Lesson: How to construct a parabola or quadratic function?
THEORETICAL PART
A parabola is a graph of a function described by the formula ax 2 +bx+c=0.
To construct a parabola you need to follow simple algorithm actions:
1) Parabola formula y=ax 2 +bx+c,
If a>0 then the branches of the parabola are directed up,
otherwise the branches of the parabola are directed down.
Free member c this point intersects the parabola with the OY axis;
2), it is found using the formula x=(-b)/2a, we substitute the found x into the parabola equation and find y;
3)Function zeros or, in other words, the points of intersection of the parabola with the OX axis, they are also called the roots of the equation. To find the roots we equate the equation to 0 ax 2 +bx+c=0;
Types of equations:
a) Complete quadratic equation looks like ax 2 +bx+c=0 and is solved by the discriminant;
b) Incomplete quadratic equation of the form ax 2 +bx=0. To solve it, you need to take x out of brackets, then equate each factor to 0:
ax 2 +bx=0,
x(ax+b)=0,
x=0 and ax+b=0;
c) Incomplete quadratic equation of the form ax 2 +c=0. To solve it, you need to move the unknowns to one side, and the knowns to the other. x =±√(c/a);
4) Find several additional points to construct the function.
PRACTICAL PART
And so now, using an example, we will analyze everything step by step:
Example #1:
y=x 2 +4x+3
c=3 means the parabola intersects OY at the point x=0 y=3. The branches of the parabola look up since a=1 1>0.
a=1 b=4 c=3 x=(-b)/2a=(-4)/(2*1)=-2 y= (-2) 2 +4*(-2)+3=4- 8+3=-1 vertex is at point (-2;-1)
Let's find the roots of the equation x 2 +4x+3=0
Using the discriminant we find the roots
a=1 b=4 c=3
D=b 2 -4ac=16-12=4
x=(-b±√(D))/2a
x 1 =(-4+2)/2=-1
x 2 =(-4-2)/2=-3
Let's take several arbitrary points that are located near the vertex x = -2
x -4 -3 -1 0
y 3 0 0 3
Substitute instead of x into the equation y=x 2 +4x+3 values
y=(-4) 2 +4*(-4)+3=16-16+3=3
y=(-3) 2 +4*(-3)+3=9-12+3=0
y=(-1) 2 +4*(-1)+3=1-4+3=0
y=(0) 2 +4*(0)+3=0-0+3=3
It can be seen from the function values that the parabola is symmetrical with respect to the straight line x = -2
Example #2:
y=-x 2 +4x
c=0 means the parabola intersects OY at the point x=0 y=0. The branches of the parabola look down since a=-1 -1 Let's find the roots of the equation -x 2 +4x=0
Incomplete quadratic equation of the form ax 2 +bx=0. To solve it, you need to take x out of brackets, then equate each factor to 0.
x(-x+4)=0, x=0 and x=4.
Let's take several arbitrary points that are located near the vertex x=2
x 0 1 3 4
y 0 3 3 0
Substitute instead of x into the equation y=-x 2 +4x values
y=0 2 +4*0=0
y=-(1) 2 +4*1=-1+4=3
y=-(3) 2 +4*3=-9+13=3
y=-(4) 2 +4*4=-16+16=0
It can be seen from the function values that the parabola is symmetrical about the straight line x = 2
Example No. 3
y=x 2 -4
c=4 means the parabola intersects OY at the point x=0 y=4. The branches of the parabola look up since a=1 1>0.
a=1 b=0 c=-4 x=(-b)/2a=0/(2*(1))=0 y=(0) 2 -4=-4 the vertex is at point (0;-4 )
Let's find the roots of the equation x 2 -4=0
Incomplete quadratic equation of the form ax 2 +c=0. To solve it, you need to move the unknowns to one side, and the knowns to the other. x =±√(c/a)
x 2 =4
x 1 =2
x 2 =-2
Let's take several arbitrary points that are located near the vertex x=0
x -2 -1 1 2
y 0 -3 -3 0
Substitute instead of x into the equation y= x 2 -4 values
y=(-2) 2 -4=4-4=0
y=(-1) 2 -4=1-4=-3
y=1 2 -4=1-4=-3
y=2 2 -4=4-4=0
It can be seen from the function values that the parabola is symmetrical about the straight line x = 0
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The presentation “Function y=ax 2, its graph and properties” is a visual aid that was created to accompany the teacher’s explanation on this topic. This presentation discusses in detail the quadratic function, its properties, features of plotting, and the practical application of the methods used for solving problems in physics.
By providing high degree clarity, this material will help the teacher to increase the effectiveness of teaching and will provide an opportunity to more rationally distribute time in the lesson. With the help of animation effects, highlighting concepts and important points in color, students’ attention is focused on the subject being studied, achieving better memorization definitions and the course of reasoning when solving problems.
The presentation begins with an introduction to the title of the presentation and the concept quadratic function. The importance of this topic is emphasized. Students are asked to remember the definition of a quadratic function as a functional dependence of the form y=ax 2 +bx+c, in which is an independent variable, and are numbers, with a≠0. Separately, on slide 4 it is noted for remembering that the domain of definition of this function is the entire axis of real values. Conventionally, this statement is denoted by D(x)=R.
An example of a quadratic function is its important application in physics - the path dependence formula for uniformly accelerated motion from time to time. At the same time, in physics lessons, students study formulas various types movements, so they will need the ability to solve such problems. On slide 5, students are reminded that when a body moves with acceleration and at the beginning of the time count the distance traveled and the speed of movement are known, then the functional dependence representing such movement will be expressed by the formula S=(at 2)/2+v 0 t+S 0 . Below is an example of turning this formula into a given quadratic function if the values of acceleration = 8, initial speed = 3 and initial path = 18. In this case, the function will take the form S=4t 2 +3t+18.
Slide 6 examines the form of the quadratic function y=ax 2, in which it is represented at. If =1, then the quadratic function has the form y=x 2. It is noted that the graph of this function will be a parabola.
The next part of the presentation is devoted to plotting a quadratic function. It is proposed to consider plotting the function y=3x 2 . First, the table indicates the correspondence between the function values and the argument values. It is noted that the difference between the constructed graph of the function y=3x 2 and the graph of the function y=x 2 is that each value will be three times greater than the corresponding one. This difference is well tracked in the table view. Nearby, in the graphical representation, the difference in the narrowing of the parabola is also clearly visible.
The next slide looks at plotting the quadratic function y=1/3 x 2. To construct a graph, you need to indicate in the table the values of the function at a number of its points. It is noted that each value of the function y=1/3 x 2 is 3 times less than the corresponding value of the function y=x 2. This difference, in addition to the table, is clearly visible in the graph. Its parabola is more expanded relative to the ordinate axis than the parabola of the function y=x 2.
Examples help you understand general rule, according to which you can then more simply and quickly construct the corresponding graphs. On slide 9, a separate rule is highlighted that the graph of the quadratic function y=ax 2 can be constructed depending on the value of the coefficient by stretching or narrowing the graph. If a>1, then the graph stretches from the x-axis by a factor. If 0 The conclusion about the symmetry of the graphs of the functions y=ax 2 and y=-ax2 (at ≠0) relative to the abscissa axis is separately highlighted on slide 12 for memorization and is clearly displayed on the corresponding graph. Next, the concept of the graph of a quadratic function y=x 2 is extended to the more general case of the function y=ax 2, stating that such a graph will also be called a parabola. Slide 14 discusses the properties of the quadratic function y=ax 2 when positive. It is noted that its graph passes through the origin of coordinates, and all points except lie in the upper half-plane. The symmetry of the graph relative to the ordinate axis is noted, specifying that opposite values of the argument correspond to the same function values. It is indicated that the interval of decrease of this function is (-∞;0], and the increase of the function is performed on the interval. The values of this function cover the entire positive part of the real axis, it is equal to zero at the point, and has no greatest value. Slide 15 describes the properties of the function y=ax 2 if negative. It is noted that its graph also passes through the origin, but all its points, except, lie in the lower half-plane. The graph is symmetrical about the axis, and opposite values of the argument correspond to equal values of the function. The function increases on the interval and decreases on. The values of this function lie in the interval, it is equal to zero at a point, and has no minimum value. Summarizing the characteristics considered, on slide 16 it is concluded that the branches of the parabola are directed downwards at, and upwards at. The parabola is symmetrical about the axis, and the vertex of the parabola is located at the point of its intersection with the axis. The vertex of the parabola y=ax 2 is the origin. Also, an important conclusion about parabola transformations is displayed on slide 17. It presents options for transforming the graph of a quadratic function. It is noted that the graph of the function y=ax 2 is transformed by symmetrically displaying the graph relative to the axis. It is also possible to compress or stretch the graph relative to the axis. The last slide makes general conclusions about transformations of the graph of a function. The conclusions are presented that the graph of a function is obtained by a symmetric transformation about the axis. And the graph of the function is obtained by compressing or stretching the original graph from the axis. In this case, tensile extension from the axis is observed in the case when. By compressing the axis by 1/a times, the graph is formed in the case. The presentation “Function y=ax 2, its graph and properties” can be used by a teacher as a visual aid in an algebra lesson. Also, this manual covers the topic well, giving an in-depth understanding of the subject, so it can be offered for independent study by students. This material will also help the teacher give explanations during distance learning. 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 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 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 a 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. Additional materials Educational aids and simulators in the Integral online store for grade 8
Guys, in the last lessons we built a large number of graphs, including many parabolas. Today we will summarize the knowledge we have gained and learn how to plot this function in its most general form. Let's look at the function $y=a*x^2+b*x+c$. This function is called "quadratic" because the highest power is second, that is, a square. The coefficients are the same as defined above. In the last lesson, in the last example, we looked at plotting a graph of a similar function. 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 proven in the most general case. Today we will look at one such evidence. Guys, you can see the full power of the mathematical apparatus, but also its complexity. Let us isolate the perfect square from the quadratic trinomial: To plot the graph $y=a(x+l)^2+m$, you need to plot the function $y=ax^2$. Moreover, the vertex of the parabola will be located at the point with coordinates $(-l;m)$. Example 1. Solution. Solution. There are many ways to simplify the construction of parabola graphs. Example 3. Solution. 
Presentation and lesson on the topic:
"Graph of the function $y=ax^2+bx+c$. Properties"
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A manual for the textbook by Dorofeev G.V. A manual for the textbook by Nikolsky S.M.
Let's look at the quadratic trinomial $a*x^2+b*x+c$. $a, b, c$ are called coefficients. They can be any numbers, but $a≠0$. $a*x^2$ is called the leading term, $a$ is 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 is equal to zero.
Let's prove that any such quadratic function can be reduced to the form: $y=a(x+l)^2+m$.
$a*x^2+b*x+c=(a*x^2+b*x)+c=a(x^2+\frac(b)(a)*x)+c=$ $= a(x^2+2\frac(b)(2a)*x+\frac(b^2)(4a))-\frac(b^2)(4a)+c=a(x+\frac(b) (2a))^2+\frac(4ac-b^2)(4a)$.
We got what we wanted.
Any quadratic function can be represented as:
$y=a(x+l)^2+m$, where $l=\frac(b)(2a)$, $m=\frac(4ac-b^2)(4a)$.
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 axis, as we can see, are calculated by the formula: $x_(c)=-\frac(b)(2a) $.
To calculate the y-axis coordinate of the vertex of a parabola, you can:
How to calculate the ordinate of a vertex? Again, the choice is yours, but usually the second method will be easier to calculate.
If you need to describe some properties or answer some specific questions, you do not always need to build a graph of the function. We will consider the main questions that can be answered without construction in the following example.
Without graphing the function $y=4x^2-6x-3$, answer the following questions:
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 abscissa of the vertex above $x_(c)=\frac(3)(4)$.
We find the ordinate of the vertex by direct substitution into the original function:
$y_(в)=4*(\frac(3)(4))^2-6*\frac(3)(4)-3=\frac(9)(4)-\frac(18)(4 )-\frac(12)(4)=-\frac(21)(4)$.
c) The graph of the required function will be obtained by parallel transfer of the graph $y=4x^2$. Its branches look up, which means the branches of the parabola of the original function will also look up.
In general, if the coefficient $a>0$, then the branches look upward, if the coefficient $a
Example 2.
Graph the function: $y=2x^2+4x-6$.
Let's find the coordinates of the vertex of the parabola:
$x_(c)=-\frac(b)(2a)=-\frac(4)(4)=-1$.
$y_(в)=2*(-1)^2+4(-1)-6=2-4-6=-8$.
Let's mark the coordinate of the vertex on the coordinate axis. At this point, as if in a new coordinate system, we will construct a parabola $y=2x^2$.
Find the largest and smallest value of the function: $y=-x^2+6x+4$ on the segment $[-1;6]$.
Let's build a graph of this function, select the required interval and find the lowest and highest points of our graph.
Let's find the coordinates of the vertex of the parabola:
$x_(c)=-\frac(b)(2a)=-\frac(6)(-2)=3$.
$y_(в)=-1*(3)^2+6*3+4=-9+18+4=13$.
At the point with coordinates $(3;13)$ we construct a parabola $y=-x^2$. 
Let's select the required interval. 
The lowest point has a coordinate of -3, the highest point has a coordinate of 13.
$y_(name)=-3$; $y_(maximum)=13$.Problems to solve independently
1. Without graphing the function $y=-3x^2+12x-4$, answer the following questions:
a) Identify the straight line that serves as the axis of the parabola.
b) Find the coordinates of the vertex.
c) Which way does the parabola point (up or down)?
2. Construct a graph of the function: $y=2x^2-6x+2$.
3. Construct a graph of the function: $y=-x^2+8x-4$.
4. Find the largest and smallest value of the function: $y=x^2+4x-3$ on the segment $[-5;2]$.
