Convenient for, having given a specific value of the independent variable x (argument), to calculate the corresponding value of the dependent variable y. For example, if the function y = x 2 is given, i.e. f(x) = x 2, then for x = 1 we get y = 1 2 = 1; In short, it is written like this: f(1) = 1. For x = 2 we get f(2) = 2 2 = 4, i.e. y = 4; for x = - 3 we get f(- 3) = (- 3) 2 = 9, i.e. y = 9, etc.
Already in the 7th grade, you and I began to understand that in the equality y = f(x) the right side, i.e. the expression f(x) is not limited to the four cases listed above (C, kx, kx + m, x 2).
For example, we have already encountered piecewise functions, i.e. functions, given by different formulas at different intervals. Here is one such function: y = f(x), where
Do you remember how to graph such functions? First you need to construct a parabola y = x 2 and take its part at x< 0 (левая ветвь параболы, рис. 1), затем надо построить прямую у = 2х и взять ее часть при х >0 (Fig. 2). And finally, you need to combine both selected parts in one drawing, i.e. build on one coordinate plane(see Fig. 3).
Now our task is the following: to replenish the stock of studied functions. IN real life There are processes described by various mathematical models of the form y = f(x), not only those that we listed above. In this section we will consider the function y = kx 2, where coefficient k is any non-zero number.
In fact, the function y = kx 2 in one case is a little familiar to you. Look: if k = 1, then we get y = x 2; You studied this function in 7th grade and probably remember that its graph is a parabola (Fig. 1). Let's discuss what happens at other values of the coefficient k.
Consider two functions: y = 2x 2 and y = 0.5x 2. Let's make a table of values for the first function y = 2x 2:
Let's construct the points (0; 0), (1; 2), (-1; 2), (2; 8), (-2; 8), (1.5; 4.5), (-1.5; 4.5) on coordinate plane(Fig. 4); they outline a certain line, let’s draw it (Fig. 5).
Let's make a table of values for the second function y = 0.5x 2:
Let's construct points (0; 0), (1; 0.5), (-1; 0.5), (2; 2), (-2; 2), C; 4.5), (-3; 4.5) on the coordinate plane (Fig. 6); they outline a certain line, let’s draw it (Fig. 7)
.
The points shown in Fig. 4 and 6 are sometimes called control points for the graph of the corresponding function.
Compare Figures 1, 5 and 7. Isn’t it true that the lines drawn are similar? Each of them is called a parabola; in this case, the point (0; 0) is called the vertex of the parabola, and the y-axis is the axis of symmetry of the parabola. The “speed of upward movement” of the parabola’s branches or, as they also say, the “degree of steepness” of the parabola depends on the value of the coefficient k. This is clearly visible in Fig. 8, where all three parabolas constructed above are located on the same coordinate plane.
The situation is exactly the same with any other function of the form y = kx 2, where k > 0. Its graph is a parabola with a vertex at the beginning coordinates, the branches of the parabola are directed upward, and the steeper the higher the coefficient k. The y-axis is the axis of symmetry of the parabola. By the way, for the sake of brevity, mathematicians often say “parabola y = kx 2” instead of the long phrase “parabola serving as a graph of the function y = kx 2”, and instead of the term “axis of symmetry of a parabola” they use the term “parabola axis”.
Do you notice that there is an analogy with the function y = kx? If k > 0, then the graph of the function y = kx is a straight line passing through the origin of coordinates (remember, we said briefly: straight line y = kx), and here, too, the “degree of steepness” of the straight line depends on the value of the coefficient k. This is clearly visible in Fig. 9, where in one coordinate system are depicted graphics linear functions y = kx for three values of the coefficient
Let's return to the function y = kx 2. Let us find out how things stand in the case of a negative coefficient ft. Let's build, for example, a graph of the function
y = - x 2 (here k = - 1). Let's create a table of values:
Mark the points (0; 0), (1; -1), (-1; -1), (2; -4), (-2; -4), (3; -9), (- 3; - 9) on the coordinate plane (Fig. 10); they outline a certain line, let’s draw it (Fig. 11). This is a parabola with its vertex at the point (0; 0), the y-axis is the axis of symmetry, but unlike the case when k > 0, this time the branches of the parabola are directed downward. The situation is similar for others negative values coefficient k.
So, the graph of a function is a parabola with its vertex at the origin; the y-axis is the axis of the parabola; the branches of the parabola are directed upward at k>0 u downward at k<0.
Let us also note that the parabola y = kx 2 touches the x axis at the point (0; 0), that is, one branch of the parabola smoothly passes into the other, as if pressing against the x axis.
If built in one coordinate system function graphs y = x 2 and y = - x2, then it is easy to see that these parabolas are symmetrical to each other about the x axis, which is clearly visible in Fig. 12. In the same way, the parabolas y = 2x 2 and y = - 2x 2 are symmetrical to each other relative to the x axis (don’t be lazy, build these
two parabolas in the same coordinate system and make sure the statement is true).
In general, the graph of the function y = - f(x) is symmetrical to the graph of the function y = f(x) relative to the x-axis.
Properties of the function y = kx 2 for k > 0
Describing the properties of this function, we will rely on its geometric model - a parabola (Fig. 13).
1. Since for any value of x the corresponding value of y can be calculated using the formula y = kx 2, the function is defined at any point x (for any value of the argument x). In short, it is written like this: the domain of definition of the function is (-oo, +oo), i.e. the entire coordinate line.
2. y = 0 at x = 0; y > O at . This can also be seen from the graph of the function (it is entirely located above the x-axis), but can be justified without the help of a graph: if
Then kx 2 > O as the product of two positive numbers k and x 2 .
3. y = kx 2 - continuous function. Let us recall that for now we consider this term as a synonym for the sentence “the graph of a function is a solid line that can be drawn without lifting the pencil from the paper.” In higher grades, a more precise mathematical interpretation of the concept of continuity of a function will be given, not relying on geometric illustration.
4.y/ naim = 0 (achieved at x = 0); nai6 does not exist.
Let us remind you that (/name is smallest value functions, and Unaib. - the greatest value of the function on a given interval; if the interval is not specified, then unaim- and y naib, - respectively, the smallest and highest value functions in the domain of definition.
5. The function y = kx 2 increases as x > O and decreases as x< 0.
Let us recall that in the 7th grade algebra course we agreed to call a function whose graph on the interval under consideration goes from left to right as if “uphill”, increasing, and function, the graph of which in the interval under consideration goes from left to right, as if “downhill”, is decreasing. More precisely, we can say this: the function y = f (x) is said to be increasing on the interval X if on this interval higher value the argument corresponds to a larger function value; a function y = f (x) is said to be decreasing on an interval X if on this interval a larger value of the argument corresponds to a smaller value of the function.
In the Algebra 7 textbook, we called the process of listing the properties of a function reading a graph. The process of reading a graph will gradually become richer and more interesting as we learn new properties of functions. We discussed the five properties listed above in 7th grade for the functions we studied there. Let's add one new property.
A function y = f(x) is called bounded below if all values of the function are greater than a certain number. Geometrically, this means that the graph of the function is located above a certain direct, parallel to the x axis.
Now look: the graph of the function y = kx 2 is located above the straight line y = - 1 (or y = - 2, it doesn’t matter) - it is shown in Fig. 13. This means that y - kx2 (k > 0) is a function bounded from below.
Along with functions bounded below, functions bounded above are also considered. A function y - f(x) is said to be bounded from above if all values of the function are less than a certain number. Geometrically, this means that the graph of the function is located below some straight line parallel to the x-axis.
Is there such a line for the parabola y = kx 2, where k > 0? No. This means that the function is not upper bounded.
So, we got one more property, let's add it to the five listed above.
6. The function y = kx 2 (k > 0) is bounded below and not bounded above.
Properties of the function y = kx 2 at k< 0
When describing the properties of this function, we rely on its geometric model- parabola (Fig. 14).
1. The domain of definition of the function is (-oo, +oo).
2. y = 0 at x = 0; at< 0 при .
Z.y = kx 2 - continuous function.
4. y nai6 = 0 (achieved at x = 0), unaim does not exist.
5. The function increases as x< 0, убывает при х > 0.
6.The function is limited from above and not limited from below.
Let us explain the last property: there is a straight line parallel to the x axis (for example, y = 1, it is drawn in Fig. 14), such that the entire parabola lies below this straight line; this means that the function is upper bounded. On the other hand, it is impossible to draw a straight line parallel to the x-axis such that the entire parabola is located above this straight line; this means that the function is not bounded below.
The order of moves used above when listing the properties of a function is not a law, as long as it has developed chronologically this way.
We will develop a more or less definite order of moves gradually and unify it in the 9th grade algebra course.
Example 1. Find the smallest and largest values of the function y = 2x 2 on the segment: a) ; b) [- 2, - 1]; c) [- 1, 1.5].
a) Let's build a graph of the function y = 2x2 and highlight its part on the segment (Fig. 15). We note that 1/name. = 0 (achieved at x = 0), and y max = 8 (achieved at x = 2).
b) Let's construct a graph of the function y = 2x2 and highlight its part on the segment [- 2, - 1] (Fig. 16). We note that 2/max = 2 (achieved at x = - 1), and y max = 8 (achieved at x = - 2).
c) Let's build a graph of the function y = 2x2 and highlight its part on the segment [- 1, 1.5] (Fig. 17). We note that unanm = 0 (achieved at x = 0), and y is most achieved at the point x = 1.5; Let's calculate this value: (1.5) = 2-1.5 2 = 2-2.25 = 4.5. So, y max =4.5.
Example 2. Solve the equation - x 2 = 2x - 3.
Solution. In the textbook "Algebra-7" we developed algorithm graphical solution of equations, let us recall it.
To solve the equation f(x) = g (x) graphically, you need:
1) consider two functions y = -x 2 and y = 2x -3;
2) construct a graph of the function i/ = / (x);
3) build a graph of the function y = g (x);
4) find the intersection points of the constructed graphs; abscis-
The sys of these points are the roots of the equation f(x) = g (x).
Let's apply this algorithm to given equation.
1) Consider two functions: y = - x2 and y = 2x - 3.
2) Let's construct a parabola - a graph of the function y = - x 2 (Fig. 18).
3) Let's construct a graph of the function y = 2x - 3. This is a straight line; to construct it, it is enough to find any two points on the graph. If x = 0, then y = - 3; if x = 1, then y = -1. So, we found two points (0; -3) and (1; -1). The straight line passing through these two points (graph of the function y = 2x - 3) is shown in the same drawing (see Fig. 18).
4) According to the drawing, we find that the straight line and the parabola intersect at two points A(1; -1) and B(-3; -9). Means, given equation has two roots: 1 and - 3 - these are the abscissas of points A and B.
Answer: 1,-3.
Comment. Of course, you cannot blindly trust graphic illustrations. Maybe it just seems to us that point A has coordinates (1; - 1), but in fact they are different, for example (0.98; - 1.01)?
Therefore, it is always useful to check yourself. So, in the example considered, you need to make sure that point A(1; -1) belongs to the parabola y = - x 2 (this is easy - just substitute the coordinates of point A into the formula y = - x 2; we get - 1 = - 1 2 - correct numerical equality) and the straight line y = 2x - 3 (and this is easy - just substitute the coordinates of point A into the formula y = 2x - 3; we get - 1 = 2-3 - the correct numerical equality). The same must be done for point 8. This check shows that in the equation considered, graphical observations led to the correct result.
Example 3. Solve the system
Solution. Let's transform the first equation of the system to the form y = - x 2. The graph of this function is a parabola shown in Fig. 18.
Let us transform the second equation of the system to the form y = 2x - 3. The graph of this function is the straight line shown in Fig. 18.
The parabola and the straight line intersect at points A (1; -1) and B (- 3; - 9). The coordinates of these points serve as solutions to a given system of equations.
Answer: (1; -1), (-3; -9).
Example 4. Given a function y - f (x), where
Required:
a) calculate f(-4), f(-2), f(0), f(1.5), f(2), f(3);
b) construct a graph of the function;
c) use a graph to list the properties of the function.
a) The value x = - 4 satisfies the condition - therefore, f(-4) must be calculated using the first line of the function definition. We have f(x) = - 0.5x2, which means f(-4) = -0.5 . (-4) 2 = -8.
Similarly we find:
f(-2) = -0.5 .
(-2) 2 =-2;
f(0) = -0.5 .
0 2 = 0.
The value satisfies the condition, so it must be calculated using the second line of the function specification. We have f(x) = x + 1, which means 
The value x = 1.5 satisfies condition 1< х < 2, т. е. f(1,5) надо вычислять по третьей строке задания функции. Имеем f (х) = 2х 2 , значит, f(1,5) = 2-1,5 2 = 4,5.
Similarly we get f(2)= 2 .
2 2 =8.
The value x = 3 does not satisfy any of the three conditions for specifying the function, and therefore f(3) in in this case cannot be calculated, the point x = 3 does not belong to the domain of definition of the function. The task of calculating f(3) is incorrect.
b) We will build the graph “piece by piece”. First, let's construct a parabola y = -0.5x 2 and select its part on the segment [-4, 0] (Fig. 19). Then we construct the straight line y = x + 1 u. Let's select its part on the half-interval (0, 1] (Fig. 20). Next, we will construct a parabola y = 2x2 and select its part on the half-interval (1, 2] (Fig. 21).
Finally, we will depict all three “pieces” in one coordinate system; we obtain a graph of the function y = f(x) (Fig. 22).
c) Let's list the properties of the function or, as we agreed to say, read the graph.
1. The domain of definition of the function is the segment [-4, 2].
2. y = 0 at x = 0; y > 0 at 0<х<2;у<0 при - 4 < х < 0.
3. The function undergoes a discontinuity at x = 0.
4. The function increases on the segment [-4, 2].
5. The function is limited both from below and from above.
6. y max = -8 (achieved at x = -4); y most6 . = 8 (achieved at x = 2).
Example 5. The function y = f(x) is given, where f(x) = 3x 2. Find.
A function of the form where is called quadratic function.
Graph of a quadratic function – parabola.
Let's consider the cases:
I CASE, CLASSICAL PARABOLA
That is , ,
To construct, fill out the table by substituting the x values into the formula:
Mark the points (0;0); (1;1); (-1;1), etc. on the coordinate plane (the smaller the step we take the x values (in this case, step 1), and the more x values we take, the smoother the curve will be), we get a parabola:
It is easy to see that if we take the case , , , that is, then we get a parabola that is symmetrical about the axis (oh). It’s easy to verify this by filling out a similar table:
II CASE, “a” IS DIFFERENT FROM UNIT
What will happen if we take , , ? How will the behavior of the parabola change? With title="Rendered by QuickLaTeX.com" height="20" width="55" style="vertical-align: -5px;"> парабола изменит форму, она “похудеет” по сравнению с параболой (не верите – заполните соответствующую таблицу – и убедитесь сами):!}
In the first picture (see above) it is clearly visible that the points from the table for the parabola (1;1), (-1;1) were transformed into points (1;4), (1;-4), that is, with the same values, the ordinate of each point is multiplied by 4. This will happen to all key points of the original table. We reason similarly in the cases of pictures 2 and 3.
And when the parabola “becomes wider” than the parabola:
Let's summarize:
1)The sign of the coefficient determines the direction of the branches. With title="Rendered by QuickLaTeX.com" height="14" width="47" style="vertical-align: 0px;"> ветви направлены вверх, при - вниз. !}
2) Absolute value coefficient (modulus) is responsible for the “expansion” and “compression” of the parabola. The larger , the narrower the parabola, the smaller |a|, the wider the parabola.
III CASE, “C” APPEARS
Now let's introduce into the game (that is, consider the case when), we will consider parabolas of the form . It is not difficult to guess (you can always refer to the table) that the parabola will shift up or down along the axis depending on the sign:
IV CASE, “b” APPEARS
When will the parabola “break away” from the axis and finally “walk” along the entire coordinate plane? When will it stop being equal?
Here to construct a parabola we need formula for calculating the vertex: , .
So at this point (as at the point (0;0) of the new coordinate system) we will build a parabola, which we can already do. If we are dealing with the case, then from the vertex we put one unit segment to the right, one up, - the resulting point is ours (similarly, a step to the left, a step up is our point); if we are dealing with, for example, then from the vertex we put one unit segment to the right, two - upward, etc.
For example, the vertex of a parabola:
Now the main thing to understand is that at this vertex we will build a parabola according to the parabola pattern, because in our case.
When constructing a parabola after finding the coordinates of the vertex veryIt is convenient to consider the following points:
1) parabola will definitely pass through the point . Indeed, substituting x=0 into the formula, we obtain that . That is, the ordinate of the point of intersection of the parabola with the axis (oy) is . In our example (above), the parabola intersects the ordinate at point , since .
2) axis of symmetry parabolas is a straight line, so all points of the parabola will be symmetrical about it. In our example, we immediately take the point (0; -2) and build it symmetrical relative to the symmetry axis of the parabola, we get the point (4; -2) through which the parabola will pass.
3) Equating to , we find out the points of intersection of the parabola with the axis (oh). To do this, we solve the equation. Depending on the discriminant, we will get one (, ), two ( title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">, ) или нИсколько () точек пересечения с осью (ох) !} . In the previous example, our root of the discriminant is not an integer; when constructing, it doesn’t make much sense for us to find the roots, but we clearly see that we will have two points of intersection with the axis (oh) (since title="Rendered by QuickLaTeX.com" height="14" width="54" style="vertical-align: 0px;">), хотя, в общем, это видно и без дискриминанта.!}
So let's work it out
Algorithm for constructing a parabola if it is given in the form
1) determine the direction of the branches (a>0 – up, a<0 – вниз)
2) we find the coordinates of the vertex of the parabola using the formula , .
3) we find the point of intersection of the parabola with the axis (oy) using the free term, construct a point symmetrical to this point with respect to the symmetry axis of the parabola (it should be noted that it happens that it is unprofitable to mark this point, for example, because the value is large... we skip this point...)
4) At the found point - the vertex of the parabola (as at the point (0;0) of the new coordinate system) we construct a parabola. If title="Rendered by QuickLaTeX.com" height="20" width="55" style="vertical-align: -5px;">, то парабола становится у’же по сравнению с , если , то парабола расширяется по сравнению с !}
5) We find the points of intersection of the parabola with the axis (oy) (if they have not yet “surfaced”) by solving the equation
Example 1
Example 2
Note 1. If the parabola is initially given to us in the form , where are some numbers (for example, ), then it will be even easier to construct it, because we have already been given the coordinates of the vertex . Why?
Let's take quadratic trinomial and highlight in it perfect square: Look, so we got that, . You and I previously called the vertex of a parabola, that is, now, .
For example, . We mark the vertex of the parabola on the plane, we understand that the branches are directed downward, the parabola is expanded (relative to ). That is, we carry out points 1; 3; 4; 5 from the algorithm for constructing a parabola (see above).
Note 2. If the parabola is given in a form similar to this (that is, presented as a product of two linear factors), then we immediately see the points of intersection of the parabola with the axis (ox). In this case – (0;0) and (4;0). For the rest, we act according to the algorithm, opening the brackets.