Solution of square inequalities, presentation. Solution of square inequalities, presentation Parabola touches the abscissa axis
Graphical method for solving square inequalities Algebra grade 8
Definition Square inequalities are inequalities of the form ax 2 + b x + c> 0, ax 2 + b x + c
Using the graph of the function y = x 2 - 6 x + 8, determine at what values of x a) y = 0, b) y> 0, c) y 0 for x 4 y
Algorithm for solving the square inequality Find the roots of the square trinomial ax 2 + b x + c Mark the found roots on the x-axis and determine where the branches of the parabola are directed (up or down), which serves as the graph of the function y = ax 2 + b x + c; sketch out the graph. Using the obtained geometric model, determine at which intervals of the x-axis the ordinates of the graph are positive (negative); include these gaps in the response.
Example 1 Solve the inequality: x 2 - 9 0 x 2 - 9 = 0, x 2 = 9, x 1,2 = 3, mark the roots on the Ox axis The branches of the parabola are directed upwards (a = 1, 1> 0) Draw sketch of the graph We are looking for the values of x at which the points of the parabola lie above or on the Ox axis (the sign of the inequality is not strict “≥”) Answer: x - 3, x 3 - 3 3 x x - 3 x 3
Example 2 Solve the inequality: х 2 - х +12> 0 х 2 - х +12 = 0, х 1 = - 4, х 2 = 3 The branches of the parabola are directed downward (a = - 1, -1 ") Answer: - 4 - 4
Example 3 Solve the inequality: x 2 + 9> 0 x 2 + 9 = 0, x 2 = 9, 9 0) Draw a sketch of the graph We are looking for the values of x at which the graph of the function is located above the Ox axis. Answer: x is any number (or (- ∞; + ∞)). x All points of the parabola lie above the Ox axis. The inequality holds for any value of x
Example 4 Solve the inequality: x 2 + 9 0) Draw a sketch of the graph We are looking for the values of x at which the graph of the function is located below the Ox axis. Answer: there are no solutions x There are no points on the parabola that lie below the Ox axis. Inequality has no solutions.
Example 5 Solve the inequality: - 4x 2 + 12x-9 0 - 4x 2 + 12x-9 = 0, D = 0, x = 1.5 The branches of the parabola are directed downward (a = 4, 4
Example 6 Solve the inequality: - 4x 2 + 12x-9> 0 - 4x 2 + 12x-9 = 0, D = 0, x = 1.5 The branches of the parabola are directed downward (a = 4, 4
Example 7 Solve the inequality: - 4x 2 + 12x-9 0 - 4x 2 + 12x-9 = 0, D = 0, x = 1.5 The branches of the parabola are directed downward (a = 4, 4
Example 8 Solve the inequality: - 4x 2 + 12x-9
On the subject: methodological developments, presentations and notes
1.Demonstration material for systematizing and generalizing knowledge on the above topic, made in the form of a multimedia presentation with video and sound, which will make it possible to use it both in the lesson and for ...
Inequalities occupy an important place in the algebra course. They are not a small part of the content of the entire course of algebra. Thanks to the ability to solve various kinds of inequalities, success in many other sciences can be achieved. In order for the material that is taught in the lesson to be better assimilated, it is recommended to use different visualizations, including presentations.
slides 1-2 (Presentation topic "Solving square inequalities. Part 1", example)
This presentation is intended to be a lesson in explaining new material that is part of the Inequalities Lesson System. Before embarking on the study of this topic "Solving square inequalities", students should receive the necessary amount of knowledge about what inequality is, the properties of numerical inequalities, and how linear inequalities are solved. Presentations on these topics are available on this resource.
At the very beginning of the presentation, the author invites students to get acquainted with the concept of square inequalities. He defines them as an inequality of the form ax2 + bx + c> 0, where a> 0. In order to learn how to solve such inequalities, it is enough to know how they look. Therefore, the author immediately proposes to study the methods of solving the problem at once by examples. And the first such example demonstrates that you need to consider the function that is on the left side of the inequality. You should build her schedule. Since the task is subdivided into four subparagraphs, and all these inequalities differ only in sign, then one schedule is sufficient for all these cases. It should now be used to determine decisions.
For the first case, you need to find all the values of the function that take only positive values. On the graph, this will correspond to all the points of the graph that lie strictly above the abscissa axis. In order to determine the solutions of the second case, it is necessary to consider all points of the graph of this function, which lie strictly below the abscissa axis. Since the inequality sign is strictly less than zero... The third case differs from the first only in that the function can also take on the value zero, therefore zero is added to the solution of the first case.
slides 3-4 (examples)
Similarly, the fourth case, which is related to the second. It has the same solutions including zero. Using this example, the author shows how the solutions to inequality are written correctly in different cases. that is, in which case the parenthesis is round, in which case it is square.
Next is the second example, which shows a slightly different way to solve the quadrant inequality. Here it is already necessary to plot the graph of the function not in the coordinate system, but on a straight line, where the points of intersection of the graph with the abscissa axis should be marked. And then, looking at the sign of inequality, you should determine which part of the graph is required as solutions, which lies below or above this line. In this case, the sections of the graph are taken that lie below the straight line.
Therefore, the decision interval will be double. On the same slide, there is another example, which shows the case when the graph does not intersect a straight line, but only touches it at one point. But since, according to the condition, the sign is less than or equal to zero, then a section must be selected that is located below the straight line. But there are no such sites, the whole schedule is higher. But since equality to zero is allowed in the condition, then the only solution there will be a variable value equal to 0.5.
slides 5-6 (solution algorithm, theorem)
Then the author comes to an algorithm for solving square inequalities. It has three points. According to the first point, the quadratic equation should be solved by equating the quadratic trinomial to zero. Then mark the resulting roots on a straight line, which is the x-axis, and draw a parabola through these points by hand, taking into account the direction of the branches. And then, using this model, find all the solutions to the inequality.
And at the end of the presentation, the author proposes to consider a theorem that connects the number of solutions to an inequality from the sign of the discriminant of a trinomial. This means that with a negative discriminant with a positive first coefficient, the inequality ax2 + bx + c, which is greater than or equal to zero, has no solutions, and if greater than zero, then the solutions are all real values of the variable x.
This presentation can become an irreplaceable part of the lesson on the topic "Solving square inequalities". But this presentation is only the first part. Therefore, the continuation of this topic follows. And you can also find a presentation that will be a continuation of this one with us. If the teacher wishes, you can add your own examples to the presentation.
This presentation can be used to explain the topic "Square Inequalities". Textbook Algebra Grade 9. Authors: G.B. Dorofeev, S.B. Suvorova, E.A. Bunimovich, L.V. Kuznetsova, S. S. Minaeva. With the help of animation effects in an accessible form, the concept of square inequality is introduced. The presentation provides an algorithm for solving a square inequality, an example of a solution by an algorithm, a slide for oral work on a finished drawing of a function graph.
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Square inequalities Mathematics teacher MOU secondary school №57 in Astrakhan Bunina NV
y 0 y> 0 Y = 0 x y 2 - 3 1 y = x + x-6 2 For x = -3 and x = 2 For -3 2 For x = -3 and x = 2 x + x-6 = 0 For -3 0 y = 0 y 0 2 2 2 Inequalities of the form ax + bx + c ≥ 0, ax + bx + c> 0 or ax + bx + c ≤0, ax + bx + c
Algorithm for solving the quadratic inequality Consider the function y = ax 2 + bx + c Find the zeros of the function (solve the equation Determine the direction of the parabola branches Schematically plot the function graph. Taking into account the inequality sign, write out the answer. Ax 2 + bx + c = 0
D> 0 D = 0 D 0 a
x 2.5 1 Solve the inequality 2x -7x + 5 0 the branches of the parabola are directed upward Answer: (1; 2.5) 1. 2x -7x + 5 = 0 D = b-4ac = (- 7) -4 * 2 * 5 = 9 x = 1, x = 2.5 1 2 2 2 2 Example
1 3 y x y = x - 2x - 3 2 Solve the inequality a) x - 2x - 3> 0 2 b) x - 2x - 3≥ 0 2 c) x - 2x - 3
Solve the inequality - 4x + 2x≥0 2 1. - 4x + 2x = 0 2 4x -2x = 0 2 2x (2x -1) = 0 X = 0 x = 0.5 1 2 2.a
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Definition Square inequalities are inequalities of the form ax 2 + bx + c> 0, ax 2 + bx + c 0, ax 2 + bx + c "> 0, ax 2 + bx + c"> 0, ax 2 + bx + c "title =" (! LANG: Definition Square inequalities are inequalities of the form ax 2 + bx + c> 0 , ax 2 + bx + c"> 0, ах 2 +bх+c" title="Definition Square inequalities are inequalities of the form ax 2 + bx + c> 0, ax 2 + bx + c"> !}
Using the graph of the function y = x 2 - 6x +8, determine at what values of x a) y = 0, b) y> 0, c) y0 for x 4 y 0, c) y0 for x 4 y "> 0, c) y0 for x 4 y"> 0, c) y0 for x 4 y "title =" (! LANG: According to the graph of the function y = x 2 - 6x +8 determine at what values of x a) y = 0, b) y> 0, c) y0 for x 4 y"> title="Using the graph of the function y = x 2 - 6x +8, determine at what values of x a) y = 0, b) y> 0, c) y0 for x 4 y"> !}
Algorithm for solving the square inequality 1. Find the roots of the square trinomial ax 2 + bx + c 2. Mark the found roots on the x axis and determine where (up or down) the branches of the parabola, which serves as the graph of the function y = ax 2 + bx + c; sketch out the graph. 3. Using the obtained geometric model, determine at which intervals of the x-axis the ordinates of the graph are positive (negative); include these gaps in the response.
0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the points of the parabola lie above or on the Ox axis (the sign of the nera "title =" (! LANG: Example 1 Solve the inequality: x 2 - 9 0 1.x 2 - 9 = 0, x 2 = 9, x 1,2 = 3, mark the roots on the Ox axis 2. The branches of the parabola are directed upwards (a = 1, 1> 0) 3. Draw a sketch of the graph 4. We look for the values of x at which the points parabolas lie above or on the Ox axis (the sign of nep" class="link_thumb"> 5 !} Example 1 Solve the inequality: x 2 - x 2 - 9 = 0, x 2 = 9, x 1,2 = 3, mark the roots on the Ox axis 2. The branches of the parabola are directed upwards (a = 1, 1> 0) 3. Draw sketch of the graph 4. We are looking for the values of x at which the points of the parabola lie above or on the Ox axis (the sign of the inequality is not strict) 5. Answer: x - 3, x x x - 3 x 3 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the points of the parabola lie above or on the Ox axis (the sign of y nera "> 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the points of the parabola lie above or on axis Ox (the sign of the inequality is not strict) 5. Answer: x - 3, x 3 - 3 3 x x - 3 x 3 "> 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the points of the parabola lie above or on Axis Ox (sign at nera "title =" (! LANG: Example 1 Solve the inequality: x 2 - 9 0 1.x 2 - 9 = 0, x 2 = 9, x 1,2 = 3, mark the roots on the Ox axis 2. The branches of the parabola are directed upwards (a = 1, 1> 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the points of the parabola lie above or on the Ox axis (the sign of the"> title="Example 1 Solve the inequality: x 2 - 9 0 1.x 2 - 9 = 0, x 2 = 9, x 1,2 = 3, mark the roots on the Ox axis 2. The branches of the parabola are directed upwards (a = 1, 1> 0 ) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the points of the parabola lie above or on the Ox axis (the sign of y"> !}
0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1 "title =" (! LANG: Example 2 Solve the inequality: x 2 - x +12> 0 1.x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1" class="link_thumb"> 6 !} Example 2 Solve the inequality: x 2 - x +12> 0 1.x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1) 5 . Answer: - 4 - 4 0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1 "> 0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1) 5. Answer: - 4 - 4 0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1 "title =" (! LANG: Example 2 Solve the inequality: x 2 - x +12> 0 1. x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1"> title="Example 2 Solve the inequality: x 2 - x +12> 0 1.x 2 - x +12 = 0, x 1 = - 4, x 2 = 3 2. The branches of the parabola are directed downward (a = - 1, -1">!}
0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. Look for the values of x at which the graph of the function is located above the axis "title =" (! LANG: Example 3 Solve the inequality: x 2 + 9> 0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the graph of the function is located above the axis" class="link_thumb"> 7 !} Example 3 Solve the inequality: x> 0 1.x = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the graph of the function is located above the Ox axis. 5. Answer: x - any number (or (-; +)). x All points of the parabola lie above the Ox axis. The inequality holds for any value of x 0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the graph of the function is located above the axis "> 0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. Look for the values of x at which the graph of the function is located above the Ox axis. 5. Answer: x - any number (or (-; +)). X All points of the parabola lie above the Ox axis. The inequality is fulfilled for any value of x "> 0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We look for the values of x at which the graph of the function is located above the axis" title = "(! LANG : Example 3 Solve the inequality: x 2 + 9> 0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We look for the values of x at which the graph of the function is located above the axis"> title="Example 3 Solve the inequality: x 2 + 9> 0 1.x 2 + 9 = 0, x 2 = 9, 9 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the graph of the function is located above the axis"> !}
0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the graph of the function is located below the axis "title =" (! LANG: Example 4 Solve the inequality: x 2 + 9 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the function graph is located below the os" class="link_thumb"> 8 !} Example 4 Solve the inequality: x 0) 3. Draw a sketch of the graph 4. Look for the values of x at which the graph of the function is located below the Ox axis. 5. Answer: there are no solutions х On the parabola there are no points lying below the axis Ox. Inequality has no solutions. 0) 3.Draw a sketch of the graph 4.Look for the values of x at which the graph of the function is located below the axis "> 0) 3.Draw the sketch of the graph 4.Look for the values of x at which the graph of the function is located below the axis. 5.Answer: there are no solutions x On the parabola has no points lying below the Ox axis. The inequality of solutions has no. "> 0) 3. Draw a sketch of the graph 4. We look for the values of x at which the graph of the function is located below the axis" title = "(! LANG: Example 4 Solve the inequality: x 2 + 9 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the graph of the function is located below the axis"> title="Example 4 Solve the inequality: x 2 + 9 0) 3. Draw a sketch of the graph 4. We are looking for the values of x at which the graph of the function is located below the axis"> !}
Example 5 Solve the inequality: - 4x 2 + 12x x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downward (a = 4, 4
Example 6 Solve the inequality: - 4x 2 + 12x-9> x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downward (a = 4, 4 0 1.- 4x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downward (a = 4, 4 "> 0 1.- 4x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downward (a = 4.4 "> 0 1.- 4x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downward (a = 4, 4 "title =" (! LANG: Example 6 Solve the inequality: - 4x 2 + 12x-9> 0 1.- 4x 2 + 12x-9 = 0, D = 0, x = 1.5 2 The branches of the parabola are directed downward (a = 4, 4"> title="Example 6 Solve the inequality: - 4x 2 + 12x-9> 0 1.- 4x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downward (a = 4, 4"> !}
Example 7 Solve the inequality: - 4x 2 + 12x x 2 + 12x-9 = 0, D = 0, x = 1.5 2. The branches of the parabola are directed downward (a = 4, 4
Skills and skills required to successfully solve square inequalities graphically. 1) Be able to solve quadratic equations. 2) be able to build a graph quadratic function and determine from the graph at what values of x the function takes positive, negative, non-positive, non-negative values. shah.ucoz.ru/load/8_klass/8_klass/postroenie_grafikov_vida_u_f_x_l_m_postroenie_grafika_kvadrati chnoj_funkcii /
0. We can solve the inequality graphically. To do this, p "title =" (! LANG: Let's build a graph and determine at what values of x the function takes positive values. A square inequality is an inequality that can be reduced to the form ax 2 + bx + c> 0. We can solve the inequality graphically method. For this p" class="link_thumb"> 3 !} Let's build a graph and determine at what values of x the function takes positive values. A square inequality is an inequality that can be reduced to the form ax 2 + bx + c> 0. We can solve the inequality graphically. To do this, consider the function 0. We can solve the inequality graphically. For this p "> 0. We can solve the inequality graphically. To do this, consider the function"> 0. We can solve the inequality graphically. To do this, p "title =" (! LANG: Let's build a graph and determine at what values of x the function takes positive values. A square inequality is an inequality that can be reduced to the form ax 2 + bx + c> 0. We can solve the inequality graphically method. For this p"> title="Let's build a graph and determine at what values of x the function takes positive values. A square inequality is an inequality that can be reduced to the form ax 2 + bx + c> 0. We can solve the inequality graphically. For this p"> !}
X Y 1 1 x 01 y a> 0 - the branches are directed upwards X x = 2 - the axis of symmetry Let's mark the symmetrical points. Let's build a graph. 0 - branches directed upwards Х х = 2 - axis of symmetry Let's mark symmetrical points. Let's build a graph. "> 0 - branches are directed upwards X x = 2 - an axis of symmetry. Let's mark symmetrical points. Let's build a graph."> 0 - branches are directed upward. X x = 2 - an axis of symmetry. Let's mark symmetric points. Let's build a graph. "Title =" (! LANG: 07/26/20154 Х У 1 1 х 01 у-5-8-2 а> 0 - branches are directed upwards Х х = 2 - axis of symmetry Let's mark the symmetrical points."> title="07/26/20154 Х У 1 1 х 01 у-5-8-2 а> 0 - branches directed upwards Х х = 2 - axis of symmetry Let's mark the symmetrical points. Let's build a graph."> !}
Let us determine at what values of x the function takes positive values X Y 1 1 X (the part of the graph that lies above Ox). 5
0 - branches directed upwards х = 2 - axis of symmetry Let's mark symmetrical points. What actions are required? Points of intersection with Ox. "Title =" (! LANG: What actions were unnecessary? 07/26/20156 Y 1 1 X 5-1 x 01 y-5-8-2 a> 0 - branches are directed upwards x = 2 - axis of symmetry Let's mark the symmetrical points What actions are necessary? Points of intersection with Ox." class="link_thumb"> 6 !} What actions were unnecessary? Y 1 1 X 5-1 x 01 y a> 0 - the branches are directed upwards х = 2 - the axis of symmetry Let's mark the symmetrical points. What actions are required? Intersection points with Oh. 0 - branches directed upwards х = 2 - axis of symmetry Let's mark the symmetrical points. What actions are required? Points of intersection with Ox. "> 0 - branches are directed upwards х = 2 - axis of symmetry Let's mark symmetric points. What actions are necessary? Points of intersection with Ox."> 0 - branches are directed upwards х = 2 - axis of symmetry Let's mark symmetric points. What actions are required? Points of intersection with Ox. "Title =" (! LANG: What actions were unnecessary? 07/26/20156 Y 1 1 X 5-1 x 01 y-5-8-2 a> 0 - branches are directed upwards x = 2 - axis of symmetry Let's mark the symmetrical points What actions are required? Points of intersection with Ox."> title="What actions were unnecessary? 07/26/20156 Y 1 1 X 5-1 x 01 y-5-8-2 a> 0 - branches directed upwards х = 2 - axis of symmetry Let's mark the symmetrical points. What actions are required? Intersection points with Oh."> !}
0 - branches are directed upwards 1) Introduce the function 3) Find the intersection points with Ox: for this we solve the square equation "title =" (! LANG: Algorithm for solving the square inequality using the example of inequality. branches of the parabola. a> 0 - the branches are directed upward 1) Introduce the function 3) Find the intersection points with Ox: for this we solve the square equation" class="link_thumb"> 7 !} Algorithm for solving a square inequality using the example of inequality X) Define the direction of the branches of the parabola. a> 0 - the branches are directed upwards 1) Introduce the function 3) Find the intersection points with Ox: for this we solve the quadratic equation 4) We schematically depict a parabola. 5) Let's look at the inequality sign, select the corresponding parts of the graph and the corresponding parts of Ox. 6) 0 - branches are directed upward 1) Let us introduce the function 3) Find the points of intersection with Ox: for this we solve the square equation "> 0 - the branches are directed upward 1) We introduce the function 3) Find the points of intersection with Ox: for this we solve the quadratic equation 4) We schematically depict parabola. 5) Let's look at the inequality sign, select the corresponding parts of the graph and the corresponding parts of Ox. 6) "> 0 - branches are directed upwards 1) Introduce the function 3) Find the intersection points with Ox: for this we solve the square equation" title = "(! LANG: Algorithm for solving a square inequality using the example of an inequality. 26.07.20157 X 5 26.07.2015 2) Determine the direction of the branches of the parabola. a> 0 - the branches are directed upward 1) Introduce the function 3) Find the intersection points with Ox: for this we solve the square equation"> title="Algorithm for solving a square inequality using the example of inequality. 07/26/2015 7 X 5 07/26/2015 2) Determine the direction of the branches of the parabola. a> 0 - the branches are directed upward 1) Introduce the function 3) Find the intersection points with Ox: for this we solve the square equation"> !}
Algorithm for solving a square inequality using the example of inequality X) Define the direction of the branches of the parabola. a 
Branches, parabola not Oh. How can the parabola y = ax 2 + bx + c be located depending on the behavior of the coefficient a and the discriminant? 1) a> 0 D> 0 Branches, two points with Ox. X 2) a 0 X 3) a> 0 D = 0 X 4) a 0 X 5) a> 0 D 0 D> 0 Branches, two points with Ox. X 2) a 0 X 3) a> 0 D = 0 X 4) a 0 X 5) a> 0 D 0 D> 0 Branches, two points with Ox. X 2) a 0 X 3) a> 0 D = 0 X 4) a 0 X 5) a> 0 D 0 D> 0 Branches, two points with Ox. X 2) a 0 X 3) a> 0 D = 0 X 4) a 0 X 5) a> 0 D 0 D> 0 Branches, two points with Ox. X 2) a 0 X 3) a> 0 D = 0 X 4) a 0 X 5) a> 0 D 
0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) In this case, D = 0. x = -2 is the touch point. "title =" (! LANG: 26.07.201510 X -2-2 26.07.2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) In this case, D = 0. x = -2 - point of tangency." class="link_thumb">
10
!} X) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) In this case, D = 0. x = -2 - point of tangency. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) In this case, D = 0. x = -2 - point of tangency. "> 0 - branches. 1) V. f. 3) Ox: 4) We schematically depict a parabola. 5) => the graph is not lower than Ox. 6) In this case D = 0. x = -2 - touch point. "> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) In this case, D = 0. x = -2 is the touch point. "title =" (! LANG: 26.07.201510 X -2-2 26.07.2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) In this case, D = 0. x = -2 - point of tangency.">
title="07/26/2015 10 X -2-2 07/26/2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) In this case, D = 0. x = -2 - point of tangency.">
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0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph above Oh. 6) In this case, D = 0. x = -2 - point of tangency. What has changed? "Title =" (! LANG: 26.07.201511 X -2-2 26.07.2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph above Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed?" class="link_thumb"> 11 !} X) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph above Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph above Oh. 6) In this case, D = 0. x = -2 - point of tangency. What has changed? "> 0 - branches. 1) V. f. 3) Ox: 4) We schematically depict the parabola. 5) => the graph is higher than Ox. 6) In this case, D = 0. x = -2 is the point of tangency. What has changed? "> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph above Oh. 6) In this case, D = 0. x = -2 - point of tangency. What has changed? "Title =" (! LANG: 26.07.201511 X -2-2 26.07.2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph above Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed?"> title="07/26/2015 11 X -2-2 07/26/2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph above Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed?"> !}
0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the graph is not higher than Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Not higher Oh no there is one dot. "Title =" (! LANG: 26.07.201512 X -2-2 26.07.2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the graph is not higher than Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? No higher Oh no there is one point." class="link_thumb"> 12 !} X) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the graph is not higher than Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? No higher Oh no there is one point. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the graph is not higher than Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? There is one point not higher than Ox no. "> 0 - branches. 1) V. f. 3) Ox: 4) We schematically depict a parabola. 5) => the graph is not higher than Ox. 6) In this case D = 0. X = -2 - point of tangency. What has changed? No higher Oh no there is one point. "> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the graph is not higher than Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Not higher Oh no there is one dot. "Title =" (! LANG: 26.07.201512 X -2-2 26.07.2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the graph is not higher than Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? No higher Oh no there is one point."> title="07/26/2015 12 X -2-2 07/26/2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the graph is not higher than Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? No higher Oh no there is one point."> !}
0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph below Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Ø There is not a single dot below Oh. "Title =" (! LANG: 26.07.201513 X -2-2 26.07.2015 2) a> 0 are branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph below Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Ø There is not a single point below Oh." class="link_thumb"> 13 !} X) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph below Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Ø There is not a single point below Oh. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph below Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Ø Below Ox there is not a single point. "> 0 - branches. 1) B. ph. 3) Ox: 4) We schematically depict a parabola. 5) => graph below Ox. 6) In this case D = 0. X = - 2 - point of tangency. What has changed? Ø Below Oh, there is not a single point. "> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph below Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Ø There is not a single dot below Oh. "Title =" (! LANG: 26.07.201513 X -2-2 26.07.2015 2) a> 0 are branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph below Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Ø There is not a single point below Oh."> title="07.26.201513 X -2-2 07.26.2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => graph below Oh. 6) In this case, D = 0. x = -2 - point of tangency. What changed? Ø There is not a single point below Oh."> !}
X) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) No points of intersection with Ox. 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) No points of intersection with Ox. "> 0 - branches. 1) V. ph. 3) Ox: 4) We schematically depict a parabola. 5) => graph is not lower than Ox. 6) No points of intersection with Ox."> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) Not intersection points with Ox. "Title =" (! LANG: 07/26/2015 X 07/26/2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) No points of intersection with Ox.">
title="07/26/201514 X 07/26/2015 2) a> 0 - branches. 1) V. f. 3) Oh: 4) Let's schematically depict a parabola. 5) => the schedule is not lower than Oh. 6) No points of intersection with Ox.">
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