Solution of square graphs. How to calculate the minimum or maximum using mathematical operations

Everyone knows what a parabola is. But how to use it correctly, competently in solving various practical problems, we will understand below.

First, let us denote the basic concepts that algebra and geometry give to this term. Consider everything possible types this chart.

We learn all the main characteristics of this function. Let's understand the basics of constructing a curve (geometry). Let's learn how to find the top, other basic values of the graph of this type.

We will find out: how the required curve is correctly constructed according to the equation, what you need to pay attention to. Let's see the main practical use this unique value in human life.

What is a parabola and what does it look like

Algebra: this term refers to a graph quadratic function.

Geometry: This is a second-order curve that has a number of specific features:

Canonical parabola equation

The figure shows a rectangular coordinate system (XOY), an extremum, the direction of the function drawing branches along the abscissa axis.

The canonical equation is:

y 2 \u003d 2 * p * x,

where the coefficient p is the focal parameter of the parabola (AF).

In algebra, it is written differently:

y = a x 2 + b x + c (recognizable pattern: y = x 2).

Properties and Graph of a Quadratic Function

The function has an axis of symmetry and a center (extremum). The domain of definition is all values of the x-axis.

The range of values of the function - (-∞, M) or (M, +∞) depends on the direction of the curve branches. The parameter M here means the value of the function at the top of the line.

How to determine where the branches of a parabola are directed

To find the direction of this type of curve from an expression, you need to specify the sign in front of the first parameter algebraic expression. If a ˃ 0, then they are directed upwards. Otherwise, down.

How to find the vertex of a parabola using the formula

Finding the extremum is the main step in solving many practical problems. Of course, you can open special online calculators but it's better to be able to do it yourself.

How to define it? There is a special formula. When b is not equal to 0, we must look for the coordinates of this point.

Formulas for finding the top:

x 0 \u003d -b / (2 * a);
y 0 = y (x 0).

Example.

There is a function y \u003d 4 * x 2 + 16 * x - 25. Let's find the vertices of this function.

For such a line:

x \u003d -16 / (2 * 4) \u003d -2;
y = 4 * 4 - 16 * 2 - 25 = 16 - 32 - 25 = -41.

We get the coordinates of the vertex (-2, -41).

Parabola offset

The classic case is when in a quadratic function y = a x 2 + b x + c, the second and third parameters are 0, and = 1 - the vertex is at the point (0; 0).

Movement along the abscissa or ordinate axes is due to a change in the parameters b and c, respectively. The shift of the line on the plane will be carried out exactly by the number of units, which is equal to the value of the parameter.

Example.

We have: b = 2, c = 3.

This means that the classic view of the curve will shift by 2 unit segments along the abscissa axis and by 3 along the ordinate axis.

How to build a parabola using a quadratic equation

It is important for schoolchildren to learn how to correctly draw a parabola according to the given parameters.

By analyzing expressions and equations, you can see the following:

The point of intersection of the desired line with the ordinate vector will have a value equal to c.
All points of the graph (along the x-axis) will be symmetrical with respect to the main extremum of the function.

In addition, the intersections with OX can be found by knowing the discriminant (D) of such a function:

D \u003d (b 2 - 4 * a * c).

To do this, you need to equate the expression to zero.

The presence of parabola roots depends on the result:

D ˃ 0, then x 1, 2 = (-b ± D 0.5) / (2 * a);
D \u003d 0, then x 1, 2 \u003d -b / (2 * a);
D ˂ 0, then there are no points of intersection with the vector OX.

We get the algorithm for constructing a parabola:

determine the direction of the branches;
find the coordinates of the vertex;
find the intersection with the y-axis;
find the intersection with the x-axis.

Example 1

Given a function y \u003d x 2 - 5 * x + 4. It is necessary to build a parabola. We act according to the algorithm:

a \u003d 1, therefore, the branches are directed upwards;
extremum coordinates: x = - (-5) / 2 = 5/2; y = (5/2) 2 - 5 * (5/2) + 4 = -15/4;
intersects with the y-axis at the value y = 4;
find the discriminant: D = 25 - 16 = 9;
looking for roots

X 1 \u003d (5 + 3) / 2 \u003d 4; (4, 0);
X 2 \u003d (5 - 3) / 2 \u003d 1; (ten).

Example 2

For the function y \u003d 3 * x 2 - 2 * x - 1, you need to build a parabola. We act according to the above algorithm:

a \u003d 3, therefore, the branches are directed upwards;
extremum coordinates: x = - (-2) / 2 * 3 = 1/3; y = 3 * (1/3) 2 - 2 * (1/3) - 1 = -4/3;
with the y-axis will intersect at the value y \u003d -1;
find the discriminant: D \u003d 4 + 12 \u003d 16. So the roots:

X 1 \u003d (2 + 4) / 6 \u003d 1; (1;0);
X 2 \u003d (2 - 4) / 6 \u003d -1/3; (-1/3; 0).

From the obtained points, you can build a parabola.

Directrix, eccentricity, focus of a parabola

Based on the canonical equation, the focus F has coordinates (p/2, 0).

Straight line AB is a directrix (a kind of parabola chord of a certain length). Her equation is x = -p/2.

Eccentricity (constant) = 1.

Conclusion

We considered the topic that students study in high school. Now you know, looking at the quadratic function of a parabola, how to find its vertex, in which direction the branches will be directed, whether there is an offset along the axes, and, having a construction algorithm, you can draw its graph.