Point Finding Tasks intersections any figures are ideologically primitive. Difficulties in them are only because of arithmetic, because it is in it that various typos and errors are made.

Instruction

1. This problem is solved analytically, therefore it is allowed not to draw graphics at all straight and parabolas. Often this gives a huge plus in solving an example, because such functions can be given in the problem that it is easier and faster not to draw them.

2. According to algebra textbooks, a parabola is given by a function of the form f(x)=ax^2+bx+c, where a,b,c are real numbers, and the exponent a is good at zero. The function g(x)=kx+h, where k,h are real numbers, defines a line in the plane.

3. Dot intersections straight and parabolas are the universal point of both curves, hence the functions in it will take identical values, i.e. f(x)=g(x). This statement allows you to write the equation: ax^2+bx+c=kx+h, which will give the probability of finding a lot of points intersections .

4. In the equation ax^2+bx+c=kx+h, you need to move all the terms to the left side and bring similar ones: ax^2+(b-k)x+c-h=0. Now it remains to solve the resulting quadratic equation.

5. All detected “xes” are not yet the result for the task, because a point on the plane is characterized by two real numbers (x, y). For the complete conclusion of the solution, it is necessary to calculate the corresponding “games”. To do this, it is necessary to substitute the “xes” either into the function f (x) or into the function g (x), tea for the point intersections correct: y=f(x)=g(x). Later on you will find all the universal points of the parabola and straight .

6. To consolidate the material, it is very important to see the solution with an example. Let the parabola be given by the function f(x)=x^2-3x+3, and the straight line - g(x)=2x-3. Write the equation f(x)=g(x), i.e. x^2-3x+3=2x-3. Transferring all the terms to the left side, and bringing similar ones, you get: x^2-5x+6=0. The roots of this quadratic equation: x1=2, x2=3. Now find the corresponding "players": y1=g(x1)=1, y2=g(x2)=3. Thus, all points are found intersections: (2,1) and (3,3).

point intersections straight lines can be approximately determined from the graph. However, the exact coordinates of this point are often needed, or it is not necessary to build a graph, then it is possible to find the point intersections knowing only the equations of the lines.

Instruction

1. Let two lines be given by the general equations of the line: A1*x + B1*y + C1 = 0 and A2*x + B2*y + C2 = 0. Point intersections belongs to one straight line and the other. We express from the first equation of the line x, we get: x = -(B1*y + C1)/A1. Substitute the resulting value into the second equation: -A2*(B1*y + C1)/A1 + B2*y + C2 = 0 A1C2)/(A1B2 – A2B1). Substitute the detected value into the equation of the first straight line: A1*x + B1(A2C1 – A1C2)/(A1B2 – A2B1) + C1 = 0.A1(A1B2 – A2B1)*x + A2B1C1 – A1B1C2 + A1B2C1 – A2B1C1 = 0(A1B2 – A2B1)*x - B1C2 + B2C1 = 0 Then x = (B1C2 - B2C1)/(A1B2 - A2B1).

2. AT school course mathematicians, straight lines are often given by an equation with an angular exponent, consider this case. Let two lines be given in this way: y1 = k1*x + b1 and y2 = k2*x + b2. Apparently, at the point intersections y1 = y2, then k1*x + b1 = k2*x + b2. We get that the ordinate of the point intersections x = (b2 – b1)/(k1 – k2). Substitute x into any equation of a straight line and get y = k1(b2 – b1)/(k1 – k2) + b1 = (k1b2 – b1k2)/(k1 – k2).

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The equation parabolas is a quadratic function. There are several options for compiling this equation. It all depends on what parameters are presented in the condition of the problem.

Instruction

1. A parabola is a curve that resembles an arc in shape and is a graph power function. Regardless of which collations the parabola has, this function is even. An even function is such a function that, for all values ​​of the argument from the domain of definition, when the sign of the argument changes, the value does not change: f (-x) \u003d f (x) Start with the most primitive function: y \u003d x ^ 2. From its appearance, it is possible to conclude that it grows both for correct and for negative values ​​of the argument x. The point at which x=0, and at the same time, y = 0 is considered the minimum point of the function.

2. Below are all the main options for constructing this function and its equation. As a first example, below is a function of the form: f(x)=x^2+a, where a is an integer In order to plot the graph of this function, you need to shift the graph of the function f(x) by a units. An example is the function y=x^2+3, where the y-axis shifts the function up by two units. Given a function with the opposite sign, say y=x^2-3, then its graph is shifted down the y-axis.

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

4. There are also parabolic dependences of the form y=x^4. In such cases, x=const, and y rises steeply. However, this applies only to even functions. Graphs parabolas often present in physical problems, say, the flight of the body describes a line similar to a parabola. Also view parabolas has a longitudinal section of the headlight reflector, lamp. Unlike a sine wave, this graph is non-periodic and progressive.

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

This task is to build a point intersections straight with a plane is a classic in the course of engineering graphics and is performed using the methods of descriptive geometry and their graphic solution in the drawing.

Instruction

1. Consider the definition of a point intersections straight with a private location plane (Figure 1). The straight line l intersects the frontal projection plane?. Point them intersections K belongs to and straight and the plane, so the common projection of K2 lies on?2 and l2. That is, K2= l2??2, and its horizontal projection K1 is determined on l1 using the projection connection line. Thus, the desired point intersections K(K2K1) is constructed freely without the use of auxiliary planes. Points are defined similarly intersections straight with all sorts of private planes.

2. Consider the definition of a point intersections straight with the general plane. In Figure 2, an arbitrarily located plane is given in space? and straight line l. To define a point intersections straight with a general location plane, the method of auxiliary cutting planes is used in the following order:

3. An auxiliary cutting plane is drawn through the straight line l?. To facilitate construction, this will be the projecting plane.

5. Point K is marked intersections straight l and the constructed line intersections MN. She is the desired point intersections straight and planes.

6. Let's apply this rule to solve a specific problem in a complex drawing. Example. Define point intersections straight l with the plane of general location, given by the triangle ABC (Figure 3).

7. An auxiliary secant plane? is drawn through the straight line l, perpendicular to the projection plane?2. Its projection?2 coincides with the projection straight l2.

8. The MN line is under construction. Plane? intersects AB at point M. Its common projection M2= ?2?A2B2 and horizontal M1 onto A1B1 are marked along the line of projection connection. Plane? crosses side AC at point N. Its common projection is N2=?2?A2C2, the horizontal projection of N1 onto A1C1. The line MN belongs to both planes at the same time, and, therefore, is their line intersections .

9. Point K1 is determined intersections l1 and M1N1, after that point K2 is built with the support of the communication line. It turns out that K1 and K2 are projections of the desired point intersections K straight l and planes? ABC:K(K1K2)=l(l1l2)? ? ABC(A1B1C1, A2B2C2). Using the competing points M,1 and 2,3, the visibility is determined straight l about a given plane? ABC.

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Note!
Use an auxiliary plane when solving a problem.

Useful advice
Perform calculations using detailed drawings that match the requirements of the problem. This will help you quickly navigate the solution.

Two lines, if they are not parallel and do not coincide, strictly intersect at one point. Finding the coordinates of this place means calculating points intersections direct. Two intersecting lines invariably lie in the same plane, so it is enough to see them in the Cartesian plane. Let's look at an example of how to find the universal point of lines.

Instruction

1. Take the equations of 2 lines, remembering that the equation of a line in the Cartesian coordinate system, the equation of a line looks like ax + y + c \u003d 0, and a, b, c are ordinary numbers, and x and y are the coordinates of points. For example, find points intersections straight lines 4x+3y-6=0 and 2x+y-4=0. To do this, find the solution to the system of these 2 equations.

2. To solve a system of equations, change any of the equations so that the y is preceded by an identical exponent. Because in one equation, the exponent in front of y is 1, then primitively multiply this equation by the number 3 (the exponent in front of y in another equation). To do this, multiply each element of the equation by 3: (2x * 3) + (y * 3) - (4 * 3) \u003d (0 * 3) and get an ordinary equation 6x + 3y-12 \u003d 0. If the exponents in front of y were wonderful from unity in both equations, both equalities would have to be multiplied.

3. Subtract the other from one equation. To do this, subtract from the left side of one the left side of the other and do the same with the right. Get this expression: (4x + 3y-6) - (6x + 3y-12) \u003d 0-0. Because there is a “-” sign in front of the bracket, change all the signs in brackets to the opposite ones. Get this expression: 4x + 3y-6 - 6x-3y + 12 = 0. Simplify the expression and you will see that the variable y has disappeared. The new equation looks like this: -2x+6=0. Transfer the number 6 to another part of the equation, and from the resulting equality -2x \u003d -6, express x: x \u003d (-6) / (-2). So you got x=3.

4. Substitute the value of x=3 in any equation, say, in the second and get the following expression: (2 * 3) + y-4 = 0. Simplify and express y: y=4-6=-2.

5. Write the resulting x and y values ​​as coordinates points(3;-2). These will be the solution to the problem. Check the value obtained by substituting into both equations.

6. If the lines are not given as equations, but are given primitively on the plane, find the coordinates points intersections graphically. To do this, extend the lines so that they intersect, then lower the perpendiculars on the x and y axes. The intersection of perpendiculars with the x and y axes will be the coordinates of this points, look at the figure and you will see that the coordinates points intersections x \u003d 3 and y \u003d -2, that is, the point (3; -2) is the solution to the problem.

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A parabola is a second-order plane curve whose canonical equation in the Cartesian coordinate system is y?=2px. Where p is the focal parameter of the parabola, equal to the distance from a fixed point F, called the focus, to a fixed line D in the same plane, called the directrix. The vertex of such a parabola passes through the preface of the coordinates, and the curve itself is symmetrical about the abscissa axis Ox. In the school course of algebra, it is customary to consider a parabola, the symmetry axis of which coincides with the ordinate axis Oy: x?=2py. And the equation is written somewhat opposite: y=ax?+bx+c, a=1/(2p). It is possible to draw a parabola by several methods, conditionally which can be called algebraic and geometric.

Instruction

1. Algebraic construction of a parabola. Find out the coordinates of the vertex of the parabola. Calculate the coordinate along the Ox axis using the formula: x0=-b/(2a), and along the Oy axis: y0=-(b?-4ac)/4a or substitute the resulting x0 value into the equation of the parabola y0=ax0?+bx0+c and calculate the value.

2. On the coordinate plane, construct the axis of symmetry of the parabola. Its formula coincides with the formula for the x0 coordinate of the parabola vertex: x=-b/(2a). Determine where the branches of the parabola point. If a>0, then the axes are directed upwards, if a

3. Take arbitrarily 2-3 values ​​for parameter x so that: x0

4. Place points 1', 2', and 3' so that they are symmetrical to points 1, 2, 3 about the axis of symmetry.

5. Unite points 1', 2', 3', 0, 1, 2, 3 with a smooth oblique line. Continue the line up or down, depending on the direction of the parabola. The parabola is built.

6. Geometric construction parabolas. This method is based on the definition of a parabola as a community of points equidistant from both the focus F and the directrix D. Therefore, first find the focal parameter of the given parabola p=1/(2a).

7. Construct the axis of symmetry of the parabola as described in step 2. On it, put a point F with a coordinate along the Oy axis equal to y \u003d p / 2 and a point D with a coordinate y \u003d -p / 2.

8. Using a square, construct a line passing through point D, perpendicular to the axis of symmetry of the parabola. This line is the directrix of the parabola.

9. Take the thread along the length equal to one of the legs of the square. Fasten one end of the thread with the button at the top of the square to which this leg adjoins, and the 2nd end at the focus of the parabola at point F. Put the ruler so that it upper edge coincided with directrix D. Place a square on the ruler with a leg free from the button.

10. Set the pencil so that with its tip it presses the thread to the leg of the square. Move the square along the ruler. The pencil will draw the parabola you need.

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Note!
Don't draw the top of the parabola as an angle. Its branches converge with each other, smoothly rounding.

Useful advice
When constructing a parabola by the geometric method, make sure that the thread is always taut.

Before proceeding to the search for the behavior of a function, it is necessary to determine the area of ​​metamorphosis of the quantities under consideration. Let us assume that the variables refer to the set of real numbers.

Instruction

1. A function is a variable that depends on the value of the argument. The argument is an independent variable. The limits of change in an argument are called the domain of possible values ​​(ROV). The behavior of the function is considered within the framework of the ODZ because, within these limits, the connection between two variables is not chaotic, but obeys certain rules and can be written as a mathematical expression.

2. Let's consider arbitrary functional connectivity F=?(x), where? is a mathematical expression. The function can have points of intersection with the coordinate axes or with other functions.

3. At the points of intersection of the function with the x-axis, the function becomes equal to zero: F(x)=0. Solve this equation. You will get the coordinates of the intersection points given function with the OX axis. There will be as many such points as there are roots of the equation in a given section of the metamorphosis of the argument.

4. At the intersection points of the function with the y-axis, the argument value is zero. Consequently, the problem turns into finding the value of the function at x=0. There will be as many points of intersection of the function with the OY axis as there are values ​​of the given function with a zero argument.

5. To find the points of intersection of a given function with another function, you need to solve the system of equations: F=?(x)W=?(x). , the intersection points with which the given function needs to be detected. Apparently, at the intersection points, both functions take equal values ​​for equal values ​​of the arguments. There will be as many universal points for 2 functions as there are solutions for the system of equations in a given area of ​​argument changes.

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At the points of intersection, the functions have equal values ​​for the identical value of the argument. To find the points of intersection of functions means to determine the coordinates of points that are universal for intersecting functions.

Instruction

1. In general, the problem of finding intersection points of functions of one argument Y=F(x) and Y?=F?(x) on the XOY plane is reduced to solving the equation Y= Y?, from the fact that at a universal point the functions have equal values. The x values ​​satisfying the equality F(x)=F?(x), (if they exist) are the abscissas of the intersection points of the given functions.

2. If the functions are given by a simple mathematical expression and depend on one argument x, then the problem of finding intersection points can be solved graphically. Plot function graphs. Determine the points of intersection with the coordinate axes (x=0, y=0). Set a few more argument values, find the corresponding function values, add the obtained points to the graphs. The more points will be used to plot, the more accurate the graph will be.

3. If the graphs of the functions intersect, determine the coordinates of the intersection points from the drawing. To check, substitute these coordinates in the formulas that define the functions. If the mathematical expressions turn out to be objective, the intersection points are found positively. If the function graphs do not intersect, try rescaling. Take a larger step between the construction points in order to determine in which part of the numerical plane the lines of the graphs converge. After that, on the identified section of the intersection, build a more detailed graph with a small step for exact definition coordinates of the intersection points.

4. If it is necessary to find the intersection points of functions not on the plane, but in three-dimensional space, it is possible to see the functions of 2 variables: Z=F(x,y) and Z?=F?(x,y). To determine the coordinates of the intersection points of functions, it is necessary to solve a system of equations with two unknown x and y at Z= Z?.

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

Consider several ways to construct a quadratic parabola. Depending on how the quadratic function is given, you can choose the most convenient one.

1 . The function is given by the formula .

Consider general algorithm plotting a quadratic parabola on the example of plotting a function graph

1 . The direction of the branches of the parabola.

Since the branches of the parabola are directed upwards.

2 . Find the discriminant of a square trinomial

The discriminant of a square trinomial is greater than zero, so the parabola has two points of intersection with the OX axis.

In order to find their coordinates, we solve the equation:

,

3 . Parabola vertex coordinates:

4 . The point of intersection of the parabola with the axis OY: (0;-5), and it is symmetrical about the axis of symmetry of the parabola.

Let's put these points on coordinate plane, and connect them with a smooth curve:

This method can be somewhat simplified.

1. Find the coordinates of the vertex of the parabola.

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

Let's use the results of plotting the function graph

Vertices of the parabola

The points closest to the top, located to the left of the top, have abscissas, respectively -1; -2; -3

The points closest to the top, located on the right, have abscissas, respectively, 0; 1; 2

Substitute the values ​​of x into the equation of the function, find the ordinates of these points and put them in the table:

Let's put these points on the coordinate plane and connect them with a smooth line:

2 . The quadratic function equation has the form - in this equation - the coordinates of the vertex of the parabola

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

For example, let's build a graph of the function .

Let's remember linear transformations function graphs. To plot a function , need

§ first plot the function ,

§ then multiply all points of the graph by 2,

§ then shift it along the OX axis by 1 unit to the right,

§ and then along the OY axis 4 units up:

Now let's look at plotting the function . In the equation of this function, and the second coefficient is an even number.