Presentation “Function y=ax2, its graph and properties. Exponential function - properties, graphs, formulas Plotting a graph of the function y ax2 bx c
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"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.
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 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 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.
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 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:
$a*x^2+b*x+c=(a*x^2+b*x)+c=a(x^2+\frac(b)(a)*x)+c=$ $= a(x^2+2\frac(b)(2a)*x+\frac(b^2)(4a))-\frac(b^2)(4a)+c=a(x+\frac(b) (2a))^2+\frac(4ac-b^2)(4a)$.
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 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 $(-l;m)$.
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:
- use the formula: $y_(в)=\frac(4ac-b^2)(4a)$,
- directly substitute the coordinate of the vertex along $x$ into the original function: $y_(в)=ax_(в)^2+b*x_(в)+c$.
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.
Example 1.
Without graphing the function $y=4x^2-6x-3$, answer the following 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 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: $y=2x^2+4x-6$.
Solution.
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 at new system coordinates we will construct a parabola $y=2x^2$.
There are many ways to simplify the construction of parabola graphs.
- We can find two symmetrical points, calculate the value of the function at these points, mark them on coordinate plane and connect them to the vertex of the curve describing the parabola.
- We can construct a branch of the parabola to the right or left of the vertex and then reflect it.
- We can build point by point.
Example 3.
Find the largest and smallest value of the function: $y=-x^2+6x+4$ on the segment $[-1;6]$.
Solution.
Let's build a graph of this function, select the required interval and find the lowest and highest points of our graph.
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. 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 segment $[-5;2]$.
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 use this formula on practice); develop the ability to determine the properties of a quadratic function from a graph (finding axis of symmetry, coordinates of the vertex of the parabola, 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: nurturing 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
Organizational moment (1-2 min)
Updating knowledge (10 min)
Presentation of new material (15 min)
Consolidating new material (12 min)
Summing up (3 min)
Homework assignment (2 min)
During the classes
Organizing time
Greeting, checking absentees, collecting notebooks.
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:
What is a quadratic function? (A function where given real numbers, , is a real variable, is called a quadratic function.)
What is the graph of a quadratic function? (The graph of a quadratic function is a parabola.)
What are the zeros of a quadratic function? (The zeros of a quadratic function are the values at which it becomes zero.)
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.)
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 .)
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: That's 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: Correct. 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: On the drawing: - the 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 of the symmetry axis 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.
Consolidating new material
Teacher: There are tasks written on the board that need to be solved in class.
Board entry: No. 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).
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.
A bad teacher presents the truth, a good teacher teaches how to obtain it.
A.Disterweg
Teacher: Netikova Margarita Anatolyevna, mathematics teacher, GBOU school No. 471, Vyborg district of St. Petersburg.
Lesson topic: “Graph of a functiony= ax 2 »
Lesson type: lesson in learning new knowledge.
Target: teach students to graph a function y= ax 2 .
Tasks:
Educational: develop the ability to construct a parabola y= ax 2 and establish a pattern between the graph of the function y= ax 2
and coefficient A.
Educational: development of cognitive skills, analytical and comparative thinking, mathematical literacy, ability to generalize and draw conclusions.
Educators: nurturing interest in the subject, accuracy, responsibility, demandingness towards oneself and others.
Planned results:
Subject: be able to use a formula to determine the direction of the branches of a parabola and construct it using a table.
Personal: be able to defend your point of view and work in pairs and in a team.
Metasubject: be able to plan and evaluate the process and result of their activities, process information.
Pedagogical technologies: elements of problem-based and advanced learning.
Equipment: interactive whiteboard, computer, handouts.
1.Formula of roots quadratic equation and decomposition quadratic trinomial by multipliers.
2. Reduction of algebraic fractions.
3.Properties and graph of the function y= ax 2 , dependence of the direction of the branches of the parabola, its “stretching” and “compression” along the ordinate axis on the coefficient a.
Lesson structure.
1.Organizational part.
2.Updating knowledge:
Examination homework
Oral work based on finished drawings
3.Independent work
4.Explanation of new material
Preparing to study new material (creating a problem situation)
Primary assimilation of new knowledge
5. Fastening
Application of knowledge and skills in a new situation.
6. Summing up the lesson.
7.Homework.
8. Lesson reflection.
Technological map of an algebra lesson in 9th grade on the topic: “Graph of a functiony=
ax 2
»
| Lesson steps | Stage tasks | Teacher activities | Student activities | UUD |
| 1.Organizational part 1 minute | Creating a working mood at the beginning of the lesson | Greets students checks their preparation for the lesson, notes those absent, writes the date on the board. | Getting ready to work in class, greeting the teacher | Regulatory: organization of educational activities. |
| 2.Updating knowledge 4 minutes | Check homework, repeat and summarize the material learned in previous lessons and create conditions for successful independent work. | Collects notebooks from six students (selectively two from each row) to check homework for assessment (Annex 1), then works with the class on interactive whiteboard (Appendix 2). | Six students hand in their homework notebooks for inspection, then answer front-end survey questions. (Appendix 2). | Cognitive: bringing knowledge into the system. Communicative: the ability to listen to the opinions of others. Regulatory: evaluating the results of your activities. Personal: assessing the level of mastery of the material. |
| 3.Independent work 10 minutes | Test your ability to factor a quadratic trinomial, reduce algebraic fractions, and describe some properties of functions using their graph. | Hands out cards to students with individual differentiated tasks (Appendix 3). and solution sheets. | Execute independent work, independently choosing the difficulty level of exercises based on points. | Cognitive: Personal: assessing the level of mastery of the material and one’s capabilities. |
| 4.Explanation of new material Preparing to study new material Primary assimilation of new knowledge | Creating a favorable environment for getting out of a problematic situation, perception and comprehension of new material, independent coming to the right conclusion | So, you know how to graph a function y= x 2 (graphs are pre-built on three boards). Name the main properties of this function: 3. Vertex coordinates 5. Periods of monotony What's in in this case equal to the coefficient at x 2 ? Using the example of the quadratic trinomial, you saw that this is not at all necessary. What sign could he be? Give examples. You will have to find out for yourself what parabolas with other coefficients will look like. The best way to study something is to discover for yourself. D.Poya We divide into three teams (in rows), choose captains who come to the board. The task for the teams is written on three boards, the competition begins! Construct function graphs in one coordinate system 1 team: a)y=x 2 b)y= 2x 2 c)y= x 2 Team 2: a)y= - x 2 b)y=-2x 2 c)y= - x 2 Team 3: a)y=x 2 b)y=4x 2 c)y=-x 2 Mission accomplished! (Appendix 4). Find functions that have the same properties. Captains consult with their teams. What does this depend on? But how do these parabolas differ and why? What determines the “thickness” of a parabola? What determines the direction of the branches of a parabola? We will conventionally call graph a) “initial”. Imagine a rubber band: if you stretch it, it becomes thinner. This means that graph b) was obtained by stretching the original graph along the ordinate. How was graph c) obtained? So, when x 2 there can be any coefficient that affects the configuration of the parabola. This is the topic of our lesson: "Graph of a functiony= ax 2 » | 1. R 4. Branches up 5. Decreases by (- Increases by ) Share with friends or save for yourself:
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