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

- (quadratic) Function having the following form: y=ax2+bx+c, where a≠0 and highest degree x is a square. The quadratic equation y=ax2 +bx+c=0 can also be solved using the following formula: x= –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, which in vectorized form has the form Q(x)=q(x)+l(x)+c, where q is a quadratic function, l is a linear function, c is a constant. Contents 1 Shifting the reference point 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

A function on a vector space defined 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 losses due to incorrect decision-making based on observed data. If the problem of estimating a signal parameter against a background of noise 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 required to be found. This… … Technical Translator's Guide

Objective function- in extremal problems, a function whose minimum or maximum needs to be found. This is a key concept in optimal programming. Having found the extremum of C.f. and, therefore, having determined the values ​​of the controlled variables that go to it... ... Economic and mathematical dictionary

Books

• Set of tables. Mathematics. Graphs of functions (10 tables), . Educational album of 10 sheets.
• Linear function. Graphical and analytical assignment of functions. Quadratic function. Transforming the graph of a quadratic function. Function y=sinx. Function y=cosx.… The most important function of school mathematics is quadratic - in problems and solutions, Petrov N.N.. The quadratic function is the main function of the school mathematics course. No wonder. On the one hand, the simplicity of this function, and on the other hand, deep meaning

. Many tasks of school...
A quadratic function is a function of the form:
y=a*(x^2)+b*x+c,
where a is the coefficient for the highest degree of unknown x,
b - coefficient for unknown x,
and c is a free member.

The graph of a quadratic function is a curve called a parabola. The general view of the parabola is shown in the figure below.

Fig.1 General view of the parabola. There are a few in various ways

plotting a quadratic function. We will look at the main and most general of them.

Algorithm for plotting a quadratic function y=a*(x^2)+b*x+c

1. Construct a coordinate system, mark a unit segment and label the coordinate axes.
2. Determine the direction of the parabola branches (up or down).

To do this, you need to look at the sign of the coefficient a. If there is a plus, then the branches are directed upward, if there is a minus, then the branches are directed downward.
3. Determine the x coordinate of the vertex of the parabola.

To do this, you need to use the formula Xvertex = -b/2*a.
4. Determine the coordinate at the vertex of the parabola.

To do this, substitute into the equation Uvershiny = a*(x^2)+b*x+c instead of x, the value of Xverhiny found in the previous step.

5. Plot the resulting point on the graph and draw an axis of symmetry through it, parallel to the Oy coordinate axis.
6. Find the points of intersection of the graph with the Ox axis. To do this you need to solve quadratic equation a*(x^2)+b*x+c = 0 one of known methods

. If the equation does not have real roots, then the graph of the function does not intersect the Ox axis.
7. Find the coordinates of the point of intersection of the graph with the Oy axis.

To do this, we substitute the value x=0 into the equation and calculate the value of y. We mark this and a point symmetrical to it on the graph.
To do this, choose an arbitrary value for the x coordinate and substitute it into our equation. We get the value of y at this point. Plot the point on the graph. And also mark a point on the graph that is symmetrical to point A(x,y).

9. Connect the resulting points on the graph with a smooth line and continue the graph beyond the extreme points, to the end of the coordinate axis. Label the graph either on the leader or, if space allows, along the graph itself.

Example of plotting

As an example, let's plot a quadratic function given by the equation y=x^2+4*x-1
1. Draw coordinate axes, label them and mark a unit segment.
2. Coefficient values ​​a=1, b=4, c= -1. Since a=1, which is greater than zero, the branches of the parabola are directed upward.
3. Determine the X coordinate of the vertex of the parabola Xvertices = -b/2*a = -4/2*1 = -2.
4. Determine the coordinate Y of the vertex of the parabola
Vertices = a*(x^2)+b*x+c = 1*((-2)^2) + 4*(-2) - 1 = -5.
5. Mark the vertex and draw the axis of symmetry.
6. Find the intersection points of the graph of the quadratic function with the Ox axis. We solve the quadratic equation x^2+4*x-1=0.
x1=-2-√3 x2 = -2+√3. We mark the obtained values ​​on the graph.
7. Find the points of intersection of the graph with the Oy axis.
x=0; y=-1
8. Choose an arbitrary point B. Let it have coordinate x=1.
Then y=(1)^2 + 4*(1)-1= 4.
9. Connect the obtained points and sign the graph.

Probably everyone knows what a parabola is. But we’ll look at how to use it correctly and competently when solving various practical problems below.

First, let us outline the basic concepts that algebra and geometry give to this term. Let's consider all possible types of this graph.

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

Let's find out: how to correctly construct the desired curve using the equation, what you need to pay attention to. Let's see the basics practical use this unique value in human life.

What is a parabola and what does it look like?

Algebra: This term refers to the graph of a 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 branches of the function drawing along the abscissa axis.

The canonical equation is:

y 2 = 2 * p * x,

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

In algebra it will be 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 abscissa axis.

The range of values ​​of the function – (-∞, M) or (M, +∞) depends on the direction of the branches of the curve. 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 a curve of this type from an expression, you need to determine the sign before the first parameter of the algebraic expression. If a ˃ 0, then they are directed upward. If it's the other way around, 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 determine it? There is a special formula. When b is not equal to 0, we need to look for the coordinates of this point.

Formulas for finding the vertex:

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

Example.

There is a function y = 4 * x 2 + 16 * x – 25. Let’s find the vertices of this function.

For a line like this:

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

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

Parabola displacement

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

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

Example.

We have: b = 2, c = 3.

This means that the classic curve shape will shift by 2 unit segment along the abscissa axis and 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 using given parameters.

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

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

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

D = (b 2 - 4 * a * c).

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

The presence of roots of a parabola depends on the result:

• D ˃ 0, then x 1, 2 = (-b ± D 0.5) / (2 * a);
• D = 0, then x 1, 2 = -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 ordinate axis;
• find the intersection with the x-axis.

Example 1.

Given the function y = x 2 - 5 * x + 4. It is necessary to construct a parabola. We follow the algorithm:

1. a = 1, therefore, the branches are directed upward;
2. extremum coordinates: x = - (-5) / 2 = 5/2; y = (5/2) 2 - 5 * (5/2) + 4 = -15/4;
3. intersects with the ordinate axis at the value y = 4;
4. let's find the discriminant: D = 25 - 16 = 9;
5. looking for roots:

• X 1 = (5 + 3) / 2 = 4; (4, 0);
• X 2 = (5 - 3) / 2 = 1; (10).

Example 2.

For the function y = 3 * x 2 - 2 * x - 1 you need to construct a parabola. We act according to the given algorithm:

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

• X 1 = (2 + 4) / 6 = 1; (1;0);
• X 2 = (2 - 4) / 6 = -1/3; (-1/3; 0).

Using the obtained points, you can construct a parabola.

Directrix, eccentricity, focus of a parabola

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

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

Eccentricity (constant) = 1.

Conclusion

We looked at a 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 a displacement along the axes, and, having a construction algorithm, you can draw its graph.

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To understand what will be written here, you need to know well what a quadratic function is and what it is used with. If you consider yourself a pro when it comes to quadratic functions, welcome. But if not, you should read the thread.

Let's start with a small checks:

1. What does a quadratic function look like in general form (formula)?
2. What is the graph of a quadratic function called?
3. How does the leading coefficient affect the graph of a quadratic function?

If you were able to answer these questions right away, continue reading. If at least one question caused difficulties, go to.

So, you already know how to handle a quadratic function, analyze its graph and build a graph by points.

Well, here it is: .

Let's briefly remember what they do odds.

1. The leading coefficient is responsible for the “steepness” of the parabola, or, in other words, for its width: the greater, the narrower the parabola (steeper), and the smaller, the wider the parabola (flatter).
2. The free term is the coordinate of the intersection of the parabola with the ordinate axis.
3. And the coefficient is somehow responsible for the displacement of the parabola from the center of coordinates. Let's talk about this in more detail now.

Where do we always start to build a parabola? What is its distinctive point?

This vertex. Do you remember how to find the coordinates of the vertex?

The abscissa is searched using the following formula:

Like this: than more, those to the left the vertex of the parabola moves.

The ordinate of the vertex can be found by substituting into the function:

Put it in and do the math yourself. What happened?

If you do everything correctly and simplify the resulting expression as much as possible, you get:

It turns out that the more modulo, those higher will vertex parabolas.

Let's finally move on to plotting the graph.
The easiest way is to build a parabola starting from the top.

Example:

Construct a graph of the function.

Solution:

First, let's determine the coefficients: .

Now let's calculate the coordinates of the vertex:

Now remember: all parabolas with the same leading coefficient look the same. This means that if we build a parabola and move its vertex to a point, we will get the graph we need:

Simple, right?

There is only one question left: how to quickly draw a parabola? Even if we draw a parabola with the vertex at the origin, we still have to build it point by point, and this is long and inconvenient. But all parabolas look the same, maybe there is a way to speed up their drawing?

When I was in school, my math teacher told everyone to cut out a parabola-shaped stencil from cardboard so they could quickly draw it. But you won’t be able to walk around with a stencil everywhere, and you won’t be allowed to take it to the exam. This means that we will not use foreign objects, but will look for a pattern.

Let's consider the simplest parabola. Let's build it point by point:

This is the pattern here. If from the vertex we shift to the right (along the axis) by, and upward (along the axis) by, then we will get to the point of the parabola. Further: if from this point we move to the right and up to, we will again get to the point of the parabola. Next: right on and up on. What's next? Right on and up on. And so on: move one to the right, and the next odd number up. Then we do the same with the left branch (after all, the parabola is symmetrical, that is, its branches look the same):

Great, this will help you construct any parabola from a vertex with a leading coefficient equal to. For example, we learned that the vertex of a parabola is at a point. Construct (yourself, on paper) this parabola.

Built?

It should look like this:

Now we connect the resulting points:

That's all.

OK, well, now we can only build parabolas with?

Of course not. Now let's figure out what to do with them, if.

Let's look at a few typical cases.

Great, you’ve learned how to draw a parabola, now let’s practice using real functions.

So, draw graphs of these functions:

Answers:

3. Top: .

Do you remember what to do if the senior coefficient is less?

We look at the denominator of the fraction: it is equal. So, we will move like this:

• right - up
• right - up
• right - up

and also to the left:

4. Top: .

Oh, what can we do about it? How to measure cells if the vertex is somewhere between the lines?..

And we'll cheat. Let's first draw a parabola, and only then move its vertex to a point. No, let’s do something even more cunning: Let’s draw a parabola, and then move the axes:- on down, a - on right:

This technique is very convenient in the case of any parabola, remember it.

Let me remind you that we can represent the function in this form:

For example: .

What does this give us?

The fact is that the number that is subtracted from in brackets () is the abscissa of the vertex of the parabola, and the term outside the brackets () is the ordinate of the vertex.

This means that, having constructed a parabola, you will simply need move the axis to the left and the axis down.

Example: Let's build a graph of a function.

Let's select a complete square:

What number deducted from in brackets? This (and not how you can decide without thinking).

So, let's build a parabola:

Now we shift the axis down, that is, up:

And now - to the left, that is, to the right:

That's all. This is the same as moving a parabola with its vertex from the origin to a point, only the straight axis is much easier to move than a curved parabola.

Now, as usual, myself:

And don’t forget to erase old axles with an eraser!

I'm as answers To check, I’ll write you the ordinates of the vertices of these parabolas:

Did everything come together?

If yes, then you are great! Knowing how to handle a parabola is very important and useful, and here we found out that it is not difficult at all.

CONSTRUCTION OF A GRAPH OF A QUADRATIC FUNCTION. BRIEFLY ABOUT THE MAIN THINGS

Quadratic function- a function of the form, where, and are any numbers (coefficients), - a free term.

The graph of a quadratic function is a parabola.

Vertex of the parabola:
, i.e. The larger \displaystyle b , the more to the left the vertex of the parabola moves.
We substitute it into the function, and we get:
, i.e. the \displaystyle b is larger in absolute value, the higher the top of the parabola will be

The free term is the coordinate of the intersection of the parabola with the ordinate axis.

Well, the topic is over. If you are reading these lines, it means you are very cool.

Because only 5% of people are able to master something on their own. And if you read to the end, then you are in this 5%!

Now the most important thing.

You have understood the theory on this topic. And, I repeat, this... this is just super! You are already better than the vast majority of your peers.

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For what?

For successfully passing the Unified State Exam, for entering college on a budget and, MOST IMPORTANTLY, for life.

I won’t convince you of anything, I’ll just say one thing...

People who received a good education, earn much more than those who did not receive it. This is statistics.

But this is not the main thing.

The main thing is that they are MORE HAPPY (there are such studies). Perhaps because many more opportunities open up before them and life becomes brighter? Don't know...

But think for yourself...

What does it take to be sure to be better than others on the Unified State Exam and ultimately be... happier?

GAIN YOUR HAND BY SOLVING PROBLEMS ON THIS TOPIC.

You won't be asked for theory during the exam.

You will need solve problems against time.

And, if you haven’t solved them (A LOT!), you’ll definitely make a stupid mistake somewhere or simply won’t have time.

It's like in sports - you need to repeat it many times to win for sure.

Find the collection wherever you want, necessarily with solutions, detailed analysis and decide, decide, decide!

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Find problems and solve them!

Many problems require calculating the maximum or minimum value of a quadratic function. The maximum or minimum can be found if the original function is written in standard form: or through the coordinates of the vertex of the parabola: f (x) = a (x − h) 2 + k (\displaystyle f(x)=a(x-h)^(2)+k). Moreover, the maximum or minimum of any quadratic function can be calculated using mathematical operations.

Steps

The quadratic function is written in standard form

Write the function in standard form. A quadratic function is a function whose equation involves a variable x 2 (\displaystyle x^(2)). The equation may or may not include a variable x (\displaystyle x). If an equation includes a variable with an exponent greater than 2, it does not describe a quadratic function. If necessary, provide similar terms and rearrange them to write the function in standard form.

• For example, given the function f (x) = 3 x + 2 x − x 2 + 3 x 2 + 4 (\displaystyle f(x)=3x+2x-x^(2)+3x^(2)+4). Add terms with variable x 2 (\displaystyle x^(2)) and members with variable x (\displaystyle x) to write the equation in standard form:
  • f (x) = 2 x 2 + 5 x + 4 (\displaystyle f(x)=2x^(2)+5x+4)

1. The graph of a quadratic function is a parabola. The branches of a parabola are directed up or down. If the coefficient a (\displaystyle a) with variable x 2 (\displaystyle x^(2)) a (\displaystyle a)

   • f (x) = 2 x 2 + 4 x − 6 (\displaystyle f(x)=2x^(2)+4x-6). Here a = 2 (\displaystyle a=2)
   • f (x) = − 3 x 2 + 2 x + 8 (\displaystyle f(x)=-3x^(2)+2x+8). Here, therefore, the parabola is directed downward.
   • f (x) = x 2 + 6 (\displaystyle f(x)=x^(2)+6). Here a = 1 (\displaystyle a=1), so the parabola is directed upward.
   • If the parabola is directed upward, you need to look for its minimum. If the parabola is pointing down, look for its maximum.

2. Calculate -b/2a. Meaning − b 2 a (\displaystyle -(\frac (b)(2a))) is the coordinate x (\displaystyle x) vertices of the parabola. If a quadratic function is written in standard form a x 2 + b x + c (\displaystyle ax^(2)+bx+c), use the coefficients for x (\displaystyle x) And x 2 (\displaystyle x^(2)) in the following way:

   • In the function coefficients a = 1 (\displaystyle a=1) And b = 10 (\displaystyle b=10)
     • x = − 10 (2) (1) (\displaystyle x=-(\frac (10)((2)(1))))
     • x = − 10 2 (\displaystyle x=-(\frac (10)(2)))
   • As a second example, consider the function. Here a = − 3 (\displaystyle a=-3) And b = 6 (\displaystyle b=6). Therefore, calculate the “x” coordinate of the vertex of the parabola as follows:
     • x = − b 2 a (\displaystyle x=-(\frac (b)(2a)))
     • x = − 6 (2) (− 3) (\displaystyle x=-(\frac (6)((2)(-3))))
     • x = − 6 − 6 (\displaystyle x=-(\frac (6)(-6)))
     • x = − (− 1) (\displaystyle x=-(-1))
     • x = 1 (\displaystyle x=1)

3. Find the corresponding value of f(x). Plug the found value of “x” into the original function to find the corresponding value of f(x). This way you will find the minimum or maximum of the function.

   • In the first example f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1) you have calculated that the x coordinate of the vertex of the parabola is x = − 5 (\displaystyle x=-5). In the original function, instead of x (\displaystyle x) substitute − 5 (\displaystyle -5)
     • f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1)
     • f (x) = (− 5) 2 + 10 (− 5) − 1 (\displaystyle f(x)=(-5)^(2)+10(-5)-1)
     • f (x) = 25 − 50 − 1 (\displaystyle f(x)=25-50-1)
     • f (x) = − 26 (\displaystyle f(x)=-26)
   • In the second example f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4) you found that the x coordinate of the vertex of the parabola is x = 1 (\displaystyle x=1). In the original function, instead of x (\displaystyle x) substitute 1 (\displaystyle 1) to find her maximum value:
     • f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4)
     • f (x) = − 3 (1) 2 + 6 (1) − 4 (\displaystyle f(x)=-3(1)^(2)+6(1)-4)
     • f (x) = − 3 + 6 − 4 (\displaystyle f(x)=-3+6-4)
     • f (x) = − 1 (\displaystyle f(x)=-1)

4. Write down your answer. Re-read the problem statement. If you need to find the coordinates of the vertex of a parabola, write down both values ​​in your answer x (\displaystyle x) And y (\displaystyle y)(or f (x) (\displaystyle f(x))). If you need to calculate the maximum or minimum of a function, write down only the value in your answer y (\displaystyle y)(or f (x) (\displaystyle f(x))). Look at the sign of the coefficient again a (\displaystyle a) to check whether you calculated the maximum or minimum.

   • In the first example f (x) = x 2 + 10 x − 1 (\displaystyle f(x)=x^(2)+10x-1) meaning a (\displaystyle a) positive, so you have calculated the minimum. The vertex of the parabola lies at the point with coordinates (− 5 , − 26) (\displaystyle (-5,-26)), and the minimum value of the function is − 26 (\displaystyle -26).
   • In the second example f (x) = − 3 x 2 + 6 x − 4 (\displaystyle f(x)=-3x^(2)+6x-4) meaning a (\displaystyle a) negative, so you have found the maximum. The vertex of the parabola lies at the point with coordinates (1 , − 1) (\displaystyle (1,-1)), and the maximum value of the function is − 1 (\displaystyle -1).

5. Determine the direction of the parabola. To do this, look at the sign of the coefficient a (\displaystyle a). If the coefficient a (\displaystyle a) positive, the parabola is directed upward. If the coefficient a (\displaystyle a) negative, the parabola is directed downward. For example:

   • . Here a = 2 (\displaystyle a=2), that is, the coefficient is positive, so the parabola is directed upward.
   • . Here a = − 3 (\displaystyle a=-3), that is, the coefficient is negative, so the parabola is directed downward.
   • If the parabola is directed upward, you need to calculate the minimum value of the function. If the parabola is directed downward, you need to find the maximum value of the function.

6. Find the minimum or maximum value of the function. If the function is written through the coordinates of the vertex of the parabola, the minimum or maximum is equal to the value of the coefficient k (\displaystyle k). In the above examples:

   • f (x) = 2 (x + 1) 2 − 4 (\displaystyle f(x)=2(x+1)^(2)-4). Here k = − 4 (\displaystyle k=-4). This is the minimum value of the function because the parabola is directed upward.
   • f (x) = − 3 (x − 2) 2 + 2 (\displaystyle f(x)=-3(x-2)^(2)+2). Here k = 2 (\displaystyle k=2). This is the maximum value of the function because the parabola is directed downward.

7. Find the coordinates of the vertex of the parabola. If the problem requires finding the vertex of a parabola, its coordinates are (h , k) (\displaystyle (h,k)). Please note that when a quadratic function is written through the coordinates of the vertex of a parabola, the subtraction operation must be enclosed in parentheses (x − h) (\displaystyle (x-h)), so the value h (\displaystyle h) is taken with the opposite sign.

   • f (x) = 2 (x + 1) 2 − 4 (\displaystyle f(x)=2(x+1)^(2)-4). Here the addition operation (x+1) is enclosed in parentheses, which can be rewritten as follows: (x-(-1)). Thus, h = − 1 (\displaystyle h=-1). Therefore, the coordinates of the vertex of the parabola of this function are equal to (− 1 , − 4) (\displaystyle (-1,-4)).
   • f (x) = − 3 (x − 2) 2 + 2 (\displaystyle f(x)=-3(x-2)^(2)+2). Here in brackets is the expression (x-2). Hence, h = 2 (\displaystyle h=2). The coordinates of the vertex are (2,2).

How to Calculate Minimum or Maximum Using Math Operations

1. First, let's look at the standard form of the equation. Write the quadratic function in standard form: f (x) = a x 2 + b x + c (\displaystyle f(x)=ax^(2)+bx+c). If necessary, add similar terms and rearrange them to obtain a standard equation.

   • For example: .

2. Find the first derivative. The first derivative of a quadratic function, which is written in standard form, is equal to f ′ (x) = 2 a x + b (\displaystyle f^(\prime )(x)=2ax+b).

   • f (x) = 2 x 2 − 4 x + 1 (\displaystyle f(x)=2x^(2)-4x+1). The first derivative of this function is calculated as follows:
     • f ′ (x) = 4 x − 4 (\displaystyle f^(\prime )(x)=4x-4)

3. Equate the derivative to zero. Recall that the derivative of a function is equal to the slope of the function at a certain point. At minimum or maximum, the slope is zero. Therefore, to find the minimum or maximum value of a function, the derivative must be set to zero. In our example.