Graph of parabola y. Plotting a graph of a quadratic function. Visual Guide (2019)

Textbook:

  • Makarychev Yu. N., Mindyuk N. R. Mathematics. 7th grade

Goals:

  • develop graphic literacy when plotting,
  • build a skill research work,
  • cultivate clarity in answering, accuracy, and responsibility.
  • I. Student survey

    1. What is a function called?
    2. (A function is the dependence of one variable on another, in which each value of the independent variable corresponds to a single value of the dependent variable)

    3. What is the domain of a function?
    4. (All values ​​that the independent variable (argument) takes form the domain of the function.)

    5. What is the range of a function?
    6. (All values ​​that the dependent variable takes are called function values)

    7. What functions did we get to know?
    8. a) with a linear function of the form y = kx + b,

      direct proportionality of the form y = kx

      b) with functions of the form y = x 2, y = x 3

    9. What is the graph of a linear function? ( straight). How many points are needed to construct this graph?

    Without performing construction, determine the relative position of the graphs of the functions given by the following formulas:

    A ) y = 3x + 2; y = 1.2x + 5;

    b) y = 1.5x + 4; y = -0.2x + 4; y = x + 4;

    With) y = 2x + 5; y = 2x - 7; y = 2x

    Figure 1

    The figure shows graphs of linear functions ( Each student is given a sheet of paper with the graphs at their desk.). Write a formula for each graph

    What function graphs are we still familiar with? ( y = x 2; y = x 3 )

    1. What is the graph of a function y = x 2 (parabola).
    2. How many points do we need to construct to depict a parabola? ( 7, one of which is the vertex of a parabola).

    Let's build a parabola given by the formula y = x 2

    x -3 -2 -1 0 1 2 3
    y = x 2 9 4 1 0 1 4 9
    y = x 2 + 2 11 6 3 2 3 6 11

    Figure 2

    What properties does the graph of a function have? y = x 3 ?

    1. If x = 0 , That y = 0 - vertex of the parabola (0;0)
    2. Scope: X - any number, D (y) = (- ?; ?) D (y) = R
    3. Range of values at ? 0
    4. E (y) =
    5. The function increases over the interval

      The function increases over the interval - at these values ​​of x, moving along the parabola from left to right, we “go down the hill” (see Fig. 55). The function y = x 2 increases along the ray;
      b) on the segment [- 3, - 1.5];
      c) on the segment [- 3, 2].

      Solution,

      a) Let's construct a parabola y = x 2 and select that part of it that corresponds to the values ​​of the variable x from the segment (Fig. 56). For the selected part of the graph we find at the name. = 1 (at x = 1), y max. = 9 (at x = 3).

      b) Let's construct a parabola y = x 2 and select that part of it that corresponds to the values ​​of the variable x from the segment [-3, -1.5] (Fig. 57). For the selected part of the graph, we find y name. = 2.25 (at x = - 1.5), y max. = 9 (at x = - 3).

      c) Let's construct a parabola y = x 2 and select that part of it that corresponds to the values ​​of the variable x from the segment [-3, 2] (Fig. 58). For the selected part of the graph, we find y max = 0 (at x = 0), y max. = 9 (at x = - 3).

      Advice. To avoid plotting the function y - x 2 point by point each time, cut out a parabola template from thick paper. With its help you will draw a parabola very quickly.

      Comment. By inviting you to prepare a parabola template, we seem to be equalizing the rights of the function y = x 2 and linear function y = kx + m. After all, the schedule linear function is a straight line, and to depict a straight line, an ordinary ruler is used - this is the template for the graph of the function y = kx + m. So let you have a template for the graph of the function y = x 2.

      Example 2. Find the intersection points of the parabola y = x 2 and the straight line y - x + 2.

      Solution. Let us construct in one coordinate system the parabola y = x 2 and the straight line y = x + 2 (Fig. 59). They intersect at points A and B, and from the drawing it is not difficult to find the coordinates of these points A and B: for point A we have: x = - 1, y = 1, and for point B we have: x - 2, y = 4.

      Answer: the parabola y = x 2 and the straight line y = x + 2 intersect at two points: A (-1; 1) and B (2; 4).

      Important note. Until now, you and I have been quite bold in drawing conclusions using the drawing. However, mathematicians do not trust drawings too much. Having discovered in Figure 59 two points of intersection of a parabola and a straight line and having determined the coordinates of these points using the drawing, the mathematician usually checks himself: whether the point (-1; 1) actually lies on both the straight line and the parabola; does the point (2; 4) really lie on both a straight line and a parabola?

      To do this, you need to substitute the coordinates of points A and B into the equation of the straight line and into the equation of the parabola, and then make sure that in both cases the correct equality is obtained. In example 2, in both cases the equalities will be true. This check is especially often carried out when there is doubt about the accuracy of the drawing.

      In conclusion, we note one interesting property of the parabola, discovered and proven jointly by physicists and mathematicians.

      If we consider the parabola y = x 2 as a screen, as a reflective surface, and place a light source at the point, then the rays, reflected from the parabola of the screen, form a parallel beam of light (Fig. 60). The point is called the focus of the parabola. This idea is used in cars: the reflective surface of the headlight has a parabolic shape, and the light bulb is placed at the focal point - then the light from the headlight spreads far enough.

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