, given by the tricycle ABC (Figure 3).

7. Through the line l, an additional area is drawn?, Perpendicular to the projection area? 2. What is the projection? 2 stand out from the projection straight l2.

8. There will be a line MN. Ploshchina? overturns AB at point M. 2? A2B2 і horizontal M1 to A1B1 on the line of projection connection. Plane? overturns the AC side at point N. What is the forward projection N2 =? 2? A2C2, horizontal projection of N1 on A1C1. The straight line MN is located one-time on the same areas, and, therefore, є line їх overflow .

9. Start point K1 overflow l1 і M1N1, when the call is on, the point K2 will be. Go, K1 and K2 - projection of the bazhany point overflow K straight l і area? ABC: K (K1K2) = l (l1l2)? ? ABC (A1B1C1, A2B2C2). With the addition of competing points M, 1 and 2.3, visibility begins straight l What is the given area? ABC.

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Brutal respect!
Freeze additional area for new tasks.

Korisna is happy
Vykonovyte counting, stopping reports of the chair, showing the minds of the head. Tse will help you to build a zorієntuvatisya in solutions.

The two are straight, as the stench is not parallel and does not go astray, it is not deafeningly fading into one point. Viyaviti coordinates tsyogo miscya - means to count points overflow straight. Two cross-over straight lines lie in the same area, if you can see them in the Cartesian area. Pick up on the butt, yak viyaviti go to the point straight.

Instructions

1. See the two straight lines, remembering those who are straight on in the Cartesian coordinate system of the straight line as ah + wu + s = 0, where a, b, c are special numbers, and x and y are the coordinates of points. For the butt to shine points overflow straight 4x + 3y-6 = 0 і 2x + y-4 = 0.

2. For the revision of the system and the rivnyan, be-yak from the ryvnyan so, before y standing an identical indicator. To that, in one, the indicator is in front of the road 1, then, primitively, multiply the price by the number 3 (the indicator in front of the first one). For a whole, be-like element of rivnyannya, multiply by 3: (2x * 3) + (y * 3) - (4 * 3) = (0 * 3) that otrimate zvychaine rіvnyannya 6x + 3y-12 = 0. miracles come from the same in both rural areas, many will need to be offended.

3. See from one іnche іnshe. For a complete view of the left part of one, the left part of the other, but do the same on the right. Cut off such a hang: (4x + 3y-6) - (6x + 3y-12) = 0-0. To that, there is a sign "-" in front of the bow, remember all the signs in the bows on the opposite side. Otrim such a hang: 4x + 3y-6 - 6x-3y + 12 = 0. Spread the viraz and wipe it, so it’s awesome. The new viglada is as follows: 2x + 6 = 0. Transfer the number 6 to the іnshu part of the іvnyаnnya, і from the discarded parity 2х = -6 іslovіt х: х = (- 6) / (- 2). In such a rank, we were given x = 3.

4. Submit the value x = 3 in whether it is equal, say, in a friend and deny such a hang: (2 * 3) + y-4 = 0. Simplify and hold y: y = 4-6 = -2.

5. Write down the values ​​of x and y at the coordinates points(3; -2). This will solve the problem. Reverting the meaning of the meaning by the way of substitution in the offense of the rivnyannya.

6. It’s not straightforward in viglyadi rivnyan, but primitive in the area, to see the coordinates points overflow graphically. For the sake of it, sell it straight, so that the stench overwhelmed, sending it down on the axis oh and oh perpendiculars. Peretin perpendiculars with axes oh and oh, if the coordinates of the center points, Marvel at the little ones, and you will shake the coordinates points overflow x = 3 і у = -2, that is the point (3; -2) і solving problems.

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A parabola is a flat curve of a different order, the canonical one in the Cartesian coordinate system of the ma viglyad y? = 2px. De p is the focal parameter of the parabola, but the road is displayed from the fixed point F, called the focus, to the fixed straight D in the same area, where the name is the director. The vertex of such a parabola passes through the front of coordinates, and the curve itself is symmetric with respect to the abscis axis Ox. The schoolchildren's course of algebra has taken to look at a parabola, while the symmetry of which is to be seen from the axis of ordinates Oy: x? = 2py. A rіvnyannya with a tsomu write down the navpaki: y = ax? + Bx + c, a = 1 / (2p). A namaluvati parabola is allowed by a number of methods that are cleverly allowed to be called algebraic and geometric.

Instructions

1. Algebraic Pobudova Parabola. Learn the coordinates of the vertex of the parabola. Calculate the coordinate along the Ox axis according to the formula: x0 = -b / (2a), and along the Oy axis: y0 = - (b? -4ac) / 4a, or if the x0 value is deducted in the equal parabola y0 = ax0? + Bx0 + c і enumerate the value.

2. On the coordinate area, see the symmetry of the parabola. The formula is based on the formula for the coordinates x0 of the vertex of the parabola: x = -b / (2a). Visnite, kudi straightened the parabola head. If a> 0, then the axis is straight up the hill, if a

3. Take 2 or 3 values ​​for the parameter x so, wb: x0

4. Put points 1 ', 2', і 3 'so that the stench will be symmetrical to points 1, 2, 3 as well as the axis of symmetry.

5. Find points 1 ', 2', 3 ', 0, 1, 2, 3 with a smooth oblique line. Continue the line uphill or downward, in the fallowness, right in the direction of the parabola. The parabola is prompted.

6. Geometric for the parabola. The Danish method of recording on a given parabola, such as the number of points, equally distant from the focus F, and from the direction D. Therefore, the focal parameter of a given parabola p = 1 / (2a) can be found.

7. Indulge in all the symmetry of the parabola, as described in 2 crots. Put the speck F with the coordinate on the Oy axis on the n_y with the y = p / 2 and point D with the coordinate y = -p / 2.

8. For the help of the kutnik, induce the line to pass through point D, perpendicular to the axis of symmetry of the parabola. Tsya liniya is the director of the parabola.

9. Cut the thread along the door to one of the legs of the cot. One end of the thread with a button should be fastened to the top of the kutnik, until the legs are attached, and the second end is at the focus of the parabola at point F. ...

10. Olivets stand up like this, so that his winds press the thread to the leg of the kutnik. Rukhayte kutnik along the line. Олівець vikreslit you need a parabola.

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Brutal respect!
Do not paint the apex of the parabola at the viglyadі kuta. Її gilki converge one by one, smoothly rounded.

Korisna is happy
When the parabola is prompted by the geometric method of stitching, the thread is invariably pulled.

Earlier, you should start before the visualization of the behavior of the function; It is acceptable to admit that there are no changes to come up to unlimited numbers.

Instructions

1. The function is a variable value, which can be stored as a value for the argument. Dovid is a change of self. Between the changes to the argument is changed to the area of ​​possible values ​​(ODZ). The behavior of the function is to be looked at within the framework of the ODZ to the fact that the interconnection between the two is more chaotic in the interchanges, and according to the singing rules and can be written in the view of the mathematical virase.

2. Is it clear enough functionality F =? (X) huh? - mathematical viraz. The function can move the point with the coordinate axes or with other functions.

3. At the points, the function changes from the abscis to zero: F (x) = 0. You deduct the coordinates of the points overturning the given function from the OX system. Such points will be the style, the roots of the rivnyannya on the given argument of metamorphosis will be known.

4. At the points, the function changes from the ordinates to the value of the argument to zero. So, the default value of the function will be transformed into the value of the function at x = 0. The point of the overturn of the function from the OY will be the style, since the value of the given function will be known if the value is zero.

5. To know the points to cross the given function from the other function, it is necessary to update the system of equals: F =? (X) W =? (X). There? (X) is a viraz describing the given function F ,? (X) is a viraz, which describes the function W, the point is overturned from a given function of the requirement to evolve. Mabut, scho, at points overretinu offensive functions, take equal values ​​with equal values ​​of arguments. The starting points for 2 functions will be styling, while the solutions for the system of equipments are based on the given argument.

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At the points, the functions of the function change to equal values ​​with the identical value of the argument. Viyaviti points overretinu functions - means rather the coordinates of the points that are out of the way for the intersecting functions.

Instructions

1. Does the zealous viglyad have the task of knowing the points to overturn the functions of one argument Y = F (x) і Y? = F? (X) on the XOY area it is built up to the date of the level Y = Y? The value of x, which is satisfied with the equality F (x) = F? (X), (how it smells) є abscissas of points overflow the given functions.

2. If the functions are given by an awkward mathematical virase and lie in one argument x, then the setting of the known points to cross is allowed graphically. Encourage graphics functions. Assign the points to be overrun with the coordinate axes (x = 0, y = 0). Give another value to the argument, show the same value of the function, add a dot on the graphs. What is more huge than the points will be vikoristano for motivation, tim virnishe will be grafik.

3. If the graphs of functions to overflow, instead of the chairs, the coordinates of the points overflow. For converting, enter the coordinates in the formula, which are set to the function. As well as mathematical virazi appear to be active, the points of overturn appear positively. As long as the graphs of functions do not change, try changing the scale. To grow the crochet between the dots, inspire it to be bigger, but on the other hand, the lines of the graphs are approaching. When writing on the spotted retinue, ask for a larger lecture graph with another croc for the exact value of the coordinates of the retinue points.

4. It is necessary to see the points of repetition of functions not on the square, but in the trivial space, to be able to see the functions of 2 ministers: Z = F (x, y) і Z? = F? (X, y). For the value of the coordinates of the points, the overturning of the functions is necessary to establish the system of equal values ​​from two unknown x and y at Z = Z?.

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Also, the main parameters of the graph of the quadratic function are shown on the small screen:

Clear a few ways to induce a square parabola. Due to the fact that a quadratic function is given by the rank, it is possible to vibrate the best value.

1 ... The function is given by the formula .

Clear a gallant algorithm to induce a graph of a quadratic parabola on the butt, motivate the graph of the function

1 ... Straight parabolic throat.

So yak, the parabolas are straight up the hill.

2 ... We know the discriminant of a square trinomial

The discrimnant of a square trinomial is greater than zero, so the parabola has two points to cross over from the axis of the OX.

In order to know the coordinates, virishimo rivnyannya:

,

3 ... Parabolic vertex coordinates:

4 ... The point of the parabolic crossover is from the axis of the OY: (0; -5), and is symmetrical as well as the axis of the symmetry of the parabola.

It is applied to the points on the coordinate area, and from the same smooth curve:

Tsey sposib can be easily simplified.

1. We know the code of the vertex of the parabola.

2. We know the coordinates of the points where the right-hander and the left-hander stand at the top.

Quickly with the results, prompting the graph of the function

Krrdinati tops of the parabola

Near to the top of the point, the expansion of the ground from the top may be abscissa to -1; -2; -3

Near the top of the point, the right-hander can be moved to the abscissa as seen from 0; 1; 2

The substitution of the values ​​of x in the equal function, we know the ordinate points and enter the x in the table:

We apply the points on the cordinate area and on the same smooth line:

2 ... Equivalent quadratic function of ma viglyad - in ts'mu pivnyannі - coordinates of the vertex of the parabola

for in equal quadratic functions , І other kofіtsієnt - guy number.

I'll stay for a butt graph of functions .

Zgadaimo line-up of graphical functions. Show the function graph , required

§ a selection of graphical functions,

§ at the same time all points of the graph are multiplied by 2,

§ then destroy the axis of the OH axis by 1 unit to the right,

§ and then the OY axis for 4 units up the hill:

Now I can see the graph of the function ... At the same level of functionality, and the other function is a pair of numbers.