Schedule square trinomial
2019-04-19
Square trinomial
We called an entire rational function of the second degree a square trinomial:
$y = ax^2 + bx + c$, (1)
where $a \neq 0$. Let us prove that the graph of a square trinomial is a parabola obtained by parallel shifts (in the directions of the coordinate axes) from the parabola $y = ax^2$. To do this, we reduce expression (1) by simple identical transformations to the form
$y = a(x + \alpha)^2 + \beta$. (2)
The corresponding transformations, written below, are known as "exact square selection":
$y = x^2 + bx + c = a \left (x^2 + \frac(b)(a) x \right) + c = a \left (x^2 + \frac(b)(a) x + \frac (b^2)(4a^2) \right) - \frac (b^2)(4a) + c = a \left (x + \frac(b)(2a) \right)^2 - \frac (b^2 - 4ac)(4a)$. (2")
We have reduced the square trinomial to the form (2); wherein
$\alpha = \frac(b)(2a), \beta = - \frac (b^2 - 4ac)(4a)$
(These expressions should not be memorized; it is more convenient to perform the transformation of the trinomial (1) to the form (2) directly each time).
Now it is clear that the graph of the trinomial (1) is a parabola equal to the parabola $y = ax^2$ and obtained by shifting the parabola $y = ax^2$ in the directions of the coordinate axes by $\alpha$ and $\beta$ (taking into account the sign $\alpha$ and $\beta$) respectively. The vertex of this parabola is placed at the point $(- \alpha, \beta)$, its axis is the straight line $x = - \alpha$. For $a > 0$ the vertex is the lowest point of the parabola, for $a
Let us now study the square trinomial, i.e., find out its properties depending on the numerical values of the coefficients $a, b, c$ in its expression (1).
Let us denote the quantity $b^2- 4ac$ in equality (2") by $d$:
$y = a \left (x + \frac(b)(2a) \right)^2 - \frac(d)(4a)$; (4)
$d = b^2 - 4ac$ is called the discriminant of the square trinomial. The properties of the trinomial (1) (and the location of its graph) are determined by the signs of the discriminant $d$ and the leading coefficient $a$.
1) $a > 0, d 0$; since $a > 0$, then the graph is located above the top $O^( \prime)$; it lies in the upper half-plane ($y > 0$ - fig. a.).
2) $a
3) $a > 0, d > 0$. The vertex $O^( \prime)$ lies below the $Ox$ axis, the parabola intersects the $Ox$ axis at two points $x_1, x_2$ (Fig. c.).
4) $a 0$. The vertex $O^( \prime)$ lies above the $Ox$ axis, the parabola again intersects the $Ox$ axis at two points $x_1, x_2$ (Fig. d).
5) $a > 0, d = 0$. The vertex lies on the $Ox$ axis itself, the parabola is located in the upper half-plane (Fig. e).
6) $a
Conclusions. If $d 0$), or lower (for $a
If $d > 0$, then the function is sign-alternating (the graph partly lies below, partly above the $Ox$ axis). A square trinomial with $d > 0$ has two roots (zeros) $x_1, x_2$. For $a > 0$ it is negative in the interval between the roots (Fig. c) and positive outside this interval. For $a
Introductory remarks and simple examples
Example 1. For what values of a does the equation ax 2 + 2x + 1 = 0 have two different roots?
Solution.
This equation is quadratic with respect to the variable x for a№ 0 and has distinct roots when its discriminant
i.e. for a< 1.
In addition, for a = 0, the equation 2x + 1 = 0 is obtained, which has one root.
Thus, a О (– Ґ ; 0) AND (0; 1).
Rule 1 If the coefficient at x 2 of a polynomial of the second degree contains a parameter, it is necessary to analyze the case when it vanishes.
Example 2. The equation ax 2 + 8x + c = 0 has a single root equal to 1. What are a and c?
Solution. Let's start solving the problem from the special case a = 0, the equation has the form 8x + c = 0. This linear equation has a solution x 0 = 1 for c = - 8.
For a no. 0 quadratic equation has a single root if 
In addition, substituting the root x 0 \u003d 1 into the equation, we get a + 8 + c \u003d 0.
Solving a system of two linear equations, we find a = c = – 4.
Theorem 1.
For a reduced square trinomial y = x 2 + px + q (under the condition p 2і
4q)
the sum of the roots x 1 + x 2 \u003d - p, the product of the roots x 1 x 2 \u003d q, the difference of the roots is 
and the sum of the squares of the roots x 1 2 + x 2 2 = p 2 - 2q.
Theorem 2.
For a square trinomial y = ax 2 + bx + c with two roots x 1 and x 2 we have
decomposition ax 2 + bx + c \u003d a (x - x 1) (x - x 2), for a trinomial with one root x 0 - decomposition
ax 2 + bx + c \u003d a (x - x 0) 2.
Comment. Often about quadratic equations with a discriminant, zero and having, respectively, one root, it is said that it has two coinciding roots (?). This is related to the factorization of the polynomial given in Theorem 2.(In this case, it is necessary to speak and understand correctly “one root of multiplicity two.” - Approx. ed.)
We will pay attention to this subtlety and single out the case of a single root of multiplicity 2.
Example 3. In the equation x 2 + ax + 12 = 0, determine a in such a way that the difference of the roots of the equation is equal to one.
Solution. Root difference 
whence a = ± 7.
Example 4. For what a is the sum of the squares of the roots of the equation 2x 2 + 4x + a = 0 equal to 6?
Solution. We write the equation in the form 
whence x 1 2 + x 2 2 = 4 - a = 6 and a = - 2.
Example 5. For all a, solve the equation ax 2 - 2x + 4 = 0.
Solution. If a = 0, then x = 2. If a№
0, then the equation becomes quadratic. Its discriminant
equals D = 4 – 16a. If D< 0, т. е. a > ,
the equation has no solutions. If D = 0, i.e. a = ,
x = 4. If D > 0, i.e. a< ,
the equation has two roots
Location of the roots of a square trinomial
The graph of the quadratic equation is a parabola, and the solutions of the quadratic equation are the abscissas of the points of intersection of this parabola with the Ox axis. The basis for solving all the problems of this section is the study of the location of parabolas with given properties on the coordinate plane.
Example 6. For which a the roots of the equation x 2 - 2ax + a 2 - a - 6 = 0 have different signs?
Solution (Fig. 1).
A quadratic equation either has no solutions (the graph is a parabola of type D), or has one or two positive roots (parabola C), or has one or two negative roots (parabola A), or has roots of different signs (parabola B).
It is easy to see that the last type of parabolas, unlike the others, is characterized by the fact that f(0)< 0. Таким образом, f(0) = a 2 – a – 6 < 0, откуда 0 < a < .
This solution admits a generalization, which we formulate as the following rule.
Rule 2. In order for the equation ax 2 + bx + c = 0
had two different roots x 1 and x 2 such that x 1< M < x 2 , необходимо и достаточно, чтобы a f(M) < 0.
Example 7. For what a does the equation x 2 - 2ax + a 2 - a - 6 \u003d 0 have two different roots of the same sign?
Solution. We are interested in parabolas of types A and C (see Fig. 1). They are characterized by
whence a О (- 6; - 2) AND (3; + Ґ ).
Example 8. For what a does the equation x 2 - 2ax + a 2 - a - 6 \u003d 0 have two different positive roots?
Solution. We are interested in the parabolas of type C in fig. 1.
For the equation to have roots, we require
Since both roots of the equation must be positive by condition, then the abscissa of the vertex of the parabola, which lies between the roots, is positive: x 0 \u003d a\u003e 0.
Vertex ordinate f(x 0)< 0 в силу того, что мы потребовали существование корней, поэтому если, кроме того, потребовать выполнение условия f(x 0) >0, then, due to the continuity of the function under study, there is a point x 1 ABOUT (0; x 0) such that f(x 1) = 0. Obviously, this is a smaller root of the equation.
So, f(0) = a 2 - a - 6 > 0, and, collecting all the conditions together, we get the system
with solution a 0 (3; + Ґ ).
Example 9. For what a does the equation x 2 - 2ax + a 2 - a - 6 have two different negative roots?
Solution. Having studied the parabolas of type A in fig. 1, we get the system
whence a О (– 6; – 2).
We generalize the solution of the previous problems in the form of the following rule.
Rule 3. In order for the equation ax 2 + bx + c = 0 to have two different roots x 1 and x 2, each of which is greater (less) than M, it is necessary and sufficient that
Example 10. The function f(x) is given by the formula
Find all values of the parameter a for which the equation f(x) = 0 has at least one solution.
Solution. All possible solutions to this equation are obtained as solutions to the quadratic equation
x 2 - (4a + 14)x + 4a 2 + 33a + 59 = 0
with the additional condition that at least one (obviously larger) root x 2 and a.
Naturally, for the equation to have roots, it must be = - 5 (a + 2) і
0,
whence a Ј – 2.
The graph of the left side of the selected equation is a parabola, the abscissa of the vertex of which is equal to x 0 = 2a + 7. Two types of parabolas give the solution to the problem (Fig. 2).
A: x 0 і a, whence a і – 7. In this case, the larger root of the polynomial x 2 i x 0 i a.
B:x0< a, f(a)
Ј
0, whence 
.
In this case also the larger root of the polynomial x 2 and a.
Finally 
.
Three solutions to one inequality
Example 11. Find all values of the parameter a for which the inequality x 2 - 2ax + a 2 + 2a - 3 > 0
performed:
1) for all values of x;
2) for all positive values x;
3) for all values of x O [– 1; 1].
Solution.
First way.
1) Obviously, this inequality holds for all x, when the discriminant is negative, i.e.
\u003d a 2 - (a 2 + 2a - 3) \u003d - 2a + 3< 0,
whence a >.
2) To better understand what is required in the condition of the problem, we apply a simple trick: draw some parabolas on the coordinate plane, and then take and close the half-plane left relative to the Oy axis. The part of the parabola that remains visible must be above the Ox axis.
The condition of the problem is fulfilled in two cases (see Fig. 3):
< 0, откуда a > ;
B: both roots (maybe one, but double) of the equation x 2 - 2ax + a 2 + 2a - 3 = 0 are to the left of the origin. By rule 3, this condition is equivalent to the system of inequalities D i 0, x 0 j 0 and f(0) i 0.
However, when solving this system, the first inequality can be omitted, since even if some value a does not satisfy the condition Dі 0, then it automatically falls into the solution of point A. Thus, we solve the system
whence a Ј – 3.
Combining the solutions of items A and B, we get
answer: 
3) The condition of the problem is fulfilled in three cases (see Fig. 4):
A: the graph of the function y = x 2 - 2ax + a 2 + 2a - 3 lies above the Ox axis, i.e. D< 0, откуда a > ;
B: both roots (maybe one of multiplicity 2) of the equation x 2 - 2ax + a 2 + 2a - 3 = 0 are to the left - 1. This condition is equivalent, as we know from rule 3, to the system of inequalities Dі 0,x0< – 1, f(– 1) > 0;
C: both roots of the equation x 2 - 2ax + a 2 + 2a - 3 = 0 are to the right of 1.
This condition is equivalent to D i 0, x 0 > 1, f(1) > 0.
However, in points B and C, as well as in the solution of the previous problem, the inequality associated with the discriminant can be omitted.
Accordingly, we obtain two systems of inequalities
After considering all cases, we get the result: a >
in point
in C.
The answer to the problem is the union of these three sets.
The second way. In order for the condition of each of the three points of the task to be fulfilled, the smallest value of the function
y \u003d x 2 - 2ax + a 2 + 2a - 3 on each of the corresponding gaps must be positive.
1) The top of the parabola y \u003d x 2 - 2ax + a 2 + 2a - 3 is at the point (a; 2a - 3), therefore the smallest value of the function on the entire real line is 2a - 3, and a >.
2) on the semiaxis x i 0 the smallest value of the function is f(0) = a 2 + 2a – 3 if a< 0, и f(a) = 2a – 3, если a
і
0. Analyzing both cases, we obtain 
3) The smallest on the segment [- 1; 1] function value is
Since the smallest value must be positive, we obtain systems of inequalities
The solution of these three systems is the set
The third way. 1) The top of the parabola y \u003d x 2 - 2ax + a 2 + 2a - 3
is at the point (a; 2a – 3). Let us draw on the coordinate plane a set formed by the vertices of all parabolas for different a (Fig. 5).
This is the line y = 2x – 3. Recall that each point of this line corresponds to its own value of the parameter, and from each point of this line a parabola corresponding to the given value of the parameter “emerges”. Parabolas that are entirely above the Ox axis are characterized by the condition 2a – 3 > 0.
2) The solutions of this paragraph are all solutions of the first paragraph, and, in addition, parabolas for which a are negative, and f(0) = a 2 + 2a - 3і 0.
3) From fig. 5 that we are interested in parabolas for which either a is negative and f(– 1) = a 2 + 4a – 2 > 0,
or a is positive and f(1) = a 2 – 2 > 0.
Equations and inequalities reducing to square ones
Example 12. For what values of a does the equation 2x 4 - 2ax 2 + a 2 - 2 = 0 have no solutions?
Solution. Making the replacement y \u003d x 2, we get the quadratic equation f (y) \u003d 2y 2 - 2ay + a 2 - 2 \u003d 0.
The resulting equation has no solution when D< 0. Кроме того, первоначальное уравнение не имеет решений, когда корни уравнения f(y) = 0 отрицательны.
These conditions can be written as a set
where 
Example 13. For each value of the parameter a, solve the equation cos x sin 2x = asin 3x.
Solution. Since 2cos x sin 2x = sin x + sin 3x and sin 3x = 3sin x - 4sin 3 x,
then the equation will be written in the form sin x (sin 2 x (4a - 2) - (3a - 2)) = 0.
Hence we obtain solutions x = p n, n O Z for any a. The equation 
has solutions 
not coinciding with the solutions of the first equation, only under the condition 
The last restrictions are equivalent
Answer: x \u003d p n, n O Z for any a; Besides,
Example 14. Find all values of the parameter a, for each of which the inequality
a 2 + 2a - sin 2 x - 2acos x > 2 holds for any number x.
Solution. Let us transform the inequality to the form cos 2 x – 2acos x + a 2 + 2a – 3 > 0
and make the change t = cos x. It is important to note that the parameter t ranges from -1 to 1, so the problem is reformulated as follows: find all a such that
t 2 - 2at + a 2 + 2a - 3 > 0
is satisfied for all t ABOUT [- 1; 1]. We have already solved this problem earlier.
Example 15. Determine for what values of a the equation log 3 (9 x + 9a 3) = x has solutions, and find them.
Solution. Let's transform the equation to the form 9 x - 3 x + 9a 3 = 0
and, making the change y = 3 x , we get y 2 – y + 9a 3 = 0.
If the discriminant is negative, then the equation has no solutions. When the discriminant
D \u003d 1 - 36a 3 \u003d 0, the equation has a single root,
and x = – log 3 2. Finally, when the discriminant is positive, i.e.,
the original equation has one root 
,
and if, in addition, expression 1 is positive,
then the equation has a second root 
.
So we finally get 
,
there are no solutions for the remaining a.
Example 16. For each value of the parameter a, solve the equation sin 4 x + cos 4 x + sin 2 x + a = 0.
Solution. Because
rewrite the equation in the form sin 2 x - 2sin x - 2a - 2 = 0.
Let y = sin 2x, then y 2 – 2y – 2a – 2 = 0 (| y | J 1).
The graph of the function on the left side of the equation is a parabola with a vertex whose abscissa is y 0 = 1; the value of the function at the point y = – 1 is equal to 1 – 2a; the discriminant of the equation is 8a + 12. This means that the larger root y 2 of the equation y 2 - 2y - 2a - 2 = 0, even if it exists, is greater than 1, and the corresponding equation sin 2x = y 2 has no solutions. 3. For what values of a does the equation 2x 2 + (3a + 1)x + a 2 + a + 2 = 0 have at least one root?
4. The equation ax 2 + bx + 5 = 0 has a single root equal to 1. What are a and b?
5. For what values of the parameter a do the roots of the quadratic equation 5x 2 - 7x + a = 0 relate as 2 to 5?
6. In the equation ax 2 + 8x + 3 = 0, determine a in such a way that the difference of the roots of the equation is equal to one.
7. For what a is the sum of the squares of the roots of the equation x 2 - 2ax + 2(a + 1) = 0 equal to 20?
8. For what b and c the equation c + bx - 2x 2 = 0 has one positive and one negative root?
9. Find all values of the parameter a for which one root of the equation x 2 - (a + 1)x + 2 = 0 is greater than a, and the other is less than a.
10. Find all values of the parameter a for which the equation x 2 + (a + 1)x + 2 = 0 has two different roots of the same sign.
11. For what values of a are all the resulting roots of the equation (a - 3)x 2 - 2ax + 6a = 0 positive?
12. For which a are all the resulting roots of the equation (1 + a)x 2 - 3ax + 4a = 0 greater than 1?
13. Find all values of the parameter a for which both different roots of the equation x 2 + x + a = 0 will be greater than a.
14. For what values of a are both roots of the equation 4x 2 - 2x + a \u003d 0 enclosed between - 1 and 1?
15. For what values of a does the equation x 2 + 2(a - 1)x + a + 5 = 0 have at least one positive root?
16. The function f(x) is given by the formula
Find all values of the parameter a for which the equation f(x) = 0 has at least one solution.
17. For what a is the inequality (a 2 – 1)x 2 + 2(a – 1)x + 2 > 0 true for all x?
18. For what values of the parameter a does the inequality ax 2 + 2x > 1 – 3a hold true for all positive x?
19. For what values of a does the equation x 4 + (1 - 2a)x 2 + a 2 - 1 \u003d 0 have no solutions?
20. For what values of the parameter a does the equation 2x 4 - 2ax 2 + a2 - 2 = 0 have one or two solutions?
21. For each value of a, solve the equation acos x cos 2x = cos 3x.
22. Find all values of the parameter a, for each of which the inequality cos 2 x + 2asin x – 2a< a 2
– 4 выполняется для любого числа x.
23. For all a, solve the equation log 2 (4 x + a) = x.
24. For each value of the parameter a, solve the equation sin 2 x + asin 2 2x = sin.
Definition
parabola is the graph of a quadratic function $y = ax^(2) + bx + c$, where $a \neq 0$.
Graph of the function $y = x^2$.
For a schematic construction of the graph of the function $y = x^2$, we find several points that satisfy this equality. For convenience, we write the coordinates of these points in the form of a table:
Graph of the function $y = ax^2$.
If the coefficient $a > 0$, then the graph $y = ax^2$ is obtained from the graph $y = x^2$ either by vertical expansion (for $a > 1$) or by compression towards the $x$ axis (for $0< a < 1$). Изобразим для примера графики $y = 2x^2$ и $y = \dfrac{x^2}{2}$:
| $y = 2x^2$ | $y = \dfrac(x^2)(2)$ |
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If $a< 0$, то график функции $y = ax^2$ можно получить из графика $y = |a|x^2$, отразив его симметрично относительно оси $x$. Построим графики функций $y = - x^2$, $y = -2x^2$ и $y = - \dfrac{x^2}{2}$:
| $y = - x^2$ | $y = -2x^2$ | $y = - \dfrac(x^2)(2)$ |
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Graph of a quadratic function.
To construct a graph of the function $y = ax^2 + bx + c$, you need to select a full square from the square trinomial $ax^2 + bx + c$, that is, represent it in the form $a(x - x_0)^2 + y_0$ . The graph of the function $y = a(x - x_0)^2 + y_0$ is obtained from the corresponding graph $y = ax^2$ by shifting $x_0$ along the $x$ axis, and $y_0$ along the $y$ axis. As a result, the point $(0;0)$ will move to the point $(x_0;y_0)$.
Definition
pinnacle parabola $y = a(x - x_0)^2 + y_0$ is a point with coordinates $(x_0;y_0)$.
Let's construct a parabola $y = 2x^2 - 4x - 6$. Selecting the full square, we get $y = 2(x - 1)^2 - 8$.
| Let's plot $y = 2x^2$ | Shift it to the right by 1 | And down by 8 |
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The result is a parabola with vertex at the point $(1;-8)$.
The graph of the quadratic function $y = ax^2 + bx + c$ intersects the $y$ axis at the point $(0; c)$ and the $x$ axis at the points $(x_(1,2);0)$, where $ x_(1,2)$ are the roots of the quadratic equation $ax^2 + bx + c = 0$ (moreover, if the equation has no roots, then the corresponding parabola does not intersect the $x$ axis).
For example, the parabola $y = 2x^2 - 4x - 6$ intersects the axes at the points $(0; -6)$, $(-1; 0)$ and $(3; 0)$.
Defined by the formula $a((x)^(2))+bx+c$ $(a\ne 0).$ The numbers $a, b$ and $c$ are the coefficients of a square trinomial, they are usually called: b - second or average coefficient, c - free term. A function of the form y = ax 2 + bx + c is called a quadratic function.
All of these parabolas have their vertex at the origin; for a > 0, this is the lowest point of the graph (the smallest value of the function), and for a< 0, наоборот, highest point (highest value functions). The axis Oy is the axis of symmetry for each of these parabolas.
As you can see, for a > 0, the parabola is directed upwards, for a< 0 - вниз.
There is a simple and convenient graphical method that allows you to build any number of points of the parabola y = ax 2 without calculations, if you know a point of the parabola other than the vertex. Let the point M(x 0 , y 0) lie on the parabola y = ax 2 (Fig. 2). If we want to build an additional n points between points O and M, then we divide the segment ON of the abscissa axis by n + 1 equal parts and at the division points we draw perpendiculars to the Ox axis. We divide the segment NM into the same number of equal parts and connect the division points with rays to the origin. The desired points of the parabola lie at the intersection of perpendiculars and rays with the same numbers (in Fig. 2, the number of division points is 9).
The graph of the function y =ax 2 + bx + c differs from the graph y = ax 2 only in its position and can be obtained simply by moving the curve in the drawing. This follows from the representation of the square trinomial in the form
whence it is easy to conclude that the graph of the function y = ax 2 + bx + c is a parabola y = ax 2 , whose vertex is moved to the point
and its symmetry axis remained parallel to the Oy axis (Fig. 3). From the resulting expression for a square trinomial, all its basic properties easily follow. The expression D \u003d b 2 - 4ac is called the discriminant of the square trinomial ax 2 + bx + c and the discriminant of the associated quadratic equation ax 2 + bx + c \u003d 0. The sign of the discriminant determines whether the graph of the square trinomial intersects the x-axis or lies on one side from her. Namely, if D< 0, то парабола не имеет общих точек с осью Ox, при этом: если a >0, then the parabola lies above the Ox axis, and if a< 0, то ниже этой оси (рис. 4). В случае D >0 the graph of a square trinomial intersects the x-axis at two points x 1 and x 2 , which are the roots of the quadratic equation ax 2 + bx + c = 0 and are equal, respectively
For D = 0, the parabola touches the Ox axis at the point
The properties of a square trinomial underlie the solution of square inequalities. Let's explain this with an example. Let it be required to find all solutions of the inequality 3x 2 - 2x - 1< 0. Найдем дискриминант квадратного трехчлена, стоящего в левой части неравенства: D = 16. Так как D >0, then the corresponding quadratic equation 3x 2 − 2x − 1 = 0 has two different roots, they are determined by the formulas given earlier:
x 1 \u003d -1/3 and x 2 \u003d 1.
In the considered square trinomial, a = 3 > 0, which means that the branches of its graph are directed upwards and the values of the square trinomial are negative only in the interval between the roots. So, all solutions of the inequality satisfy the condition
−1/3 < x < 1.
TO quadratic inequalities various inequalities can be reduced by the same substitutions by which various equations are reduced to a quadratic one.
Lesson: how to build a parabola or a quadratic function?
THEORETICAL PART
A parabola is a graph of a function described by the formula ax 2 +bx+c=0.
To build 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,
and then the branches of the parabola are directed down.
free member c this point intersects the parabola with the OY axis;
2) , it is found by 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 ax2+bx+c=0;
Types of equations:
a) The complete quadratic equation is ax2+bx+c=0 and is solved by the discriminant;
b) Incomplete quadratic equation of the form ax2+bx=0. To solve it, you need to take x out of brackets, then equate each factor to 0:
ax2+bx=0,
x(ax+b)=0,
x=0 and ax+b=0;
c) Incomplete quadratic equation of the form ax2+c=0. To solve it, you need to move the unknown to one side, and the known to the other. x =±√(c/a);
4) Find some additional points to build the function.
PRACTICAL PART
And so now, with an example, we will analyze everything by actions:
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 because 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 the top is at the point (-2;-1)
Find the roots of the equation x 2 +4x+3=0
We find the roots by the discriminant
a=1 b=4 c=3
D=b 2 -4ac=16-12=4
x=(-b±√(D))/2a
x1=(-4+2)/2=-1
x2=(-4-2)/2=-3
Let's take some arbitrary points that are near the top x=-2
x -4 -3 -1 0
y 3 0 0 3
We substitute instead of x in the equation y \u003d 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 values of the function that the parabola is symmetrical about the straight line x \u003d -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 because a=-1 -1 Find the roots of the equation -x 2 +4x=0
An 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 some arbitrary points that are near the vertex x=2
x 0 1 3 4
y 0 3 3 0
We substitute instead of x in the equation y \u003d -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 values of the function that the parabola is symmetrical about the straight line x \u003d 2
Example #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 because a=1 1>0.
a=1 b=0 c=-4 x=(-b)/2a=0/(2*(1))=0 y=(0) 2 -4=-4 vertex is at point (0;-4 )
Find the roots of the equation x 2 -4=0
An incomplete quadratic equation of the form ax 2 +c=0. To solve it, you need to move the unknown to one side, and the known to the other. x =±√(c/a)
x2=4
x1=2
x 2 \u003d -2
Let's take some arbitrary points that are near the top x=0
x -2 -1 1 2
y 0 -3 -3 0
We substitute instead of x in the equation y \u003d 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 values of the function that the parabola is symmetrical about the straight line x=0
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