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Equation of a parabola whose branches are directed downwards. Parabola - properties and graph of a quadratic function

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Probably everyone knows what a parabola is. Here's how to use it correctly and competently when solving various practical problems, we'll figure it out below.

First, let us outline the basic concepts that algebra and geometry give to this term. Let's consider everything possible types this chart.

Let's find out all the main characteristics of this function. Let's understand the basics of curve construction (geometry). Let's learn how to find the top and other basic values ​​of a graph of this type.

Let's find out: how to correctly construct the desired curve using the equation, what you need to pay attention to. Let's see the basics practical use this unique value in human life.

What is a parabola and what does it look like?

Algebra: This term refers to the graph of a quadratic function.

Geometry: this is a second-order curve that has a number of specific features:

Canonical parabola equation

The figure shows a rectangular coordinate system (XOY), an extremum, the direction of the branches of the function drawing along the abscissa axis.

The canonical equation is:

y 2 = 2 * p * x,

where coefficient p is the focal parameter of the parabola (AF).

In algebra it will be written differently:

y = a x 2 + b x + c (recognizable pattern: y = x 2).

Properties and graph of a quadratic function

The function has an axis of symmetry and a center (extremum). The domain of definition is all values ​​of the abscissa axis.

The range of values ​​of the function – (-∞, M) or (M, +∞) depends on the direction of the branches of the curve. The parameter M here means the value of the function at the top of the line.

How to determine where the branches of a parabola are directed

To find the direction of a curve of this type from an expression, you need to determine the sign before the first parameter of the algebraic expression. If a ˃ 0, then they are directed upward. If it's the other way around, down.

How to find the vertex of a parabola using the formula

Finding the extremum is the main step in solving many practical problems. Of course, you can open special online calculators, but it’s better to be able to do it yourself.

How to determine it? There is a special formula. When b is not equal to 0, we need to look for the coordinates of this point.

Formulas for finding the vertex:

  • x 0 = -b / (2 * a);
  • y 0 = y (x 0).

Example.

There is a function y = 4 * x 2 + 16 * x – 25. Let’s find the vertices of this function.

For a line like this:

  • x = -16 / (2 * 4) = -2;
  • y = 4 * 4 - 16 * 2 - 25 = 16 - 32 - 25 = -41.

We get the coordinates of the vertex (-2, -41).

Parabola displacement

The classic case is when in a quadratic function y = a x 2 + b x + c, the second and third parameters are equal to 0, and = 1 - the vertex is at the point (0; 0).

Movement along the abscissa or ordinate axes is due to changes in the parameters b and c, respectively. The line on the plane will be shifted by exactly the number of units equal to the value of the parameter.

Example.

We have: b = 2, c = 3.

It means that classic look the curve will shift by 2 unit segments along the abscissa axis and by 3 along the ordinate axis.

How to build a parabola using a quadratic equation

It is important for schoolchildren to learn how to correctly draw a parabola according to given parameters.

By analyzing expressions and equations, you can see the following:

  1. The point of intersection of the desired line with the ordinate vector will have a value equal to c.
  2. All points of the graph (along the x-axis) will be symmetrical relative to the main extremum of the function.

In addition, the intersection points with OX can be found by knowing the discriminant (D) of such a function:

D = (b 2 - 4 * a * c).

To do this, you need to equate the expression to zero.

The presence of roots of a parabola depends on the result:

  • D ˃ 0, then x 1, 2 = (-b ± D 0.5) / (2 * a);
  • D = 0, then x 1, 2 = -b / (2 * a);
  • D ˂ 0, then there are no points of intersection with the vector OX.

We get the algorithm for constructing a parabola:

  • determine the direction of the branches;
  • find the coordinates of the vertex;
  • find the intersection with the ordinate axis;
  • find the intersection with the x-axis.

Example 1.

Given the function y = x 2 - 5 * x + 4. It is necessary to construct a parabola. We follow the algorithm:

  1. a = 1, therefore, the branches are directed upward;
  2. extremum coordinates: x = - (-5) / 2 = 5/2; y = (5/2) 2 - 5 * (5/2) + 4 = -15/4;
  3. intersects with the ordinate axis at the value y = 4;
  4. let's find the discriminant: D = 25 - 16 = 9;
  5. looking for roots:
  • X 1 = (5 + 3) / 2 = 4; (4, 0);
  • X 2 = (5 - 3) / 2 = 1; (10).

Example 2.

For the function y = 3 * x 2 - 2 * x - 1 you need to construct a parabola. We act according to the given algorithm:

  1. a = 3, therefore, the branches are directed upward;
  2. extremum coordinates: x = - (-2) / 2 * 3 = 1/3; y = 3 * (1/3) 2 - 2 * (1/3) - 1 = -4/3;
  3. will intersect with the y-axis at the value y = -1;
  4. let's find the discriminant: D = 4 + 12 = 16. So the roots are:
  • X 1 = (2 + 4) / 6 = 1; (1;0);
  • X 2 = (2 - 4) / 6 = -1/3; (-1/3; 0).

Using the obtained points, you can construct a parabola.

Directrix, eccentricity, focus of a parabola

Based on the canonical equation, the focus of F has coordinates (p/2, 0).

Straight line AB is a directrix (a kind of chord of a parabola of a certain length). Its equation: x = -p/2.

Eccentricity (constant) = 1.

Conclusion

We looked at a topic that schoolchildren study in high school. Now you know, looking at the quadratic function of a parabola, how to find its vertex, in which direction the branches will be directed, whether there is a displacement along the axes, and, having a construction algorithm, you can draw its graph.

Definition. Parabola is a set of points on a plane, each of which is at the same distance from a given point, called the focus, and from a given line, called the directrix and not passing through the focus.

Let's place the origin of coordinates in the middle between the focus and the directrix.

Magnitude R(distance from focus to directrix) is called parameter parabolas. Let us derive the canonical equation of the parabola.

From geometric relationships: A.M. = M.F.; A.M. = x + p/2;

M.F. 2 = y 2 + (x – p/2) 2

(x + p/2) 2 = y 2 + (x – p/2) 2

x 2 +xp+p 2 /4 = y 2 +x 2 – xp + p 2 /4

y 2 = 2px(3.7)

Directrix equation: x = - p/2 , focus coordinates F(p/2;0), OhOh ( right ) .

A beam of rays with a source located at the focus, after reflection from a parabola, will turn into a parallel beam of rays. Parabolic mirror antennas are built on this principle.

Depending on the choice of the position of the origin point and the coordinate axes relative to the focus and directrix, three more canonical equations of the parabola can be obtained:

y 2 = -2 px: focus coordinates F(- p/2;0), the center of the parabola is at the origin. Axis of symmetry – axis OhOh(left).

X 2 = 2 py: focus coordinates F(0; p/2), the center of the parabola is at the origin. Axis of symmetry – axis OU, the branches of the parabola are directed in the positive direction of the axis OU(up).

X 2 = -2 py: focus coordinates F(0;- p/2), the center of the parabola is at the origin. Axis of symmetry – axis OU, the branches of the parabola are directed in the negative direction of the axis OU(down).

However, more often you have to deal with the usual parabola equation, known from school:

y = ax 2 + bx + c(3.8) , Where a, b, c – parabola parameters. Graphs for various values ​​of these parameters:

a < 0

a > 0

Typically, several parabolas are used to plot a graph. key points: roots, axis of symmetry, vertex of the parabola, where (up or down) the branches of the parabola are directed, etc. It is assumed that finding these key points from the parabola equation

Example. On a parabola at 2 = 8x find a point whose distance from the directrix is ​​4.

From the parabola equation we find that p = 4.

r = x + p/2 = 4; hence:

x = 2;y 2 = 16;y =4. Searched points: M 1 (2; 4),M 2 (2; -4).

§4. Coordinate systems.

Any point on the plane can be uniquely determined using various coordinate systems, the choice of which is determined by various factors.

Setting method initial conditions to solve any specific practical problem can determine the choice of one or another coordinate system. For ease of calculation, it is often preferable to use coordinate systems other than the Cartesian rectangular system. In addition, the clarity of the presentation of the final answer often also strongly depends on the choice of coordinate system.

Let's consider the so-called polar coordinate system; it is very convenient and is used quite often.

How to build a parabola? There are several ways to graph a quadratic function. Each of them has its pros and cons. Let's consider two ways.

Let's start by plotting a quadratic function of the form y=x²+bx+c and y= -x²+bx+c.

Example.

Graph the function y=x²+2x-3.

Solution:

y=x²+2x-3 is a quadratic function. The graph is a parabola with branches up. Parabola vertex coordinates

From the vertex (-1;-4) we build a graph of the parabola y=x² (as from the origin of coordinates. Instead of (0;0) - vertex (-1;-4). From (-1;-4) we go to the right by 1 unit and up by 1 unit, then left by 1 and up by 1; then: 2 - right, 4 - up, 2 - left, 3 - up; 9 - up, 3 - left, 9 - up If. these 7 points are not enough, then 4 to the right, 16 to the top, etc.).

The graph of the quadratic function y= -x²+bx+c is a parabola, the branches of which are directed downward. To construct a graph, we look for the coordinates of the vertex and from it we construct a parabola y= -x².

Example.

Graph the function y= -x²+2x+8.

Solution:

y= -x²+2x+8 is a quadratic function. The graph is a parabola with branches down. Parabola vertex coordinates

From the top we build a parabola y= -x² (1 - to the right, 1- down; 1 - left, 1 - down; 2 - right, 4 - down; 2 - left, 4 - down, etc.):

This method allows you to build a parabola quickly and does not cause difficulties if you know how to graph the functions y=x² and y= -x². Disadvantage: if the vertex coordinates are fractional numbers, building a graph is not very convenient. If you need to know exact values points of intersection of the graph with the Ox axis, you will have to additionally solve the equation x²+bx+c=0 (or -x²+bx+c=0), even if these points can be directly determined from the drawing.

Another way to construct a parabola is by points, that is, you can find several points on the graph and draw a parabola through them (taking into account that the line x=xₒ is its axis of symmetry). Usually for this they take the vertex of the parabola, the points of intersection of the graph with the coordinate axes and 1-2 additional points.

Draw a graph of the function y=x²+5x+4.

Solution:

y=x²+5x+4 is a quadratic function. The graph is a parabola with branches up. Parabola vertex coordinates

that is, the vertex of the parabola is the point (-2.5; -2.25).

Are looking for . At the point of intersection with the Ox axis y=0: x²+5x+4=0. The roots of the quadratic equation x1=-1, x2=-4, that is, we got two points on the graph (-1; 0) and (-4; 0).

At the point of intersection of the graph with the Oy axis x=0: y=0²+5∙0+4=4. We got the point (0; 4).

To clarify the graph, you can find an additional point. Let's take x=1, then y=1²+5∙1+4=10, that is, another point on the graph is (1; 10). We mark these points on the coordinate plane. Taking into account the symmetry of the parabola relative to the straight line passing through its vertex, we mark two more points: (-5; 6) and (-6; 10) and draw a parabola through them:

Graph the function y= -x²-3x.

Solution:

y= -x²-3x is a quadratic function. The graph is a parabola with branches down. Parabola vertex coordinates

The vertex (-1.5; 2.25) is the first point of the parabola.

At the points of intersection of the graph with the abscissa axis y=0, that is, we solve the equation -x²-3x=0. Its roots are x=0 and x=-3, that is (0;0) and (-3;0) - two more points on the graph. The point (o; 0) is also the point of intersection of the parabola with the ordinate axis.

At x=1 y=-1²-3∙1=-4, that is (1; -4) is an additional point for plotting.

Constructing a parabola from points is a more labor-intensive method compared to the first one. If the parabola does not intersect the Ox axis, more additional points will be required.

Before continuing to construct graphs of quadratic functions of the form y=ax²+bx+c, let us consider the construction of graphs of functions using geometric transformations. It is also most convenient to construct graphs of functions of the form y=x²+c using one of these transformations—parallel translation.

Category: |

As practice shows, tasks on the properties and graphs of a quadratic function cause serious difficulties. This is quite strange, because they study the quadratic function in the 8th grade, and then throughout the first quarter of the 9th grade they “torment” the properties of the parabola and build its graphs for various parameters.

This is due to the fact that when forcing students to construct parabolas, they practically do not devote time to “reading” the graphs, that is, they do not practice comprehending the information received from the picture. Apparently, it is assumed that, after constructing a dozen or so graphs, a smart student himself will discover and formulate the relationship between the coefficients in the formula and appearance graphic arts. In practice this does not work. For such a generalization, serious experience in mathematical mini-research is required, which most ninth-graders, of course, do not possess. Meanwhile, the State Inspectorate proposes to determine the signs of the coefficients using the schedule.

We will not demand the impossible from schoolchildren and will simply offer one of the algorithms for solving such problems.

So, a function of the form y = ax 2 + bx + c called quadratic, its graph is a parabola. As the name suggests, the main term is ax 2. That is A should not be equal to zero, the remaining coefficients ( b And With) can equal zero.

Let's see how the signs of its coefficients affect the appearance of a parabola.

The most simple dependency for the coefficient A. Most schoolchildren confidently answer: “if A> 0, then the branches of the parabola are directed upward, and if A < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой A > 0.

y = 0.5x 2 - 3x + 1

IN in this case A = 0,5

And now for A < 0:

y = - 0.5x2 - 3x + 1

In this case A = - 0,5

Impact of the coefficient With It's also pretty easy to follow. Let's imagine that we want to find the value of a function at a point X= 0. Substitute zero into the formula:

y = a 0 2 + b 0 + c = c. It turns out that y = c. That is With is the ordinate of the point of intersection of the parabola with the y-axis. Typically, this point is easy to find on the graph. And determine whether it lies above zero or below. That is With> 0 or With < 0.

With > 0:

y = x 2 + 4x + 3

With < 0

y = x 2 + 4x - 3

Accordingly, if With= 0, then the parabola will necessarily pass through the origin:

y = x 2 + 4x


More difficult with the parameter b. The point at which we will find it depends not only on b but also from A. This is the top of the parabola. Its abscissa (axis coordinate X) is found by the formula x in = - b/(2a). Thus, b = - 2ax in. That is, we proceed as follows: we find the vertex of the parabola on the graph, determine the sign of its abscissa, that is, we look to the right of zero ( x in> 0) or to the left ( x in < 0) она лежит.

However, that's not all. We also need to pay attention to the sign of the coefficient A. That is, look at where the branches of the parabola are directed. And only after that, according to the formula b = - 2ax in determine the sign b.

Let's look at an example:

The branches are directed upwards, which means A> 0, the parabola intersects the axis at below zero, that is With < 0, вершина параболы лежит правее нуля. Следовательно, x in> 0. So b = - 2ax in = -++ = -. b < 0. Окончательно имеем: A > 0, b < 0, With < 0.