Algebra lesson notes for 8th grade secondary school

Lesson topic: Function

The purpose of the lesson:

· Educational: define the concept of a quadratic function of the form (compare graphs of functions and ), show the formula for finding the coordinates of the vertex of a parabola (teach how to apply this formula in practice); to develop the ability to determine the properties of a quadratic function from a graph (finding the axis of symmetry, the coordinates of the vertex of a parabola, the coordinates of the points of intersection of the graph with the coordinate axes).

· Developmental: development of mathematical speech, the ability to correctly, consistently and rationally express one’s thoughts; developing the skill of correctly writing mathematical text using symbols and notations; development of analytical thinking; development of students’ cognitive activity through the ability to analyze, systematize and generalize material.

· Educational: fostering independence, the ability to listen to others, developing accuracy and attention in written mathematical speech.

Lesson type: learning new material.

Teaching methods:

generalized reproductive, inductive heuristic.

Requirements for students' knowledge and skills

know what a quadratic function of the form is, the formula for finding the coordinates of the vertex of a parabola; be able to find the coordinates of the vertex of a parabola, the coordinates of the points of intersection of the graph of a function with the coordinate axes, and use the graph of a function to determine the properties of a quadratic function.

Equipment:

Lesson Plan

I. Organizational moment (1-2 min)

II. Updating knowledge (10 min)

III. Presentation of new material (15 min)

IV. Consolidating new material (12 min)

V. Summing up (3 min)

VI. Homework assignment (2 min)

During the classes

I. Organizational moment

Greeting, checking absentees, collecting notebooks.

II. Updating knowledge

Teacher: In today's lesson we will study a new topic: "Function". But first, let's repeat the previously studied material.

Frontal survey:

1) What is called a quadratic function? (A function where given real numbers, , is a real variable, is called a quadratic function.)

2) What is the graph of a quadratic function? (The graph of a quadratic function is a parabola.)

3) What are the zeros of a quadratic function? (The zeros of a quadratic function are the values ​​at which it becomes zero.)

4) List the properties of the function. (The values ​​of the function are positive at and equal to zero at; the graph of the function is symmetrical with respect to the ordinate axes; at - the function increases, at - decreases.)

5) List the properties of the function. (If , then the function takes positive values ​​at , if , then the function takes negative values ​​at , the value of the function is only 0; the parabola is symmetrical about the ordinate axis; if , then the function increases at and decreases at , if , then the function increases at , decreases – at .)

III. Presentation of new material

Teacher: Let's start learning new material. Open your notebooks, write down the date and topic of the lesson. Pay attention to the board.

Writing on the board: Number.

Function.

Teacher: On the board you see two graphs of functions. The first graph, and the second. Let's try to compare them.

You know the properties of the function. Based on them, and comparing our graphs, we can highlight the properties of the function.

So, what do you think will determine the direction of the branches of the parabola?

Students: The direction of the branches of both parabolas will depend on the coefficient.

Teacher: Absolutely right. You can also notice that both parabolas have an axis of symmetry. In the first graph of the function, what is the axis of symmetry?

Students: For a parabola, the axis of symmetry is the ordinate axis.

Teacher: Right. What is the axis of symmetry of a parabola?

Students: The axis of symmetry of a parabola is the line that passes through the vertex of the parabola, parallel to the ordinate axis.

Teacher: Right. So, the axis of symmetry of the graph of a function will be called a straight line passing through the vertex of the parabola, parallel to the ordinate axis.

And the vertex of a parabola is a point with coordinates . They are determined by the formula:

Write the formula in your notebook and circle it in a frame.

Writing on the board and in notebooks

Coordinates of the vertex of the parabola.

Teacher: Now, to make it more clear, let's look at an example.

Example 1: Find the coordinates of the vertex of the parabola .

Solution: According to the formula

Teacher: As we have already noted, the axis of symmetry passes through the vertex of the parabola. Look at the blackboard. Draw this picture in your notebook.

Write on the board and in notebooks:

Teacher: In the drawing: - equation of the axis of symmetry of a parabola with the vertex at the point where the abscissa is the vertex of the parabola.

Let's look at an example.

Example 2: Using the graph of the function, determine the equation for the axis of symmetry of the parabola.

The equation for the axis of symmetry has the form: , which means the equation for the axis of symmetry of this parabola is .

Answer: - equation of the axis of symmetry.

IV. Consolidation of new material

Teacher: The tasks that need to be solved in class are written on the board.

Writing on the board: № 609(3), 612(1), 613(3)

Teacher: But first, let's solve an example not from the textbook. We will decide at the board.

Example 1: Find the coordinates of the vertex of a parabola

Solution: According to the formula

Answer: coordinates of the vertex of the parabola.

Example 2: Find the coordinates of the intersection points of the parabola with coordinate axes.

Solution: 1) With axis:

Those.

According to Vieta's theorem:

The points of intersection with the x-axis are (1;0) and (2;0).

2) With axle:

The point of intersection with the ordinate axis (0;2).

Answer: (1;0), (2;0), (0;2) – coordinates of the points of intersection with the coordinate axes.

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 is known as the quadratic trinomial.

Recall that ax 2 is the leading term of this quadratic trinomial, and 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 brackets for 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 adding similar terms, we get the expression: -3 multiplied by the square of the sum (x+1) add 4.

Let's build 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 that can be proven by isolating the perfect square of the 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 by parallel transfer of 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 second one is the most difficult, since, in addition to calculations, you also need knowledge of the formulas by which x is zero and y is zero.

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

We determine right away that the graph is a parabola, the branches are directed upward, since the leading coefficient is 2, and this is a 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 points x=0 and x=3. y=1. Let's mark these points. Based on three known points, we construct the desired 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 the following example we will learn how 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 build a parabola and determine that the smallest value of the function is -5 at x=0, and the largest is 3 at x=2.