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Presentation and lesson on the topic:
"Graph of the function $y=ax^2+bx+c$. Properties"

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Teaching aids and simulators in the online store "Integral" for grade 8
Manual for the textbook Dorofeeva G.V. Manual for the textbook Nikolsky S.M.

Guys, in the last lessons we built a large number of graphs, including many parabolas. Today we will summarize the knowledge gained and learn how to build graphs of this function in the most general form.
Let's consider the square trinomial $a*x^2+b*x+c$. $a, b, c$ are called coefficients. They can be any number, but $a≠0$. $a*x^2$ is called the leading term, $a$ is called the leading coefficient. It is worth noting that the coefficients $b$ and $c$ can be equal to zero, that is, the trinomial will consist of two terms, and the third one is equal to zero.

Let's consider the function $y=a*x^2+b*x+c$. This function is called "quadratic" because the highest power is the second, that is, a square. The coefficients are the same as defined above.

In the last lesson in the last example, we analyzed the construction of a graph of a similar function.
Let's prove that any such quadratic function can be reduced to the form: $y=a(x+l)^2+m$.

The graph of such a function is constructed using an additional coordinate system. In big mathematics, numbers are quite rare. Almost any problem needs to be proved in the general case. Today we will analyze one such evidence. Guys, you can see all the power of the mathematical apparatus, but also its complexity.

We select the full square from the square trinomial:
$a*x^2+b*x+c=(a*x^2+b*x)+c=a(x^2+\frac(b)(a)*x)+c=$ $= a(x^2+2\frac(b)(2a)*x+\frac(b^2)(4a))-\frac(b^2)(4a)+c=a(x+\frac(b) (2a))^2+\frac(4ac-b^2)(4a)$.
We got what we wanted.
Any quadratic function can be represented as:
$y=a(x+l)^2+m$, where $l=\frac(b)(2a)$, $m=\frac(4ac-b^2)(4a)$.

To plot $y=a(x+l)^2+m$, you need to plot the function $y=ax^2$. Moreover, the top of the parabola will be at the point with coordinates $(-l;m)$.
So, our function $y=a*x^2+b*x+c$ is a parabola.
The axis of the parabola will be the straight line $x=-\frac(b)(2a)$, and the coordinates of the vertex of the parabola along the abscissa, as we can see, is calculated by the formula: $x_(c)=-\frac(b)(2a) $.
To calculate the coordinate of the vertex of a parabola along the y-axis, you can:

  • use the formula: $y_(c)=\frac(4ac-b^2)(4a)$,
  • directly substitute the $x$ coordinate of the vertex into the original function: $y_(c)=ax_(c)^2+b*x_(c)+c$.
How to calculate the ordinate of a vertex? Again, the choice is yours, but usually the second way will be easier to calculate.
If you want to describe some properties or answer some specific questions, you do not always need to plot a function. The main questions that can be answered without construction will be considered in the following example.

Example 1
Without plotting the function $y=4x^2-6x-3$ answer to next questions:


Solution.
a) The axis of the parabola is the straight line $x=-\frac(b)(2a)=-\frac(-6)(2*4)=\frac(6)(8)=\frac(3)(4)$ .
b) We found the vertex abscissa above $x_(c)=\frac(3)(4)$.
We find the ordinate of the vertex by direct substitution into the original function:
$y_(v)=4*(\frac(3)(4))^2-6*\frac(3)(4)-3=\frac(9)(4)-\frac(18)(4 )-\frac(12)(4)=-\frac(21)(4)$.
c) The graph of the required function will be obtained by parallel transfer of the graph $y=4x^2$. Its branches look up, and hence the branches of the parabola original function will also look up.
In general, if the coefficient $a>0$, then the branches look up, if the coefficient $a
Example 2
Graph the function: $y=2x^2+4x-6$.

Solution.
Find the coordinates of the vertex of the parabola:
$x_(b)=-\frac(b)(2a)=-\frac(4)(4)=-1$.
$y_(v)=2*(-1)^2+4(-1)-6=2-4-6=-8$.
Note the coordinate of the vertex on the coordinate axis. At this point, as if new system coordinates, we construct a parabola $y=2x^2$.

There are many ways to simplify the construction of parabola graphs.

  • We can find two symmetric points, calculate the value of the function at these points, mark them on the coordinate plane and connect them to the vertex of the curve that describes the parabola.
  • We can build a parabola branch to the right or left of the top and then reflect it.
  • We can build by points.

Example 3
Find the largest and smallest value functions: $y=-x^2+6x+4$ on the interval $[-1;6]$.

Solution.
Let's build a graph of this function, select the required interval and find the lowest and highest points of our graph.
Find the coordinates of the vertex of the parabola:
$x_(b)=-\frac(b)(2a)=-\frac(6)(-2)=3$.
$y_(v)=-1*(3)^2+6*3+4=-9+18+4=13$.
At the point with coordinates $(3;13)$ we construct a parabola $y=-x^2$. Select the required interval. The lowest point has a coordinate of -3, the highest point has a coordinate of 13.
$y_(name)=-3$; $y_(naib)=13$.

Tasks for independent solution

1. Without plotting the function $y=-3x^2+12x-4$, answer the following questions:
a) Indicate the straight line that serves as the axis of the parabola.
b) Find the coordinates of the vertex.
c) Where does the parabola point (up or down)?
2. Construct a graph of the function: $y=2x^2-6x+2$.
3. Graph the function: $y=-x^2+8x-4$.
4. Find the largest and smallest value of the function: $y=x^2+4x-3$ on the interval $[-5;2]$.

A quadratic function is a function of the form:
y=a*(x^2)+b*x+c,
where a is the coefficient at the highest degree of the unknown x,
b - coefficient at unknown x,
and c is a free member.
The graph of a quadratic function is a curve called a parabola. General form parabola is shown in the figure below.

Fig.1 General view of the parabola.

There are a few various ways plotting a quadratic function. We will consider the main and most general of them.

Algorithm for plotting a graph of a quadratic function y=a*(x^2)+b*x+c

1. Build a coordinate system, mark a single segment and label the coordinate axes.

2. Determine the direction of the branches of the parabola (up or down).
To do this, you need to look at the sign of the coefficient a. If plus - then the branches are directed upwards, if minus - then the branches are directed downwards.

3. Determine the x-coordinate of the top of the parabola.
To do this, you need to use the formula Tops = -b / 2 * a.

4. Determine the coordinate at the top of the parabola.
To do this, substitute the value of the Top found in the previous step in the equation of the Top = a * (x ^ 2) + b * x + c instead of x.

5. Put the resulting point on the graph and draw an axis of symmetry through it, parallel to the coordinate axis Oy.

6. Find the points of intersection of the graph with the x-axis.
This requires solving the quadratic equation a*(x^2)+b*x+c = 0 with one of known ways. If the equation has no real roots, then the graph of the function does not intersect the x-axis.

7. Find the coordinates of the point of intersection of the graph with the Oy axis.
To do this, we substitute the value x = 0 into the equation and calculate the value of y. We mark this and the point symmetrical to it on the graph.

8. Find the coordinates of an arbitrary point A (x, y)
To do this, we choose an arbitrary value of the x coordinate, and substitute it into our equation. We get the value of y at this point. Put a point on the graph. And also mark a point on the graph that is symmetrical to the point A (x, y).

9. Connect the obtained points on the graph with a smooth line and continue the graph beyond the extreme points, to the end of the coordinate axis. Sign the graph either on the callout, or, if space permits, along the graph itself.

An example of plotting a graph

As an example, let's plot a quadratic function given by the equation y=x^2+4*x-1
1. Draw coordinate axes, sign them and mark a single segment.
2. The values ​​of the coefficients a=1, b=4, c= -1. Since a \u003d 1, which is greater than zero, the branches of the parabola are directed upwards.
3. Determine the X coordinate of the top of the parabola Tops = -b/2*a = -4/2*1 = -2.
4. Determine the coordinate At the top of the parabola
Tops = a*(x^2)+b*x+c = 1*((-2)^2) + 4*(-2) - 1 = -5.
5. Mark the vertex and draw an axis of symmetry.
6. We find the points of intersection of the graph of a quadratic function with the Ox axis. We solve the quadratic equation x^2+4*x-1=0.
x1=-2-√3 x2 = -2+√3. We mark the obtained values ​​on the graph.
7. Find the points of intersection of the graph with the Oy axis.
x=0; y=-1
8. Choose an arbitrary point B. Let it have a coordinate x=1.
Then y=(1)^2 + 4*(1)-1= 4.
9. We connect the received points and sign the chart.

Tasks on the properties and graphs of a quadratic function, as practice shows, cause serious difficulties. This is rather strange, because the quadratic function is passed in the 8th grade, and then the entire first quarter of the 9th grade is "tortured" by the properties of the parabola and its graphs are built for various parameters.

This is due to the fact that forcing students to build parabolas, they practically do not devote time to "reading" graphs, that is, they do not practice comprehending the information received from the picture. Apparently, it is assumed that, having built two dozen graphs, a smart student himself will discover and formulate the relationship between the coefficients in the formula and appearance graphic arts. In practice, this does not work. For such a generalization, serious experience in mathematical mini-research is required, which, of course, most ninth-graders do not have. Meanwhile, in the GIA they propose to determine the signs of the coefficients precisely according to the schedule.

We will not demand the impossible from schoolchildren and simply offer one of the algorithms for solving such problems.

So, a function of the form y=ax2+bx+c is called quadratic, its graph is a parabola. As the name suggests, the main component is ax 2. That is A should not be equal to zero, the remaining coefficients ( b And With) can be equal to zero.

Let's see how the signs of its coefficients affect the appearance of the parabola.

The simplest dependence for the coefficient A. Most schoolchildren confidently answer: "if A> 0, then the branches of the parabola are directed upwards, and if A < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой A > 0.

y = 0.5x2 - 3x + 1

In this case A = 0,5

And now for A < 0:

y = - 0.5x2 - 3x + 1

In this case A = - 0,5

Influence of coefficient With also easy enough to follow. 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. As a rule, this point is easy to find on the chart. And determine whether it lies above zero or below. That is With> 0 or With < 0.

With > 0:

y=x2+4x+3

With < 0

y = x 2 + 4x - 3

Accordingly, if With= 0, then the parabola will necessarily pass through the origin:

y=x2+4x


More difficult with the parameter b. The point by 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 \u003d - b / (2a). Thus, b = - 2ax in. That is, we act as follows: on the graph we find the top of the parabola, 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, this is not all. We must also pay attention to the sign of the coefficient A. That is, to see where the branches of the parabola are directed. And only after that, according to the formula b = - 2ax in determine sign b.

Consider an example:

Branches pointing upwards A> 0, the parabola crosses the axis at below zero means 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.

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

Recall that ax 2 is the leading term of this square trinomial, and is its leading coefficient.

But the square trinomial does not always have all three terms. Take for example the expression 3x 2 + 2x, where a=3, b=2, c=0.

Let's move on to the quadratic function y \u003d 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 plot a quadratic function, for example, you can use the full square method.

Consider an example of plotting a function y equals -3x 2 - 6x + 1.

To do this, the first thing to remember is the scheme for highlighting the full square in the trinomial -3x 2 - 6x + 1.

We take out -3 from the first two terms in brackets. We have -3 times the sum of x squared plus 2x and add 1. Adding and subtracting the unit in brackets, we get the formula for the square of the sum, which can be collapsed. We get -3 times the sum (x + 1) squared minus 1, add 1. Expanding the brackets and adding like terms, the expression comes out: -3 times the square of the sum (x + 1) add 4.

Let's plot the resulting function by going to auxiliary system coordinates with origin at the point with coordinates (-1; 4).

In the figure from the video, this system is indicated by dotted lines. We bind the function y equals -3x 2 to the constructed coordinate system. For convenience, we take control points. For example, (0;0), (1;-3), (-1;-3), (2;-12), (-2;-12). At the same time, we set aside them in the constructed coordinate system. The parabola obtained during the construction is the graph we need. In the figure, this is a red parabola.

Applying the full square selection method, we have a quadratic function of the form: y = a * (x + 1) 2 + m.

The graph of the parabola y \u003d ax 2 + bx + c is easy to obtain from the parabola y \u003d ax 2 by parallel translation. This is confirmed by a theorem that can be proved by taking the full 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 \u003d ax 2, combining the vertex with the point with coordinates (-l; m). The important thing is that x = -l, which means -b / 2a. So this line is the axis of the parabola ax 2 + bx + c, its vertex is at the point with the abscissa x, zero is equal to minus b divided by 2a, and the ordinate is calculated by the cumbersome formula 4ac - b 2 /. But this formula is not necessary to memorize. Since, by substituting the value of the abscissa into the function, we get the ordinate.

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

Let's take the function y \u003d -3x 2 - 6x + 1. Having drawn up the equation for the axis of the parabola, we have that x \u003d -1. And this value is the x-coordinate of the top of the parabola. It remains to find only the ordinate. Substituting the value -1 into the function, we get 4. The top of the parabola is at the point (-1; 4).

The graph of the function y \u003d -3x 2 - 6x + 1 was obtained with parallel transfer graph of the function y \u003d -3x 2, which means that it behaves similarly. The leading coefficient is negative, so the branches are directed downwards.

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

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

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

Let's plot the function y \u003d 2x 2 - x + 1.

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

Top - (1.5; -3.5). Axis - x=1.5. Take the points x=0 and x=3. y=1. Note these points. Based on three known points, we build the required graph.

To plot 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 x-axis, take two points that are symmetrical about the axis of the parabola, find the value of the function at these points and mark them on the coordinate plane;

Through three points, construct a parabola, if necessary, you can take a few more points and build 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 \u003d -2, b \u003d 8, then x zero is 2, and zero y is 3, (2; 3) is the top of the parabola, and x \u003d 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.