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.
Square Trinomial Plot
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
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 $(3; 0)$.
