Plotting a quadratic function. Visual Guide (2019). We schematically draw a parabola according to the original expression

A parabola was built on a sheet of paper - a graph of the function y=ax 2 +bx+c for a>0, b>0 and c>0, and the coordinate axes were erased. How could they be located? (Draw any example corresponding to the indicated signs of the coefficients, without changing the position of the parabola itself.)

Answer: see fig. 10.1.

Decision

Since a>0, then the branches of the parabola are “opened” along the positive direction of the y-axis. Since c>0, the point of intersection of the graph with the ordinate axis has a negative ordinate. Since -b/2a<0, то вершина параболы находится в полуплоскости x<0.

Verification Criteria

• “+” - the correct figure is given without explanations, or the correct figure with correct explanations
• “±” - the correct figure is shown, to which explanations are given that contain errors
• “±” – a correct figure is shown without explanations, or a correct figure with correct explanations, but the orientation of the coordinate system is changed on it (the rotation from the OX beam to the OY beam is clockwise)
• “ ” - the correct figure is shown, but the position of the parabola has been changed (it is upside down)
• “ ” - the figure is incorrect, but the y-axis is correctly directed

Task 2

The sum of two integers is equal to S . Masha multiplied the left number by an integer a, the right one by an integer b, added these products and found that the resulting sum is divisible by S. Alyosha, on the contrary, multiplied the left number by b, and the right one by a. Prove that his sum is also divisible by S .

Decision

Let x be the left number and y the right number; by condition: x+y=S. Then Masha got the number ax+by, and Alyosha got the number bx+ay. The sum of these numbers is ax+by+bx+ay=(a+b)(x+y)=(a+b)S, that is, it is divisible by S . Since one of the two terms (Masha's number) is divisible by S, the other (Alyosha's number) is also divisible by S, as required.

Verification Criteria

• “–” – problem not solved or solved incorrectly

Task 3

The zoo has 10 elephants and a huge weighbowl. It is known that if any four elephants stand on the left bowl and any three on the right, the left bowl will outweigh. Three elephants stood on the left bowl and two on the right. Does the left bowl necessarily outweigh?

Answer : necessarily.

Decision

First way

Let three elephants stand on the left pan of the scale, and two on the right, and at the same time the left pan did not outweigh the right one. Let us then ask the lightest of the five elephants not standing on the scales to stand on the left bowl, and the heaviest on the right. In this case, the left bowl still cannot outweigh the right one, which contradicts the condition. Therefore, the left bowl will definitely outweigh.

Second way

Let's write the masses of elephants in ascending order: m1 ≤ m2 ≤ … ≤ m10. By condition: m1 + m2 + m3 + m4 > m8 + m9 + m10. Since m4 ≤ m8, then m1 + m2 + m3 > m9 + m10. Thus, the three lightest elephants are heavier than the two heaviest ones, therefore, any three elephants are heavier than any two of the remaining ones.

Verification Criteria

• “+” – a complete justified solution is given (by any means)
• “±” – generally correct reasoning is given, containing minor gaps or inaccuracies
• “–” – only special cases or specific examples are considered
• “–” – problem not solved or solved incorrectly

Task 4

From the vertex of the obtuse angle A of the triangle ABC the height AD is lowered. A circle with center D and radius DA is drawn, which intersects sides AB and AC for the second time at points M and N, respectively. Find AC if AB = c, AM = m and AN = n.

∙Answer: mc/n.

Decision

Let us prove that AM∙AB = AN∙AC. This can be done in different ways.

First way

In right-angled triangles ADB and ADC, we draw the heights DP and DQ, respectively (see Fig. 10.4a). Then АР∙AB = AD2 = AQ∙AC. Since triangles ADM and ADN are isosceles, AP = 12AM and AQ = 12AN.

Replacing АР and АQ in the equality АР∙AB = AQ∙AC, we obtain the required.

Second way

Let us prove that the BMNC quadrilateral is an inscribed one, then the required equality will follow from the secant segment theorem applied to the point A and the circle circumscribed around the BMNC quadrilateral (see Fig. 10.4b).

Let ∠ANM = α, then ∠AOM = 2α (inscribed and central angles based on the same arc). Also, from the isosceles triangle ADM: ∠MAD = 90° - α, so ∠ABC = α. It follows from the equality ∠ABC = ∠ANM that BMNC is an inscribed one.

After the indicated equality is proved, it is enough to substitute the data from the condition of the problem into it and get the answer.

Third way

Let this circle intersect segments BD and CD at points K and L, respectively, and its radius is equal to R (see Fig. 10.4c). Then, according to the secant segment theorem: BA∙BM = BL∙BK, that is, c(c – m) = BK(BK + 2R). From the triangle ABD by the Pythagorean theorem: с2 = (BK + R)2 + R2 = 2R2 + BK2 +2BK∙R. Therefore, c(c – m) = с2 – 2R2, whence c∙m = 2R2.

Carrying out a similar argument for the side AC, we obtain that AC∙n = 2R2. Then AC = mcn.

Note that with this method of solution, instead of the Pythagorean theorem, one can apply the cosine theorem for the triangle BAK.

Verification Criteria

• “+” - complete justified solution is given
• “±” – generally correct reasoning is given, containing minor gaps or inaccuracies (for example, m and n are mixed up)
• “±” – the solution plan is correct and the correct answer has been obtained, but some of the facts used have not been proved (for example, it has been used, but not proved, that the BMNC quadrilateral is inscribed)
• “±” – the solution plan is correct, but the solution itself contains errors or is not completed
• “±” - there is no clear solution plan, but some significant facts are substantiated from which a solution can be obtained
• “–” - only the answer is given
• “–” – problem not solved or solved incorrectly

Task 5

Vasya has dismantled the frame of a triangular pyramid in the mathematics classroom and wants to make two triangles from its six edges so that each edge is a side of exactly one triangle. Will Vasya always be able to do this?

Answer: always.

Decision

Note that if Vasya manages to fold a triangle out of edges coming out of one vertex of the tetrahedron, then the second triangle is already folded, and the problem is solved.

Let AB be the longest edge of the tetrahedron DABC (see Figure 10.5).

Suppose that neither from the triple of edges with a common vertex A, nor from the triple of edges with a common vertex B, Vasya can form a triangle. This means that AB ≥ AC + AD and AB ≥ BC + BD. Then 2AB ≥ AC + AD + BC + BD.

On the other hand, according to the triangle inequality for faces ABD and ABC, we get: AB< AD + BD и АВ < AC + BC. Тогда 2АВ < AC + AD + BC + BD – противоречие.

Verification Criteria

• “+” - complete justified solution is given
• “ ” – there is a correct idea of ​​the solution, but it has not been completed or a mistake has been made
• “–” - only some special cases are analyzed (for example, a regular tetrahedron is considered)
• “–” – problem not solved or solved incorrectly

Task 6

100 on and 100 off flashlights are randomly placed in two boxes. Each flashlight has a button, pressing which turns off the burning flashlight and lights the off one.

Your eyes are blindfolded and you cannot see if the flashlight is on. But you can shift flashlights from box to box and press buttons on them. Think of a way to ensure that the burning lanterns in the boxes were evenly distributed.

Decision

First, let's move all the flashlights to the right box without touching the switches. Next, we shift any hundred flashlights from the right box to the left one, switching each one at the same time, and the goal will be achieved. Let's prove it.

When shifting (with switching) one flashlight, the difference between the numbers of burning flashlights on the right and left decreases by 1. Indeed, if we took a flashlight that was not burning, lit it and shifted it to the left, then the number of burning flashlights on the right did not change, and on the left it increased by 1. If we took a burning flashlight, extinguished it and shifted it to the left, then the number of burning ones on the right decreased by 1, and on the left it remained the same. At that moment, when all the lanterns were in the right box, the considered difference is 100, which means that after a hundred shifts it will become equal to zero, which is required.

There are other algorithms of actions.

Verification Criteria

• “+” - complete justified solution is given
• “±” - the correct algorithm is given, but its justification is incomplete (for example, it is said that the difference between burning lanterns will decrease by 1, but it is not explained why
• “±” - only the correct algorithm is given without any explanation
• “–” – problem not solved or solved incorrectly

ALGEBRA-8

Workshop Lesson

TOPIC: "Function

y \u003d ax 2 + b x + c "

The lesson was conducted using a mobile computer class

Lesson - practicum.

"Drawing with graphs of functions."

(Supported by computer program Advanced grapher .)

At school, in mathematics lessons, tasks are widely used in which students build points according to coordinates and connect them in series, while receiving a drawing of an object. Children love these activities. They diversify the activities of students during the period of development of knowledge, introduce an element of entertainment into the lesson, honing the skill.

Similar work can be done in the 8th grade, but using the graphs of a quadratic function given on segments. The topic is very suitable for this work:"Function".

GOALS:

This lesson is intended as the final lesson on this topic.

The lesson consists of 6 stages.

Equipment for the lesson:

1. 1. 1. 1. 1. 1. computer program Advancedgrapher, with the help of which the study of the topic of this lesson takes place.

                  Projector.

                  Screen.

                  Handout (cards with individual tasks).

Detailed description of each stage.

Introduction to the interface of the Advanced Grapher program.

The + button is displayed on the toolbar F - Add graph. We will use this button every time we start working with a new function. Click on this button. In the opened dialog box Graph Properties we can set the function you are interested in, as well as set the appearance of the future chart (thickness, line color, etc.).

When displaying a tab Additional properties check the interval box. Now you can set the scope of the function.

AT use button -Document properties.Or with the commandGraphs
Document properties incall the dialog boxDocument properties.

In the opened windowon the left, on the tree, you can select one of the properties you are interested in to configure(Construction, Axes, Legend, Grid).Click on the Build tab. Here you can set the maximum and minimum intervals for each of the axes separately. This can be useful when constructing those graphs in which the displacement of vertices along the axes is significant.

By clicking on the Plot List button, you will have access to any function that you have previously used.

theoretical survey.

Function Graph
is the parabola obtained by shifting the parabola
along the coordinate axes.

Collective work on creating a picture from parabolas. ("Umbrella")

Each child is given a card with a list of quadratic functions of the form
.

When working with each of the list formulas, children answer the following questions:

What is the graph of this function?

How are the branches of the parabola directed?

What are the coordinates of the vertex of the parabola?

Students open the Add Graph dialog box and enter the formula. By clicking on the OK button, the image of the graph of the function is obtained.

What should be kept in mind when plotting a function graph? (about the scope of a function)

By double-clicking on the function you are currently interested in in the Plot List window, you will have access to any function that you have previously used, i.e. you will be returned to the dialog box Graph Properties. By displaying the tab Additional properties, check the interval box, and specify the domain of definition required by the condition for this function. After making the required settings, click the OK button. The graph of the function will change its appearance in accordance with the domain of definition.

T How each following function is discussed. The constructions are performed in parallel on the students' computers and on the teacher's laptop connected to the projector.

Analyzing the future form of the graph, children have the opportunity to immediately verify the correctness of their judgments. A holistic view of the picture will convince the doubting student of the correctness of the actions performed by him.

Independent work.

Students are given various cards with a list of functions. Each child builds his drawing on his own, receiving an assessment at the end of the lesson.

Card options.

« Glasses"

"Whale"

« Chess King"

« Frog"

Summarizing and grading.

What did you learn in class today?

The criterion for your assimilation of the material will be the drawing created by each of you.

"Excellent" - The task was carried out independently. The drawing is finished. There are no comments on the charts. The domains for each graph are set correctly.

"OK" - The drawing is finished. There are some comments on finding the domain of definition, or the student turned to the teacher for help during independent work.

“Satisfactory” - The drawing has flaws. The student was unsure of his knowledge, constantly turning to the teacher for help.

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 at 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!

You can use our tasks (not necessary) and we certainly recommend them.

In order to get a hand with the help of our tasks, you need to help extend the life of the YouClever textbook that you are currently reading.

How? There are two options:

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Access to all hidden tasks is provided for the entire lifetime of the site.

In conclusion...

If you don't like our tasks, find others. Just don't stop with theory.

“Understood” and “I know how to solve” are completely different skills. You need both.

Find problems and solve!