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

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

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

Consider general algorithm for constructing a graph of a quadratic parabola by the example of plotting the function

1 ... The direction of the branches of the parabola.

Since the branches of the parabola are directed upwards.

2 ... Find the discriminant of the square trinomial

The discrimnant of the 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 OY axis: (0; -5), and it is symmetric about the axis of symmetry of the parabola.

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

This method can be simplified somewhat.

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

Crdinates of the vertex of the parabola

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

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

Let us substitute the values ​​of x in the equation of the function, find the ordinates of these points and enter them into the table:

Let's put these points on the coordinate plane and connect 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, for example, a graph of the function .

Let us recall the linear transformations of the graphs of functions. To plot a function , necessary

§ first plot the function graph,

§ then multiply the odds of all points on 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.

Point Finding Tasks intersections 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.

Instructions

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

2. According to textbooks on algebra, 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 straight line on the plane.

3. Point intersections straight and parabolas are the universal point of both curves, therefore, the functions in it will take identical values, that is, 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 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 the detected “xes” are not yet a result for the problem, because a point on the plane is characterized by two real numbers (x, y). For a complete conclusion of the solution, it is necessary to calculate the corresponding “games”. To do this, it is necessary to substitute “xes” either in the function f (x), or in 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, the main thing is to see the solution on 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), that is, x ^ 2-3x + 3 = 2x-3. Moving all the terms to the left, 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 “games”: y1 = g (x1) = 1, y2 = g (x2) = 3. Thus, all points are found intersections: (2.1) and (3.3).

Point intersections straight lines are allowed to be approximately determined according to the schedule. However, the exact coordinates of this point are often necessary, or the graph is not required to be built, then it is allowed to find the point intersections knowing only the equations of the straight lines.

Instructions

1. Let two straight lines be given by the universal equations of a straight line: A1 * x + B1 * y + C1 = 0 and A2 * x + B2 * y + C2 = 0. Point intersections belongs to both one straight line and another. Let us 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, hence y = (A2C1 - 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. In a school mathematics course, straight lines are often given by an equation with an angular index, let's consider this case. Let two straight 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 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 a graph of a power function. Regardless of what collations the parabola has, this function is even. An even function is a function whose value does not change for all values ​​of the argument from the domain when the sign of the argument changes: f (-x) = f (x) Start with the most primitive function: y = x ^ 2. From its form, it is allowed to conclude that it grows with both 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 build a 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 moves the function 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 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 only applies to even functions. parabolas are often present in physical problems, for example, the flight of a body describes a line similar to a parabola. Also view parabolas has a longitudinal section of a headlamp reflector, a lantern. Unlike a sinusoid, this graph is non-periodic and increasing.

Tip 4: How to determine the point of intersection of a straight 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 by means of descriptive geometry and their graphic solution in the drawing.

Instructions

1. Let's see the definition of a point intersections straight with a plane of a private location (Figure 1). Straight l intersects the front-projection plane?. Point them intersections K belongs to and straight and the plane, therefore, 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 link line. Thus, the desired point intersections K (K2K1) is easily constructed without using auxiliary planes. Points intersections straight with all sorts of planes of private location.

2. Let's see the definition of a point intersections straight with a plane of general disposition. In Figure 2, an arbitrarily located plane is given in space? and line l. To locate a point intersections straight with a general location plane, the method of auxiliary cut planes is used in the following order:

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

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

6. Let's apply this rule to solve a specific problem on a complex drawing. Define a point intersections straight l with a plane of general location defined by 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. Line MN is under construction. Plane? intersects AB at point M. Its general projection is marked M2 =? 2? A2B2 and horizontal M1 on A1B1 along the line of projection connection. intersects the side AC at point N. Its general projection is N2 =? 2? A2C2, the horizontal projection of N1 onto A1C1. The straight line MN belongs simultaneously to both 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 plane? ABC: K (K1K2) = l (l1l2)? ? ABC (A1B1C1, A2B2C2) Competing points M, 1 and 2,3 are used to determine visibility straight l relative to a given plane? ABC.

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Note!
Use a construction plane when solving a problem.

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

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

Instructions

1. Take the equations of 2 straight lines, remembering that the equation of a straight line in a Cartesian coordinate system, the equation of a straight line looks like ax + wu + 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 an identical indicator stands in front of y. 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) = (0 * 3) and get the ordinary equation 6x + 3y-12 = 0. If the exponents in front of y were wonderful from unity in both equations, then 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) = 0-0. Because the "-" sign is in front of the parenthesis, change all the characters in the parentheses to the opposite. 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. Move the number 6 to the other side of the equation, and from the resulting equality -2x = -6 express x: x = (- 6) / (- 2). So you got x = 3.

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

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

6. If the straight lines are not given in the form of equations, but are given primitively on a plane, find the coordinates points intersections graphically. To do this, extend the straight lines so that they intersect, then lower the perpendiculars on the oh and oh axes. The intersection of the perpendiculars with the axes oh and oh will be the coordinates of this points, look at the picture 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, which is called the directrix. The vertex of such a parabola passes through the preface of the coordinates, and the curve itself is symmetric about the abscissa axis Ox. In the school algebra course, 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, and = 1 / (2p). It is allowed to draw a parabola by several methods, which can be conventionally called algebraic and geometric.

Instructions

1. Algebraic construction of a parabola. Know the coordinates of the vertex of the parabola. Calculate the coordinate along the Ox axis by the formula: x0 = -b / (2a), and along the Oy axis: y0 = - (b? -4ac) / 4a, or substitute the obtained value x0 into the equation of the parabola y0 = ax0? + Bx0 + c and calculate the value.

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

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

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

5. Combine 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 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, put a point F with a coordinate along the Oy axis equal to y = p / 2 and a point D with a coordinate y = -p / 2.

8. Using a square, draw a line through point D, perpendicular to the axis of symmetry of the parabola. This line is the headmistress of the parabola.

9. Take a thread along a length equal to one of the legs of the square. Fix one end of the thread with a button on the top of the square, to which this leg is adjacent, and the second end - in 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 with the leg free from the button ...

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

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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 invariably taut.

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

Instructions

1. A function is a variable that depends on the value of the argument. The argument is an independent variable. The limits of variation of an argument are referred to as the region of possible values ​​(RVO). The behavior of a function is considered within the framework of the ODV because, within these limits, the relationship between two variables is not chaotic, but obeys certain rules and can be written in the form of a mathematical expression.

2. Consider an arbitrary functional connection F =? (X), where? - mathematical expression. The function can have points of intersection with coordinate axes or with other functions.

3. At the intersection points of the function with the abscissa axis, the function becomes equal to zero: F (x) = 0 Solve this equation. You will get 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 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 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 a given function F,? (X) is an expression describing the function W , the intersection point with which the given function must be found. 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 a 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 the identical value of the argument. To find points of intersection of functions means to determine the coordinates of points that are universal for intersecting functions.

Instructions

1. In its general form, the problem of finding the 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?, Because at the universal point the functions have equal values. The values ​​of x 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. Plot function graphs. Determine the points of intersection with the coordinate axes (x = 0, y = 0). Specify a few more values ​​of the argument, find the corresponding values ​​of the functions, add the obtained points to the graphs. The larger the points will be used for plotting, 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 into the formulas that define the functions. If the mathematical expressions prove to be objective, the intersection points are found positively. If the function graphs do not overlap, try changing the scale. Make the step between the construction points larger in order to determine at which part of the number plane the graph lines converge. After that, on the identified intersection, build a more detailed graph with a small step to accurately determine the 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 discern 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 the system of equations with two unfamiliar x and y at Z = Z ?.

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