Tasks for searching points crossing Any figures are ideologically primitive. Difficulties in them are only due to arithmetic, because it is in it that various typos and errors are allowed.

Instruction

1. This task is solved analytically, it is always allowed to do not draw graphs at all. straight and parabola. Often it gives a huge plus in solving the example, because the task may be given such functions that they are easier and rapidly do not draw.

2. According to the textbooks on the parabol algebra, the function of the form f (x) \u003d AX ^ 2 + BX + C, where a, b, c is real numbers, while the indicator A is good is zero. The function G (x) \u003d kx + h, where k, h is real numbers, determines the direct on the plane.

3. Point crossing straight And parabolas are a universal point of both curves, it is still an identical value in it, that is, f (x) \u003d g (x). This statement allows you to record the equation: AX ^ 2 + BX + C \u003d KX + H, which will probably detect a lot of points crossing .

4. In the equation AX ^ 2 + BX + C \u003d KX + H, it is necessary to transfer all the components to the left side and bring similar: AX ^ 2 + (B - K) X + C - H \u003d 0. Now it remains to solve the resulting square equation.

5. All detected "Xers" is not yet a result of the task, because the point on the plane is characterized by two real numbers (x, y). To fully conclude a solution, you need to calculate the corresponding "playing". To do this, it is necessary to substitute the "Xers" either to the function f (x), or in the function G (x), tea for a point crossing That's right: y \u003d f (x) \u003d g (x). Later, you will find all the universal points of parabola and straight .

6. To secure the material, the main thing is to see the solution on the example. Let the parabola be defined by the function f (x) \u003d x ^ 2-3x + 3, and the straight - G (x) \u003d 2x-3. Make the equation f (x) \u003d g (x), that is, x ^ 2-3x + 3 \u003d 2x-3. Transferring all the components to the left side, and leading similar, get: x ^ 2-5x + 6 \u003d 0. The roots of this square equation: x1 \u003d 2, x2 \u003d 3. Now detect the corresponding "playing": y1 \u003d g (x1) \u003d 1, y2 \u003d g (x2) \u003d 3. Thus, all points were found. crossing : (2,1) and (3.3).

Point crossing Direct is allowed to approximately determine the schedule. However, the exact coordinates of this point are often necessary, or the schedule is not required, then it is allowed to detect a point crossing , knowing only equations of direct.

Instruction

1. Let two straight lines are given by universal direct equations: a1 * x + b1 * y + c1 \u003d 0 and a2 * x + b2 * y + c2 \u003d 0. point crossing Belongs and one direct, and other. Express from the first equation is straight x, we obtain: x \u003d - (B1 * Y + C1) / A1. We substitute the value obtained into the second equation: -A2 * (B1 * Y + C1) / A1 + B2 * Y + C2 \u003d 0. Either -A2B1 * y - A2C1 + A1B2 * Y + A1C2 \u003d 0, selector Y \u003d (A2C1 - A1C2) / (A1B2 - A2B1). We substitute the value discovered in the first direct equation: A1 * X + B1 (A2C1 - A1C2) / (A1B2 - A2B1) + C1 \u003d 0.A1 (A1B2 - A2B1) * X + A2B1C1 - A1B1C2 + A1B2C1 - A2B1C1 \u003d 0 (A1B2 - A2B1) * X - B1C2 + B2C1 \u003d 0DH \u003d (B1C2 - B2C1) / (A1B2 - A2B1).

2. In the school course of mathematics, the direct is often given by the equation with an angular figure, we will see this case. Let two straight lines are given in this way: y1 \u003d k1 * x + b1 and y2 \u003d k2 * x + b2. Apparently at the point crossing y1 \u003d y2, then k1 * x + b1 \u003d k2 * x + b2. We get that the ordinate point crossing X \u003d (B2 - B1) / (K1 - K2). We substitute X into any equation direct and obtain Y \u003d k1 (B2 - B1) / (k1 - k2) + b1 \u003d (k1b2 - b1k2) / (k1 - k2).

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The equation parabola It is a quadratic function. There are several options for compiling this equation. It all depends on which parameters are presented in the Terk Condition.

Instruction

1. Parabola is a curve that is in its form resembles an arc and is a graph of a power function. Alone on which polerabol has a parabola, this function is even. An even referred to as this function, which, with all the values \u200b\u200bof the argument, from the definition area, when the arbitrariness change changes, the value does not change: f (-x) \u003d f (x) Start with the most primitive functions: y \u003d x ^ 2. From its species, it is allowed to make the result that it increases both with the right and with negative values \u200b\u200bof the X. The point in which X \u003d 0, and at the same time, y \u003d 0 is considered a point of a minimum function.

2. Below are all major options for building this function and its equation. As the first example, the function of the form: F (x) \u003d x ^ 2 + A, where A is a whole size of that in order to build a graph of this function, you need to move the graph function f (x) to A units. An example is the function y \u003d x ^ 2 + 3, where, along the Y axis, the function is shifted up into two units. If a function is given to the opposite sign, say y \u003d x ^ 2-3, then its graph is shifted down along the Y axis.

3. Another type of function that parabol can be set is f (x) \u003d (x + a) ^ 2. In such cases, the schedule, on the contrary, is shifted along the abscissa axis (x axis) to A units. For example, it is allowed to see the functions: y \u003d (x +4) ^ 2 and y \u003d (x-4) ^ 2. In the first case, where there is a feature with a plus sign, the graph is shifted along the X axis left, and in the second case - right. All these cases are shown in the figure.

4. There are also parabolic dependencies of the form y \u003d x ^ 4. Under such cases, x \u003d const, and y rises cool. However, it concerns even even functions. Marchors parabola Often are present in physical problems, say, the flight of the body describes the line similar to the parabola. Also species parabola It has a longitudinal section of the headlight reflector, a lantern. In the difference from sinusoid, this schedule is non-periodic and increasing.

Tip 4: How to determine the intersection point with a plane

This task is to build a point crossing straight The plane is classic in the course of engineering graphics and is performed by ways of descriptive geometry and their graphic solution in the drawing.

Instruction

1. We will see the definition of a point crossing straight With the plane of the private location (Figure 1). Praise l crosses the frontal-design plane?. Point of them crossing K belongs to I. straight and the plane, which means, the general projection of K2 lies at? 2 and L2. That is, k2 \u003d l2 ?? 2, and its horizontal projection of K1 is determined on L1 using a projection line. In the way, the desired point crossing K (k2k1) is built at ease without using the auxiliary planes. Points are determined crossing straight With all sorts of private location planes.

2. We will see the definition of a point crossing straight with the plane of the universal location. Figure 2 in space is given arbitrarily located plane? and straight l. To determine the point crossing straight With the plane of the universal location, the method of auxiliary secant planes is used in the further order:

3. Through the straight line L, auxiliary secant plane is carried out?. For facilitating constructions, it will be a design plane.

5. The point K is noted crossing straight l and built line crossing Mn. She is the desired point crossing straight and plane.

6. Apply this rule to solve a specific task on a comprehensive drawing. Example. Determine the point crossing straight l with a universal layout plane given by the ABC triangle (Figure 3).

7. Through the straight line l, auxiliary secant plane is carried out?, Perpendicular to the plane of the projection? 2. Her projection? 2 coincides with the projection straight L2.

8. Mn line is built. Plane? Crossing AB at point M. It marks its total projection M2 \u003d? 2? A2B2 and horizontal M1 on A1B1 along the projection link. intersects the side of the AC at point n. its total projection n2 \u003d? 2? A2C2, horizontal projection N1 on A1C1.prase Mn belongs to the total planes, and, it means, is the line of them crossing .

9. The point K1 is defined crossing L1 and M1N1, after that, with support for the communication line, a point of K2 is built. It turns out, K1 and K2 - the projection of the desired point crossing K. straight l and plane? ABC: k (k1k2) \u003d L (L1l2)? ? ABC (A1B1C1, A2B2C2). In the help of competing points M, 1 and 2.3, visibility is determined straight L regarding this plane? ABC.

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Note!
Use auxiliary plane when solving the problem.

Helpful advice
Execute calculations by applying detailed drawings that meet the conditions of the task. This will help rapidly navigate in the decision.

Two straight, if they are not parallel and do not coincide, strictly intersect at one point. Detect the coordinates of this place - it means to calculate points crossing straight. Two intersecting straight lone invariably lie in the same plane, it is decided to pretrate them in a decartian plane. We will analyze on the example, how to detect the universal point of direct.

Instruction

1. Take the 2-straight equations, remembering that the equation is direct in the Cartesian coordinate system The equation direct looks like ah + v / c \u003d 0, and A, B, C - ordinary numbers, and x and y coordinate points. For example, detect points crossing straight 4x + 3ow-6 \u003d 0 and 2x + y-4 \u003d 0. To do this, detect the solution of the system of these 2 equations.

2. To solve the system of equations, change any of the equations so in order to be faced with an identical figure. Because in one equation, the indicator in front of it is 1, then it is primitive to multiply this equation to the number 3 (the indicator in front of the other equation). For this, every element of the equation multiply by 3: (2x * 3) + (y * 3) - (4 * 3) \u003d (0 * 3) and get an ordinary equation 6x + 3ow-12 \u003d 0. If the indicators in front of the u were wonderful from the unit in both equations, it would be necessary to multiply both equalities.

3. Substitute another other equation. To do this, deduct from the left side of one left part of another and correctly also come with the right. Get such an expression: (4x + 3ow-6) - (6x + 3ow-12) \u003d 0-0. Because in front of the brace is a sign "-", all signs in brackets change to the opposite. Get such an expression: 4 + 3ow-6 - 6x-3ow + 12 \u003d 0. Simplify the expression and you will see that the variable has disappeared. The new equation looks like this: -2x + 6 \u003d 0. Transfer the number 6 to another part of the equation and from the received equality -2x \u003d -6 express X: x \u003d (- 6) / (- 2). Thus, you got x \u003d 3.

4. Substitute the value x \u003d 3 in any equation, let's say, second and get such an expression: (2 * 3) + y-4 \u003d 0. Simplify and express y: y \u003d 4-6 \u003d -2.

5. Write down the obtained values \u200b\u200bof x and y in the form of coordinates points (3; -2). These will solve the problem. Check the value obtained by the substitution method in both equations.

6. If the straight lines are not given in the form of equations, and given primitively on the plane, detect the coordinates points crossing graphically. To do this, extend the direct so that they crossed, then drop on the axis OH and OU perpendicular. Intersection of perpendiculars with axes OH and OU, will be coordinates of this points , look at the drawing and you will see that the coordinates points crossing x \u003d 3 and y \u003d -2, that is, the point (3; -2) is the solution of the problem.

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Parabola is a flat curve of a second order, the canonical equation of which in the Cartesian coordinate system has the form y? \u003d 2px. Where p is the focal parameter of the parabola, equal to the distance from the fixed point F, called the focus, to the fixed direct d in the same plane, which is the name - the director. The top of such a parabola passes through the preface of the coordinates, and the curve itself is symmetrical about the abscissa axis oh. In the school year, algebra is customary to consider parabola, the axis of symmetry of which coincides with the owner of the ordinate OU: X? \u003d 2PY. And the equation is recorded slightly opposite: y \u003d ax? + Bx + c, a \u003d 1 / (2p). Draw a parabola is permitted by several methods, conditionally allowed to call algebraic and geometric.

Instruction

1. Algebraic construction of parabola. We know the coordinates of the vertices of Parabola. Coordinate on the axis OH calculate according to the formula: x0 \u003d -b / (2a), and on the axis OY: Y0 \u003d - (b? -4ac) / 4a or substitute the resulting x0 value to the parabolla equation y0 \u003d ax0? + BX0 + C and Calculate the value.

2. On the coordinate plane built the axis of the plabeara symmetry. Its formula coincides with the formula coordinate X0 vertex parabola: x \u003d -b / (2a). Determine where the branches of parabola are directed. If a\u003e 0, then the axes are directed up, if a

3. Take arbitrarily 2-3 values \u200b\u200bfor the X so, in order to: x0

4. Put points 1 ', 2', and 3 'so that they have been symmetrical to points 1, 2, 3 regarding the axis of symmetry.

5. Combine points 1 ', 2', 3 ', 0, 1, 2, 3 smooth oblique line. Continue the line up either down, depending on the direction of the parabola. Parabola is built.

6. Geometric parabola construction. This method is based on the definition of parabola, as the generality of points that are equidistant to both the focus F and the director of D. D. It first, at the beginning, detect the focal parameter of the given parabola p \u003d 1 / (2a).

7. Build the axis of symmetry Parabola, as described in 2 step. On it, put the point F with the coordinate along the OS axis equal to y \u003d p / 2 and the d point D with the coordinate y \u003d -r / 2.

8. With the help of the kitchen, build a line passing through the point D, perpendicular to the axis of the parabola symmetry. This line is a parabola director.

9. Take the thread in length equal to one of the carbon cathets. One end of the thread button to fix on the top of the square, to which this catat adjacent, and the 2nd end - in the focus of parabola at point F. Put the line so that its upper edge coincided with the director of D. On the range of the carbon free from the knitting button .

10. Install the pencil so so that he pressed the thread to the carbon cathelet. Move the carbon carbon. Pencil will draw the parabola you need.

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Note!
Do not draw the top of the parabola in the form of an angle. Her branches converge with another, smoothly sprawling.

Helpful advice
When constructing a parabola with a geometric method, follow, in order to the thread, it was always stretched.

Earlier than proceeding to finding the behavior of the function, it is necessary to determine the area of \u200b\u200bmetamorphosis of the values \u200b\u200bunder consideration. We accept the assumption that variables relate to a variety of valid numbers.

Instruction

1. Function is a variable value depending on the value of the argument. Argument is a variable independent. The limits of the change changes are referred to as the area of \u200b\u200bpossible values \u200b\u200b(OTZ). The behavior of the function is considered within the framework of OTZ because, within these limits, the association between two variables is not chaotic, but is subordinate to certain rules and can be recorded in the form of a mathematical expression.

2. We will see arbitrary functional connectedness f \u003d? (X), where? - Mathematical expression. The function may have intersection points with coordinate axes either with other functions.

3. At the points of intersection of the function with the axis of the abscissa, the function becomes equal to zero: f (x) \u003d 0.Shiste this equation. You will get the coordinates of the intersection points of the specified function with the axis oh. There will be so many such points as the roots of the equation on the specified section of the metamorphosis of the argument.

4. At the points of intersection of the function with the ordinate axis, the value of the argument is zero. Consequently, the task turns into finding the value of the function at x \u003d 0. Points of intersection of the function with the Oy axis will be as much as the values \u200b\u200bof the specified function are found at zero argument.

5. To find the intersection points of a given function with another function, it is necessary to solve the system of equations: f \u003d? (X) w \u003d? (X). (X) - an expression describing the specified function f ,? (x) - an expression describing the function W The intersection points with which the specified function is to be detected. Apparently, at the intersection points both functions take equal values \u200b\u200bat equal values \u200b\u200bof the arguments. Universal points in 2 functions will be as many as solutions in the system of equations in the specified section of the definition of the argument.

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At the points of intersection of the function have equal values \u200b\u200bwith the identical value of the argument. Detect points of intersection of functions - it means to determine the coordinates of universal for intersecting functions of points.

Instruction

1. In the general form, the task of finding points of intersection of the functions of one arion y \u003d f (x) and y? \u003d F? (X) on the Xoy plane is reduced to solving the equation y \u003d y?, From the fact that in the universal function of the function have equal values. The values \u200b\u200bof x that satisfy the equality f (x) \u003d f? (X) (if they exist) are the abscissions of the intersection points of the specified functions.

2. If the functions are specified by an uncomplicated mathematical expression and depend on one argument, the task of finding the intersection points is allowed to solve graphically. Build graphics of functions. Determine the intersection points with the coordinate axes (x \u003d 0, y \u003d 0). Set a few more arguments, detect the appropriate values \u200b\u200bof the functions, add the received points on the graphics. The more the points will be used to build, the more right there will be a schedule.

3. If the graphs of the functions intersect, determine the coordinates of the intersection points according to the drawing. To check, substitute these coordinates in the formulas that are set functions. If mathematical expressions are objective, the intersection points are discovered positively. If the functions graphics do not intersect, try to change the scale. Make a step between the points of construction is more kind, in order to determine which section of the lines of the line of graphs is brought together. After that, at the detected crossing section, build more detailed graph with a small step to accurately determine the coordinates of the intersection points.

4. If it is necessary to detect the intersection points of the functions not on the plane, but in the three-dimensional space, it is brought to see the functions of 2 variables: z \u003d f (x, y) and z? \u003d F? (X, y). To determine the coordinates of the points of intersection of functions, it is necessary to solve the system of equations with two unfamiliar x and y at z \u003d z?

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So, the main parameters of the graph of the quadratic function are shown in the figure:

Consider several ways to build an kaardic parabola.Depending on how the quadratic function is specified, you can choose the most convenient one.

1 . The function is defined by formula .

Consider common algorithm for building a chart of a quadratic parabola On the example of building a graph

1 . Direction of parabola branches.

Since, the branches of parabola are directed up.

2 . We find the discriminant square three

The discriminant of square three declared is greater than zero, so Parabola has two intersection points with the axis oh.

In order to find their coordinates, solving the equation:

,

3 . The coordinates of the vertex parabola:

4 . Parabola intersection point with OY axis: (0; -5), and it is symmetrical with a parabola symmetry axis.

We appline these points to the coordinate plane, and connect their smooth curve:

This method can be somewhat simplified.

1. Find the coordinates of the pearabol vertex.

2. Find the coordinates of the points standing on the right and to the left of the vertex.

We use the results of building a schedule

Krrtinates of the peaks parabola

The closest to the top of the point, located on the left of the vertices, have abscissa, respectively, -1; -2; -3

The top points close to the top of the right have abscissa, respectively, 0; 1; 2

Attacking the values \u200b\u200bof the equation of the function, we will find the ordents of these points and bring them to the table:

We will apply these points on the Cordnate Plane and connect the smooth line:

2 . The quadratic function equation is - In this equation - the coordinates of the vertex of parabola

or in the quadratic equation , and the second coefficient is an even number.

Build for example sample function .

Recall the linear conversion of functions graphs. To build a schedule function , need to

§ first build a graph of the function,

§ The same identicals of all points of graphics multiply by 2,

§ then move it along the axis Oh by 1 unit to the right,

§ And then along the Oy axis by 4 units up:

Now consider building a function schedule . In the equation of this function, and the second coefficient is an even number.