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Lesson topic: Function y=a and its properties.

Lesson type: Learning new material.

Lesson Objectives:

Lesson objectives:

Shape:

ability to apply properties quadratic function;

ability to graph functions;

the ability to formulate the properties of a quadratic function;

the ability to express one’s opinion and draw conclusions;

Develop: thinking, memory, ability to carry out independent activity in class.

Teaching methods

by source of knowledge: conversation, exercises;

by nature cognitive activity: search, explanatory and illustrative, reproductive.

Forms of training: frontal.

Lesson steps:

Organizational moment(1 min).

Update background knowledge and methods of action (5 min).

Learning new material (15 min).

Initial application of a new material (20 min).

Setting homework (1 min).

Summing up the lesson (3 min).

Consider an expression of the form ax 2 + bx + c, where a, b, c are real numbers, and a is different from zero. This mathematical expression known as a quadratic trinomial.

Recall that ah 2 is the highest term of this quadratic trinomial, a is its leading coefficient.

But a quadratic trinomial does not always have all three terms. Let's take for example the expression 3x 2 + 2x, where a=3, b=2, c=0.

Let's move on to the quadratic function y=ax 2 +in+c, where a, b, c are any arbitrary numbers. This function is quadratic because it contains a term of the second degree, that is, x squared.

It is quite easy to construct a graph of a quadratic function; for example, you can use the method of isolating a perfect square.

Let's consider an example of constructing a graph of the function y equals -3x 2 - 6x + 1.

To do this, the first thing we remember is the scheme for isolating a complete square in the trinomial -3x 2 - 6x + 1.

Let's take -3 out of the first two terms. We have -3 times the sum x squared plus 2x and add 1. By adding and subtracting one in parentheses, we get the sum squared formula, which can be collapsed. We get -3 multiplied by the sum (x+1) squared minus 1 add 1. Opening the brackets and bringing similar terms, the expression comes out: -3 multiplied by the square of the sum (x+1) add 4.

Let's construct a graph of the resulting function by moving to an auxiliary coordinate system with the origin at the point with coordinates (-1; 4).

In the figure from the video, this system is indicated by dotted lines. Let us associate the function y equals -3x2 to the constructed coordinate system. For convenience, let's take control points. For example, (0;0), (1;-3), (-1;-3), (2;-12), (-2;-12). At the same time, we will put them aside in the constructed coordinate system. The parabola obtained during construction is the graph we need. In the picture it is a red parabola.

Using the method of isolating a complete square, we have a quadratic function of the form: y = a*(x+1) 2 + m.

The graph of the parabola y = ax 2 + bx + c can be easily obtained from the parabola y = ax 2 by parallel translation. This is confirmed by a theorem, which can be proven by isolating perfect square binomial. The expression ax 2 + bx + c after successive transformations turns into an expression of the form: a*(x+l) 2 + m. Let's draw a graph. Let's perform a parallel movement of the parabola y = ax 2, aligning the vertex with the point with coordinates (-l; m). The important thing is that x = -l, which means -b/2a. This means that this straight line is the axis of the parabola ax 2 + bx + c, its vertex is at the point with the abscissa x zero equals minus b divided by 2a, and the ordinate is calculated using the cumbersome formula 4ac - b 2 /. But you don’t have to remember this formula. Since, by substituting the abscissa value into the function, we get the ordinate.

To determine the equation of the axis, the direction of its branches and the coordinates of the vertex of the parabola, consider the following example.

Let's take the function y = -3x 2 - 6x + 1. Having composed the equation for the axis of the parabola, we have that x = -1. And this value is the x coordinate of the vertex of the parabola. All that remains is to find the ordinate. Substituting the value -1 into the function, we get 4. The vertex of the parabola is at the point (-1; 4).

The graph of the function y = -3x 2 - 6x + 1 was obtained when parallel transfer the graph of the function y = -3x 2, which means it behaves similarly. The leading coefficient is negative, so the branches are directed downward.

We see that for any function of the form y = ax 2 + bx + c, the easiest question is the last question, that is, the direction of the branches of the parabola. If the coefficient a is positive, then the branches are upward, and if negative, then the branches are downward.

The next most difficult question is the first question, because it requires additional calculations.

And the most difficult second, since, in addition to calculations, knowledge of the formulas by which x is zero and y is zero is also necessary.

Let's build a graph of the function y = 2x 2 - x + 1.

We determine right away - the graph is a parabola, the branches are directed upward, since the leading coefficient is 2, and this positive number. Using the formula, we find the abscissa x is zero, it is equal to 1.5. To find the ordinate, remember that y zero is equal to a function of 1.5; when calculating, we get -3.5.

Top - (1.5;-3.5). Axis - x=1.5. Let's take the points x=0 and x=3. y=1. Let's mark these points. In three known points We build the required graph.

To plot a graph of the function ax 2 + bx + c you need:

Find the coordinates of the vertex of the parabola and mark them in the figure, then draw the axis of the parabola;

On the oh-axis, take two points that are symmetrical relative to the axis of the parabola, find the value of the function at these points and mark them on the coordinate plane;

Construct a parabola through three points; if necessary, you can take several more points and construct a graph based on them.

IN following example we will learn to find the largest and smallest values ​​of the function -2x 2 + 8x - 5 on the segment.

According to the algorithm: a=-2, b=8, which means x zero is 2, and y zero is 3, (2;3) is the vertex of the parabola, and x=2 is the axis.

Let's take the values ​​x=0 and x=4 and find the ordinates of these points. This is -5. We construct a parabola and determine that smallest value functions -5 at x=0, and the greatest 3, at x=2.

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 graphics. 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 simplest dependence 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.