Problems of finding points intersections any figures are ideologically primitive. Difficulties in them occur only because of arithmetic, because it is in this that various typos and errors are made.

Instructions

1. This problem is solved analytically, and therefore it is possible not to draw graphs at all straight and parabolas. Often this gives a huge advantage in solving an example, because the problem may contain such functions that it would be 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, it is zero. The function g(x)=kx+h, where k,h are real numbers, defines a line on the plane.

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

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

5. All detected “X’s” are not yet a result for the problem, because a point on the plane is characterized by two real numbers (x,y). To fully conclude the solution, you need to calculate the corresponding “players”. To do this, you need to substitute “x’s” either into the function f(x) or into the function g(x), tea for the point intersections correct: y=f(x)=g(x). Later you will discover all the universal points of the parabola and straight .

6. To consolidate the material, it is very important to see the solution using an example. Let the parabola be defined by the function f(x)=x^2-3x+3, and the straight line – g(x)=2x-3. Make up the equation f(x)=g(x), that is, x^2-3x+3=2x-3. Moving all the terms to the left side and bringing similar ones, you get: x^2-5x+6=0. The roots of this quadratic equation are: x1=2, x2=3. Now find the corresponding “players”: y1=g(x1)=1, y2=g(x2)=3. Thus, all points are detected intersections: (2,1) and (3,3).

Full stop intersections straight lines can be approximately determined from the graph. However, the exact coordinates of this point are often needed or there is no need to build a graph, then you can detect the point intersections, knowing only the equations of lines.

Instructions

1. Let two lines be defined by the universal equations of the line: A1*x + B1*y + C1 = 0 and A2*x + B2*y + C2 = 0. Point intersections belongs to both one line and another. Let's express the straight line x from the first equation, we get: x = -(B1*y + C1)/A1. Substitute the resulting value into the second equation: -A2*(B1*y + C1)/A1 + B2*y + C2 = 0. Or -A2B1*y – A2C1 + A1B2*y + A1C2 = 0, otherwise y = (A2C1 – A1C2)/(A1B2 – A2B1). Let's 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 = 0Then x = (B1C2 – B2C1)/(A1B2 – A2B1).

2. In school mathematics courses, straight lines are often given by an equation with an angular exponent; let’s look at 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 find that the ordinate of the point intersections x = (b2 – b1)/(k1 – k2). Substitute x into any equation of the 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 constructing this equation. It all depends on what parameters are presented in the problem statement.

Instructions

1. A parabola is a curve that resembles an arc in shape and is the graph of a power function. Regardless of what collations the parabola has, this function is even. An even function is one in which, 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) = f (x) Start with the most primitive function: y = x^2. From its appearance we can conclude that it increases both with correct and 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 we consider 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 along the y-axis the function is shifted up by two units. If a function with the opposite sign is given, say y=x^2-3, then its graph is shifted down along the y-axis.

3. Another type of function that can be given a parabola is f(x)=(x +a)^2. In such cases, the graph, on the contrary, shifts along the abscissa (x-axis) by a units. For example, you can 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 dependencies of the form y=x^4. In such cases, x=const, and y rises steeply. However, this only applies to even functions.Graphics parabolas are often present in physical problems, for example, the flight of a body describes a line that is similar to a parabola. Also view parabolas has a longitudinal section of the headlight reflector, lantern. Unlike a sinusoid, this graph is non-periodic and increasing.

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

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

Instructions

1. Let's look at the definition of a point intersections straight with the plane of particular location (Figure 1). The straight line l intersects the frontal projecting plane?. Point them intersections K belongs to straight and plane, which means that the general 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) can be constructed easily without using auxiliary planes. Points are determined in a similar way intersections straight with all sorts of planes of particular arrangement.

2. Let's look at the definition of a point intersections straight with the plane of universal location. In Figure 2, arbitrarily located planes are given in space? and straight line l. To determine a point intersections straight with a plane of universal location, the method of auxiliary cutting planes is used in the following order:

3. An auxiliary cutting plane is drawn through 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 universal location defined by the triangle ABC (Figure 3).

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

8. The MN line is being built. Plane? intersects AB at point M. Its general projection M2 = ?2?A2B2 and horizontal M1 on A1B1 along the line of projection connection are noted. Plane? intersects side AC at point N. Its general projection is N2 =? intersections .

9. Point K1 is determined intersections l1 and M1N1, after which 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). Visibility is determined using competing points M,1 and 2,3 straight l tangent to this plane? ABC.

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Note!
Use an auxiliary plane when solving a problem.

Helpful advice
Perform calculations using detailed drawings appropriate to the conditions of the problem. This will help you quickly navigate the decision.

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 straight Two intersecting lines invariably lie in the same plane, therefore it is enough to see them in the Cartesian plane. Let's look at an example of how to find the universal point of a straight line.

Instructions

1. Take the equations of 2 straight lines, remembering that the equation of a straight line in the Cartesian coordinate system looks like ax + y + c = 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 each of the equations so that the y is preceded by an identical exponent. Because in one equation the exponent before y is 1, then simply multiply this equation by the number 3 (the exponent before y in another equation). To do this, multiply each element of the equation by 3: (2x*3)+(y*3)-(4*3)=(0*3) and get the ordinary equation 6x+3y-12=0. If the exponents in front of y were wonderful from one in both equations, both equalities would need to be multiplied.

3. Subtract one equation from the other. To do this, subtract from the left side of one the left side of the other and do the same with the right. Get the following expression: (4x+3y-6) – (6x+3y-12)=0-0. Because there is a “-” sign in front of the bracket, change all the signs in the brackets to the opposite. Get the following 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. Move the number 6 to another part of the equation, and from the resulting equality -2x=-6, express x: x=(-6)/(-2). So you get x=3.

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

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

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

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A parabola is a plane curve of the second order, the canonical equation of which in the Cartesian coordinate system has the form 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 straight 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 x-axis Ox. In a school algebra course, it is customary to consider a parabola whose axis of symmetry coincides with the ordinate axis Oy: x? = 2py. And the equation is written somewhat opposite: y=ax?+bx+c, a=1/(2p). You can draw a parabola using several methods, which can be called algebraic and geometric.

Instructions

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 parabola equation 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 coordinate x0 of the vertex of the parabola: x=-b/(2a). Determine where the branches of the parabola are directed. If a>0, then the axes are directed upward, 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 regarding the axis of symmetry.

5. Connect 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 has been built.

6. Geometric construction of a parabola. 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, place point F with a coordinate along the Oy axis equal to y=p/2 and point D with coordinate y=-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 a thread of length equal to one of the legs of the square. Fasten one end of the thread with a button to the top of the square to which this leg is adjacent, and the 2nd end - at the focus of the parabola at point F. Place the ruler so that its upper edge coincides with the directrix D. Place the square on the ruler, the leg free from the button .

10. Position the pencil so that its tip presses the thread against the side of the square. Move the square along the ruler. The pencil will draw the parabola you need.

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Note!
Do not draw the vertex of the parabola as an angle. Its branches converge with each other, smoothly rounding.

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

Before starting to study the behavior of a function, it is necessary to determine the region of metamorphosis of the quantities under consideration. Let us accept the assumption that the variables belong to the set of real numbers.

Instructions

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

2. Let's consider an arbitrary functional connection F=?(x), where? – mathematical expression. A function can have intersection points with 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 receive the coordinates of the points of intersection of the 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 points of intersection of the function with the y-axis, the value of the argument 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 at zero argument.

5. To find the intersection points of a given function with another function, you need to solve the system of equations: F=?(x)W=?(x). Here?(x) is an expression describing the given function F, ?(x) is an expression describing the function W , the intersection points with which a given function must be detected. Apparently, at the points of intersection, both functions take equal values ​​with 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 ​​changes in the argument.

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At the points of intersection, the functions have equal values ​​with an identical value of the argument. To discover the points of intersection of functions means to determine the coordinates of points common to intersecting functions.

Instructions

1. In general terms, the problem of finding the intersection points of the functions of one argument Y=F(x) and Y?=F?(x) on the XOY plane is reduced to solving the equation Y=Y?, since at the 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 the intersection points can be solved graphically. Construct graphs of functions. Determine the intersection points with the coordinate axes (x=0, y=0). Set a few more argument values, find the corresponding function values, and add the resulting points to the graphs. The more points are used for construction, the more accurate the graph will be.

3. If the graphs of functions intersect, determine the coordinates of the intersection points from the drawing. To check, substitute these coordinates into the formulas that define the functions. If the mathematical expressions turn out to be objective, the intersection points are positively detected. If the function graphs do not intersect, try changing the scale. Take a larger step between the construction points in order to determine in which part of the numerical plane the graph lines come closer together. After this, on the identified intersection area, construct a more detailed graph with small steps to accurately determine the coordinates of the intersection points.

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

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

Let's consider several ways to construct a quadartic parabola. Depending on how the quadratic function is specified, you can choose the most convenient one.

1 . The function is given by the formula .

Let's consider general algorithm for plotting a quadratic parabola using the example of plotting a function

1 . The direction of the branches of the parabola.

Since the branches of the parabola are directed upward.

2 . Let's find the discriminant of a quadratic trinomial

The discriminant of a quadratic 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 OY axis: (0;-5), and it is symmetrical with respect to the symmetry axis of the parabola.

Let's plot these points on the 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

Vertex coordinates of a parabola

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

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

Let's plug the x values ​​into the equation of the function, find the ordinates of these points and enter them into the table:

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

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

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

Let's build a graph of the function as an example .

Let us recall linear transformations of function graphs. To plot a function , need to

§ first build a graph of the function,

§ then multiply the same values ​​of all points on the graph by 2,

§ then move it along the OX axis 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.