Find a b c of a quadratic function. Graph of a quadratic function

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To understand what will be written here, you need to know well what a quadratic function is and what it is eaten with. If you consider yourself a pro at quadratic functions, welcome. But if not, you should read the thread.

Let's start with a small checks:

1. What does a quadratic function look like in general form (formula)?
2. What is the name of the graph of a quadratic function?
3. How does the leading coefficient affect the graph of a quadratic function?

If you can answer these questions right off the bat, keep reading. If at least one question caused difficulties, go to.

So, you already know how to handle a quadratic function, analyze its graph and build a graph by points.

Well, here it is: .

Let's take a quick look at what they do. odds.

1. The senior coefficient is responsible for the “steepness” of the parabola, or, in other words, for its width: the larger, the narrower (steeper) the parabola, and the smaller, the wider (flatter) parabola.
2. The free term is the coordinate of the intersection of the parabola with the y-axis.
3. And the coefficient is somehow responsible for the displacement of the parabola from the center of coordinates. Here's more about that now.

Why do we always start building a parabola? What is her distinguishing point?

This is vertex. And how to find the coordinates of the vertex, remember?

The abscissa is searched for by the following formula:

Like this: what more, topics to the left the top of the parabola moves.

The ordinate of a vertex can be found by substituting into the function:

Substitute yourself and count. What happened?

If you do everything right and simplify the resulting expression as much as possible, you get:

It turns out that the more modulo, topics higher will vertex parabolas.

Finally, let's move on to plotting.
The easiest way is to build a parabola starting from the top.

Example:

Plot the function.

Decision:

First, let's define the coefficients: .

Now let's calculate the vertex coordinates:

And now remember: all parabolas with the same leading coefficient look the same. So, if we build a parabola and move its vertex to a point, we get the graph we need:

Simple, right?

There is only one question left: how to quickly draw a parabola? Even if we draw a parabola with a vertex at the origin, we still have to build it point by point, which is long and inconvenient. But all parabolas look the same, maybe there is a way to speed up their drawing?

When I was in school, my math teacher told everyone to cut out a parabola-shaped stencil out of cardboard so they could draw it quickly. But you won’t be able to walk everywhere with a stencil, and they won’t be allowed to take it to the exam. So, we will not use foreign objects, but we will look for a pattern.

Consider the simplest parabola. Let's build it by points:

The rule here is this. If we move from the top to the right (along the axis) to, and upwards (along the axis) to, then we will get to the point of the parabola. Further: if from this point we move to the right by and up by, we will again get to the point of the parabola. Next: right on and up on. What's next? Right on and up on. And so on: move to the right, and the next odd number up. Then we do the same with the left branch (after all, the parabola is symmetrical, that is, its branches look the same):

Great, this will help build any parabola from the vertex with the highest coefficient equal to. For example, we have learned that the vertex of a parabola is at a point. Construct (on your own, on paper) this parabola.

Built?

It should turn out like this:

Now we connect the obtained points:

That's all.

OK, well, now build only parabolas with?

Of course not. Now let's figure out what to do with them, if.

Let's consider some typical cases.

Great, we learned how to draw a parabola, now let's practice on real functions.

So, draw graphs of such functions:

Answers:

3. Top: .

Do you remember what to do if the senior coefficient is less?

We look at the denominator of the fraction: it is equal. So we'll move like this:

• right - up
• right - up
• right - up

and also to the left:

4. Top: .

Oh, what to do with it? How to measure cells if the vertex is somewhere between the lines?..

And we cheat. First, let's draw a parabola, and only then move its vertex to a point. Not even, let's do it even more tricky: Let's draw a parabola, and then move axes:- on the down, a - on right:

This technique is very convenient in the case of any parabola, remember it.

Let me remind you that we can represent the function in this form:

For example: .

What does this give us?

The fact is that the number that is subtracted from in brackets () is the abscissa of the vertex of the parabola, and the term outside the brackets () is the ordinate of the vertex.

This means that, having built a parabola, you just need to move the axis to the left and the axis to down.

Example: let's plot a function graph.

Let's select a full square:

What number subtracted from in brackets? This (and not how you can decide without thinking).

So, we build a parabola:

Now we shift the axis down, that is, up:

And now - to the left, that is, to the right:

That's all. This is the same as moving a parabola with its vertex from the origin to a point, only the straight axis is much easier to move than a crooked parabola.

Now, as usual, myself:

And do not forget to erase the old axles with an eraser!

I am as answers for verification, I will write you the ordinates of the vertices of these parabolas:

Did everything fit?

If yes, then you are great! Knowing how to handle a parabola is very important and useful, and here we have found that it is not difficult at all.

GRAPHING A QUADRATIC FUNCTION. BRIEFLY ABOUT THE MAIN

quadratic function is a function of the form, where, and are any numbers (coefficients), is a free member.

The graph of a quadratic function is a parabola.

Top of the parabola:
, i.e. the larger \displaystyle b , the more left the top of the parabola moves.
Substitute in the function, and get:
, i.e. the greater \displaystyle b modulo , the higher the top of the parabola will be

The free term is the coordinate of the intersection of the parabola with the y-axis.

Well, the topic is over. If you are reading these lines, then you are very cool.

Because only 5% of people are able to master something on their own. And if you have read to the end, then you are in the 5%!

Now the most important thing.

You've figured out the theory on this topic. And, I repeat, it's ... it's just super! You are already better than the vast majority of your peers.

The problem is that this may not be enough ...

For what?

For the successful passing of the exam, for admission to the institute on the budget and, MOST IMPORTANTLY, for life.

I will not convince you of anything, I will just say one thing ...

People who have received a good education earn much more than those who have not received it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because much more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the exam and be ultimately ... happier?

FILL YOUR HAND, SOLVING PROBLEMS ON THIS TOPIC.

On the exam, you will not be asked theory.

You will need solve problems on time.

And, if you haven’t solved them (LOTS!), you will definitely make a stupid mistake somewhere or simply won’t make it in time.

It's like in sports - you need to repeat many times to win for sure.

Find a collection anywhere you want necessarily with solutions, detailed analysis and decide, decide, decide!

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Tasks on the properties and graphs of a quadratic function, as practice shows, cause serious difficulties. This is rather strange, because the quadratic function is passed in the 8th grade, and then the entire first quarter of the 9th grade is "extorted" by the properties of the parabola and its graphs are built for various parameters.

This is due to the fact that forcing students to build parabolas, they practically do not devote time to "reading" graphs, that is, they do not practice comprehending the information received from the picture. Apparently, it is assumed that, having built two dozen graphs, a smart student himself will discover and formulate the relationship between the coefficients in the formula and the appearance of the graph. In practice, this does not work. For such a generalization, serious experience in mathematical mini-research is required, which, of course, most ninth-graders do not have. Meanwhile, in the GIA they propose to determine the signs of the coefficients precisely according to the schedule.

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

So, a function of the form y=ax2+bx+c is called quadratic, its graph is a parabola. As the name suggests, the main component is ax 2. I.e a should not be equal to zero, the remaining coefficients ( b and with) can be equal to zero.

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

The simplest dependence for the coefficient a. Most schoolchildren confidently answer: "if a> 0, then the branches of the parabola are directed upwards, and if a < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой a > 0.

y = 0.5x2 - 3x + 1

In this case a = 0,5

And now for a < 0:

y = - 0.5x2 - 3x + 1

In this case a = - 0,5

Influence of coefficient with also easy enough to follow. 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. I.e with is the ordinate of the point of intersection of the parabola with the y-axis. As a rule, this point is easy to find on the chart. And determine whether it lies above zero or below. I.e with> 0 or with < 0.

with > 0:

y=x2+4x+3

with < 0

y = x 2 + 4x - 3

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

y=x2+4x

More difficult with the parameter b. The point by 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 \u003d - b / (2a). Thus, b = - 2ax in. That is, we act as follows: on the graph we find the top of the parabola, 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, this is not all. We must also pay attention to the sign of the coefficient a. That is, to see where the branches of the parabola are directed. And only after that, according to the formula b = - 2ax in determine sign b.

Consider an example:

Branches pointing upwards a> 0, the parabola crosses the axis at below zero means 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.

In the lessons of mathematics at school, you have already become acquainted with the simplest properties and the graph of a function y=x2. Let's expand our knowledge quadratic function.

Exercise 1.

Plot a function y=x2. Scale: 1 = 2 cm. Mark a point on the Oy axis F(0; 1/4). Using a compass or strip of paper, measure the distance from the point F to some point M parabolas. Then pin the strip at point M and rotate it around this point so that it becomes vertical. The end of the strip will fall slightly below the x-axis (Fig. 1). Mark on the strip how far it goes beyond the x-axis. Take now another point on the parabola and repeat the measurement again. How much has the edge of the strip now dropped beyond the x-axis?

Result: no matter what point on the parabola y \u003d x 2 you take, the distance from this point to the point F (0; 1/4) will be greater than the distance from the same point to the x-axis always by the same number - by 1/4.

It can be said differently: the distance from any point of the parabola to the point (0; 1/4) is equal to the distance from the same point of the parabola to the line y = -1/4. This wonderful point F(0; 1/4) is called focus parabolas y \u003d x 2, and the straight line y \u003d -1/4 - headmistress this parabola. Each parabola has a directrix and a focus.

Interesting properties of a parabola:

1. Any point of the parabola is equidistant from some point, called the focus of the parabola, and some line, called its directrix.

2. If you rotate a parabola around the axis of symmetry (for example, a parabola y \u003d x 2 around the Oy axis), you get a very interesting surface, which is called a paraboloid of revolution.

The surface of a liquid in a rotating vessel has the shape of a paraboloid of revolution. You can see this surface if you stir hard with a spoon in an incomplete glass of tea, and then remove the spoon.

3. If you throw a stone in the void at a certain angle to the horizon, then it will fly along a parabola (Fig. 2).

4. If you intersect the surface of the cone with a plane parallel to any one of its generators, then in the section you get a parabola (Fig. 3).

5. In amusement parks, they sometimes arrange a funny attraction called the Paraboloid of Wonders. To each of those standing inside the rotating paraboloid, it seems that he is standing on the floor, and the rest of the people, by some miracle, keep on the walls.

6. In mirror telescopes, parabolic mirrors are also used: the light of a distant star, traveling in a parallel beam, falling on the telescope mirror, is collected in focus.

7. For spotlights, the mirror is usually made in the form of a paraboloid. If you place a light source at the focus of a paraboloid, then the rays, reflected from the parabolic mirror, form a parallel beam.

Plotting a Quadratic Function

In the lessons of mathematics, you studied how to get graphs of functions of the form from the graph of the function y \u003d x 2:

1) y=ax2– expansion of the graph y = x 2 along the Oy axis in |a| times (for |a|< 0 – это сжатие в 1/|a| раз, rice. 4).

2) y=x2+n– graph shift by n units along the Oy axis, and if n > 0, then the shift is up, and if n< 0, то вниз, (или же можно переносить ось абсцисс).

3) y = (x + m)2– graph shift by m units along the Ox axis: if m< 0, то вправо, а если m >0, then to the left, (Fig. 5).

4) y=-x2- symmetrical display about the Ox axis of the graph y = x 2 .

Let's dwell on plotting a function graph in more detail. y = a(x - m) 2 + n.

A quadratic function of the form y = ax 2 + bx + c can always be reduced to the form

y \u003d a (x - m) 2 + n, where m \u003d -b / (2a), n \u003d - (b 2 - 4ac) / (4a).

Let's prove it.

Really,

y = ax 2 + bx + c = a(x 2 + (b/a) x + c/a) =

A(x 2 + 2x (b/a) + b 2 /(4a 2) - b 2 /(4a 2) + c/a) =

A((x + b/2a) 2 - (b 2 - 4ac)/(4a 2)) = a(x + b/2a) 2 - (b 2 - 4ac)/(4a).

Let us introduce new notation.

Let be m = -b/(2a), a n \u003d - (b 2 - 4ac) / (4a),

then we get y = a(x - m) 2 + n or y - n = a(x - m) 2 .

Let's make some more substitutions: let y - n = Y, x - m = X (*).

Then we get the function Y = aX 2 , whose graph is a parabola.

The vertex of the parabola is at the origin. x=0; Y = 0.

Substituting the coordinates of the vertex in (*), we obtain the coordinates of the vertex of the graph y = a(x - m) 2 + n: x = m, y = n.

Thus, in order to plot a quadratic function represented as

y = a(x - m) 2 + n

by transformation, you can proceed as follows:

a) build a graph of the function y = x 2 ;

b) by parallel translation along the Ox axis by m units and along the Oy axis by n units - transfer the top of the parabola from the origin to the point with coordinates (m; n) (Fig. 6).

Write transformations:

y = x 2 → y = (x - m) 2 → y = a(x - m) 2 → y = a(x - m) 2 + n.

Example.

Using transformations, construct a graph of the function y = 2(x - 3) 2 in the Cartesian coordinate system – 2.

Decision.

Chain of transformations:

y=x2 (1) → y = (x - 3) 2 (2) → y = 2(x – 3) 2 (3) → y = 2(x - 3) 2 - 2 (4) .

The construction of the graph is shown in rice. 7.

You can practice quadratic function plotting by yourself. For example, build a graph of the function y = 2(x + 3) 2 + 2 in one coordinate system using transformations. If you have any questions or want to get advice from a teacher, then you have the opportunity to conduct free 25-minute lesson with an online tutor after registration . For further work with the teacher, you can choose the tariff plan that suits you.

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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 like y=x²+bx+c and y= -x²+bx+c.

Example.

Plot the function y=x²+2x-3.

Decision:

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. Instead of (0;0) - the vertex (-1;-4). From (-1;-4) we go to the right by 1 unit and up by 1, then left by 1 and up by 1, then: 2 - right, 4 - up, 2 - left, 4 - up, 3 - right, 9 - up, 3 - left, 9 - up. these 7 points are not enough, then - 4 to the right, 16 - up, etc.).

The graph of the quadratic function y= -x²+bx+c is a parabola whose branches are directed downwards. To build a graph, we are looking for the coordinates of the vertex and from it we build a parabola y= -x².

Example.

Plot the function y= -x²+2x+8.

Decision:

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 - 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 plot the functions y=x² and y= -x². Disadvantage: if the vertex coordinates are fractional numbers, plotting is not very convenient. If you want to know the exact values ​​of the intersection points of the graph with the x-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 figure.

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

Plot the function y=x²+5x+4.

Decision:

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

that is, the top 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 \u003d -1, x2 \u003d -4, that is, they received two points on the graph (-1; 0) and (-4; 0).

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

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

Plot the function y= -x²-3x.

Decision:

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

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

At the points of intersection of the graph with the x-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) are two more points on the graph. The point (o; 0) is also the point of intersection of the parabola with the y-axis.

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

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

Before we continue plotting quadratic functions of the form y=ax²+bx+c, let's consider plotting functions using geometric transformations. Graphs of functions of the form y=x²+c are also most convenient to build using one of these transformations - parallel translation.