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 a simple algorithm of 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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- — [] quadratic function A function of the form y= ax2 + bx + c (a ? 0). Graph K.f. is a parabola whose vertex has coordinates [ b / 2a, (b2 4ac) / 4a], for a> 0 branches of the parabola ... ...

QUADRATIC FUNCTION, a mathematical FUNCTION whose value depends on the square of the independent variable, x, and is given, respectively, by a quadratic POLYNOMIAL, for example: f(x) = 4x2 + 17 or f(x) = x2 + 3x + 2. see also SQUARE THE EQUATION … Scientific and technical encyclopedic dictionary

quadratic function - quadratic function is a function of the form y= ax2 + bx + c (a ≠ 0). Graph K.f. is a parabola whose vertex has coordinates [b/ 2a, (b2 4ac) /4a], for a> 0 the branches of the parabola are directed upwards, for a< 0 –вниз… …

- (quadratic) A function having the following form: y=ax2+bx+c, where a≠0 and highest degree x is a square. Quadratic equation y=ax2 +bx+c=0 can also be solved using following formula: x \u003d -b + √ (b2-4ac) / 2a. These roots are real... Economic dictionary

An affine quadratic function on an affine space S is any function Q: S→K that has the form Q(x)=q(x)+l(x)+c in vectorized form, where q is a quadratic function, l is a linear function, and c is a constant. Contents 1 Transfer of the origin 2 ... ... Wikipedia

An affine quadratic function on an affine space is any function that has the form in vectorized form, where is a symmetric matrix, a linear function, a constant. Contents ... Wikipedia

Function on vector space, given by a homogeneous polynomial of the second degree in the coordinates of the vector. Contents 1 Definition 2 Related definitions ... Wikipedia

- is a function that, in the theory of statistical decisions, characterizes the losses due to incorrect decision making based on the observed data. If the problem of estimating the signal parameter against the background of interference is being solved, then the loss function is a measure of the discrepancy ... ... Wikipedia

objective function- — [Ya.N. Luginsky, M.S. Fezi Zhilinskaya, Yu.S. Kabirov. English Russian Dictionary of Electrical Engineering and Power Engineering, Moscow, 1999] objective function In extremal problems, a function whose minimum or maximum is to be found. This is… … Technical Translator's Handbook

objective function- in extremal problems, the function, the minimum or maximum of which is required to be found. This is the key concept of optimal programming. Having found the extremum of the C.f. and, therefore, by determining the values ​​of the controlled variables that are to it ... ... Economic and Mathematical Dictionary

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If you want to participate in the big life, fill your head with math while you can. She will be of great help to you later in all your work.

M.I. Kalinin

One of the main functions of school mathematics, for which a complete theory has been constructed and all properties have been proven, is quadratic function. Students must clearly understand and know all these properties. At the same time, there are a great many problems for a quadratic function - from very simple ones, which follow directly from the theory and formulas, to the most complex ones, the solution of which requires analysis and a deep understanding of all the properties of the function.

When solving problems on a quadratic function, a large practical value has a correspondence between the algebraic description of the problem and its geometric interpretation - the image on coordinate plane function graph thumbnail. It is thanks to this feature that you always have the opportunity to check the correctness and consistency of your theoretical reasoning.

Let's consider several tasks on the topic "Quadratic function" and dwell on their detailed solution.

Task 1.

Find the sum of integer values ​​of p for which the vertex of the parabola y = 1/3x 2 – 2px + 12p is located above the Ox axis.

Decision.

The branches of the parabola are directed upwards (a = 1/3 > 0). Since the vertex of the parabola lies above the Ox axis, the parabola does not intersect the abscissa axis (Fig. 1). So the function

y = 1/3x 2 - 2px + 12p has no zeros,

and the equation

1/3x 2 - 2px + 12p = 0 has no roots.

This is possible if the discriminant of the last equation is negative.

Let's calculate it:

D / 4 \u003d p 2 - 1/3 12p \u003d p 2 - 4p;

p2-4p< 0;

p(p - 4)< 0;

p belongs to the interval (0; 4).

The sum of integer values ​​of the number p from the interval (0; 4): 1 + 2 + 3 = 6.

Answer: 6.

Note that to answer the question of the problem it was possible to solve the inequality

y in > 0 or (4ac - b 2) / 4a > 0.

Task 2.

Find the number of integer values ​​of a for which the abscissa and ordinate of the vertex of the parabola y = (x – 9a) 2 + a 2 + 7a + 6 are negative.

Decision.

If the quadratic function has the form

y = a(x – n) 2 + m, then the point with coordinates (m; n) is the vertex of the parabola.

In our case

x in = 9a; y in \u003d a 2 + 7a + 6.

Since both the abscissa and the ordinate of the top of the parabola must be negative, we will compose a system of inequalities:

(9a< 0,
(a 2 + 7a + 6< 0;

Let's solve the resulting system:

(a< 0,
((a + 1)(a + 6)< 0;

Let us depict the solution of inequalities on the coordinate lines and give the final answer:

a belongs to the interval (-6; -1).

Integer values ​​of number a: -5; -4; -3; -2. Their number: 4.

Answer: 4.

Task 3.

Find the largest integer value of m for which the quadratic function
y = -2x 2 + 8x + 2m only accepts negative values.

Decision.

The branches of the parabola are directed downwards (a = -2< 0). Данная функция будет принимать только отрицательные значения, если ее график не будет иметь общих точек с осью абсцисс, т.е. уравнение -2x 2 + 8x + 2m = 0 не будет иметь корней. Это возможно, если дискриминант последнего уравнения будет отрицательным.

2x2 + 8x + 2m = 0.

We divide the coefficients of the equation by -2, we get:

x 2 - 4x - m = 0;

D / 4 \u003d 2 2 - 1 1 (-m) \u003d 4 + m;

The largest integer value of the number m: -5.

Answer: -5.

To answer the question of the problem, it was possible to solve the inequality y in< 0 или

(4ac–b 2) / 4a< 0.

Task 4.

To find smallest value quadratic function y = ax 2 - (a + 6)x + 9, if it is known that the straight line x = 2 is the axis of symmetry of its graph.

Decision.

1) Since the line x \u003d 2 is the axis of symmetry of this graph, then x in \u003d 2. We use the formula

x in = -b / 2a, then x in = (a + 6) / 2a. But x in = 2.

Let's make an equation:

(a + 6) / 2a = 2;

Then the function takes the form

y \u003d 2x 2 - (2 + 6)x + 9;

y \u003d 2x 2 - 8x + 9.

2) Parabola branches

The smallest value of this function is equal to the ordinate of the top of the parabola (Fig. 2), which is easy to find using the formula

y in \u003d (4ac - b 2) / 4a.

y in \u003d (4 2 9 - 8 2) / 4 2 \u003d (72 - 64) / 8 \u003d 8/8 \u003d 1.

The smallest value of the considered function is 1.

Answer: 1.

Task 5.

Find the smallest integer value of the number a, at which the sets of values ​​of the function y \u003d x 2 - 2x + a and y \u003d -x 2 + 4x - a do not intersect.

Decision.

Find the set of values ​​for each function.

I way.

y 1 \u003d x 2 - 2x + a.

Let's apply the formula

y in \u003d (4ac - b 2) / 4a.

y in \u003d (4 1 a - 2 2) / 4 1 \u003d (4a - 4) / 4 \u003d 4 (a - 1) / 4 \u003d a - 1.

Since the branches of the parabola are directed upwards, then

E(y) = .

E(y 2) = (-∞; 4 - a].

Let us depict the obtained sets on the coordinate lines (Fig. 3).

The resulting sets will not intersect if the point with coordinate 4 - a is located to the left of the point with coordinate a - 1, i.e.

4-a< a – 1;

The smallest integer value of a: 3.

Answer: 3.

Problems on the location of the roots of a quadratic function, problems with parameters, and problems that reduce to quadratic functions are very popular on the exam. Therefore, when preparing for exams, you should pay close attention to them.

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