"Critical points of the function" - Critical points. Among the critical points there are extremum points. A necessary condition for an extremum. Answer: 2. Definition. But, if f "(x0) = 0, then it is not necessary that the point x0 will be an extremum point. Extreme points (repetition). Critical points of the function. Extreme points.
"Coordinate plane Grade 6" - Mathematics Grade 6. 1. X. 1. Find and write down the coordinates of points A, B, C, D: -6. Coordinate plane. O. -3. 7. W.
"Functions and their graphs" - Continuity. The largest and smallest value of the function. The concept of an inverse function. Linear. Logarithmic. Monotone. If k > 0, then the formed angle is acute, if k< 0, то угол тупой. В самой точке x = a функция может существовать, а может и не существовать. Х1, х2, х3 – нули функции у = f(x).
"Functions Grade 9" - Permissible arithmetic operations on functions. [+] - addition, [-] - subtraction, [*] - multiplication, [:] - division. In such cases, one speaks of a graphical specification of a function. Formation of a class of elementary functions. Power function y=x0.5. Iovlev Maxim Nikolaevich, a student of the 9th grade of the RIOU Raduzhskaya school.
"Lesson Tangent Equation" - 1. Clarify the concept of a tangent to a function graph. Leibniz considered the problem of drawing a tangent to an arbitrary curve. ALGORITHM FOR DEVELOPING THE EQUATION OF THE FUNCTION tangent to the GRAPH y=f(x). Lesson topic: Test: find the derivative of a function. Tangent equation. Fluxion. Grade 10. Decipher how Isaac Newton called the derivative of a function.
"Build a graph of the function" - Given the function y=3cosx. Graph of the function y=m*sin x. Plot the function graph. Content: Given a function: y=sin (x+?/2). Stretching the graph y=cosx along the y axis. To continue press L. Mouse button. The function y=cosx+1 is given. Graph offsets y=sinx vertically. The function y=3sinx is given. Graph offset y=cosx horizontally.
There are 25 presentations in total in the topic
In this article, we will look at linear function, the graph of a linear function and its properties. And, as usual, we will solve several problems on this topic.
Linear function is called a function of the form
In the function equation, the number we multiply by is called the slope factor.
For example, in the function equation ;
in the function equation ;
in the function equation ;
in the function equation.
The graph of a linear function is a straight line.
one . To plot a function, we need the coordinates of two points belonging to the graph of the function. To find them, you need to take two x values, substitute them into the equation of the function, and calculate the corresponding y values from them.
For example, to plot the function , it is convenient to take and , then the ordinates of these points will be equal to and .
We get points A(0;2) and B(3;3). Let's connect them and get the graph of the function:
2 . In the function equation, the coefficient is responsible for the slope of the function graph:
Title="(!LANG:k>0">!}
The coefficient is responsible for shifting the graph along the axis:
Title="(!LANG:b>0">!}
The figure below shows the graphs of functions; ;
Note that in all these functions the coefficient Above zero right. Moreover, the larger the value, the steeper the straight line goes.
In all functions - and we see that all graphs intersect the OY axis at the point (0;3)
Now consider the function graphs; ;
This time in all functions the coefficient less than zero, and all function graphs are skewed to the left.
Note that the larger |k|, the steeper the line goes. The coefficient b is the same, b=3, and the graphs, as in the previous case, cross the OY axis at the point (0;3)
Consider the graphs of functions ; ;
Now in all equations of functions the coefficients are equal. And we got three parallel lines.
But the coefficients b are different, and these graphs intersect the OY axis at different points:
The graph of the function (b=3) crosses the OY axis at the point (0;3)
The graph of the function (b=0) crosses the OY axis at the point (0;0) - the origin.
The graph of the function (b=-2) crosses the OY axis at the point (0;-2)
So, if we know the signs of the coefficients k and b, then we can immediately imagine what the graph of the function looks like.
If a k<0 и b>0 , then the graph of the function looks like:
If a k>0 and b>0 , then the graph of the function looks like:
If a k>0 and b<0 , then the graph of the function looks like:
If a k<0 и b<0 , then the graph of the function looks like:
If a k=0 , then the function turns into a function and its graph looks like:
The ordinates of all points of the graph of the function are equal
If a b=0, then the graph of the function passes through the origin:
it direct proportionality graph.
3 . Separately, I note the graph of the equation. The graph of this equation is a straight line parallel to the axis, all points of which have an abscissa.
For example, the equation graph looks like this:
Attention! The equation is not a function, since different values of the argument correspond to the same function value, which does not correspond to .
4 . Condition for parallelism of two lines:
Function Graph parallel to the graph of the function, if
5. The condition of perpendicularity of two lines:
Function Graph perpendicular to the graph of the function if or
6. Intersection points of the graph of the function with the coordinate axes.
with OY axis. The abscissa of any point belonging to the OY axis is equal to zero. Therefore, to find the point of intersection with the OY axis, you need to substitute zero instead of x in the equation of the function. We get y=b. That is, the point of intersection with the OY axis has coordinates (0;b).
With OX axis: The ordinate of any point belonging to the OX axis is zero. Therefore, to find the point of intersection with the OX axis, you need to substitute zero instead of y in the equation of the function. We get 0=kx+b. From here. That is, the point of intersection with the OX axis has coordinates (; 0):
Consider problem solving.
one . Build a graph of the function if it is known that it passes through the point A (-3; 2) and is parallel to the line y \u003d -4x.
There are two unknown parameters in the function equation: k and b. Therefore, in the text of the problem there should be two conditions that characterize the graph of the function.
a) From the fact that the graph of the function is parallel to the straight line y=-4x, it follows that k=-4. That is, the equation of the function has the form
b) It remains for us to find b. It is known that the graph of the function passes through the point A (-3; 2). If the point belongs to the function graph, then when substituting its coordinates into the function equation, we get the correct equality:
hence b=-10
Thus, we need to plot the function
Point A(-3;2) is known to us, take point B(0;-10)
Let's put these points in the coordinate plane and connect them with a straight line:
2. Write the equation of a straight line passing through the points A(1;1); B(2;4).
If the line passes through points with given coordinates, then the coordinates of the points satisfy the equation of the line. That is, if we substitute the coordinates of the points into the equation of a straight line, we will get the correct equality.
Substitute the coordinates of each point in the equation and get a system of linear equations.
We subtract the first equation from the second equation of the system, and we get . Substitute the value of k in the first equation of the system, and get b=-2.
So, the equation of a straight line.
3 . Plot Equation 
To find at what values of the unknown the product of several factors is equal to zero, you need to equate each factor to zero and take into account each multiplier.
This equation has no restrictions on ODZ. Let us factorize the second bracket and equate each factor to zero. We get a set of equations:
We construct graphs of all equations of the set in one coordinate plane. This is the graph of the equation :
four . Build a graph of the function if it is perpendicular to the straight line and passes through the point M (-1; 2)
We will not build a graph, we will only find the equation of a straight line.
a) Since the graph of the function, if it is perpendicular to the straight line, therefore, from here. That is, the equation of the function has the form
b) We know that the graph of the function passes through the point M (-1; 2). Substitute its coordinates into the equation of the function. We get:
From here.
Therefore, our function looks like: .
5 . Plot the Function 
Let's simplify the expression on the right side of the function equation.
Important! Before simplifying the expression, let's find its ODZ.
The denominator of a fraction cannot be zero, so title="(!LANG:x1">, title="x-1">.!}
Then our function becomes:
Title="(!LANG:delim(lbrace)(matrix(3)(1)((y=x+2) (x1) (x-1)))( )">!}
That is, we need to build a function graph and poke out two points on it: with abscissas x=1 and x=-1:
LINEAR EQUATIONS AND INEQUALITIES I
§ 3 Linear functions and their graphs
Consider the equality
at = 2X + 1. (1)
Each value of a letter X this equality associates a well-defined meaning of the letter at . If, for example, x = 0, then at = 20 + 1 = 1; if X = 10, then at = 2 10 + 1 = 21; at X \u003d - 1 / 2 we have y \u003d 2 (- 1 / 2) + 1 \u003d 0, etc. Let's turn to one more equality:
at = X 2 (2)
Each value X this equality, like equality (1), associates a well-defined value at . If, for example, X = 2, then at = 4; at X = - 3 we get at = 9, etc. Equalities (1) and (2) connect the two quantities X and at so that each value of one of them ( X ) is associated with a well-defined value of another quantity ( at ).
If each value of quantity X corresponds to a well-defined value of the quantity at, then this value at is called a function of X. Value X is called a function argument at.
Thus, formulas (1) and (2) define two different functions of the argument X .
Argument function X , having the form
y = ax + b , (3)
where a and b - some given numbers, called linear. Any of the functions can serve as an example of a linear function:
y = x
+ 2 (a
= 1, b
= 2);
at
= - 10 (a
= 0, b
= - 10);
at
= - 3X
(a
= - 3, b
= 0);
at
= 0 (a = b
= 0).
As is known from the course of the VIII class, function graph y = ax + b is a straight line. That is why this function is called linear.
Recall how the graph of a linear function is constructed y = ax + b .
1. Function Graph y = b . At a = 0 linear function y = ax + b has the form y = b . Its graph is a straight line parallel to the axis X and cross axis at at the point with the ordinate b . In figure 1 you see the graph of the function y = 2 ( b > 0), and in figure 2 - graph of the function at = - 1 (b < 0).
If not only a , but also b equals zero, then the function y=ax+b has the form at = 0. In this case, its graph coincides with the axis X (Fig. 3.)
2. Function Graph y=ah . At b = 0 linear function y = ax + b has the form y=ah .
If a a =/= 0, then its graph is a straight line passing through the origin and inclined to the axis X at an angle φ , whose tangent is a (Fig. 4). To build a straight line y=ah it suffices to find some one of its points, different from the origin. Assuming, for example, in equality y=ah X = 1, we get at = a . Therefore, point M with coordinates (1; a ) lies on our line (Fig. 4). Now drawing a straight line through the origin and point M, we obtain the desired straight line y = ax .
Figure 5 shows a straight line as an example. at = 2X (a > 0), and in figure 6 - a straight line y = - x (a < 0).
3. Function Graph y = ax + b .
Let b > 0. Then the line y = ax + b y=ah on the b units up. As an example, Figure 7 shows the construction of a straight line at = x / 2 + 3.
If a b < 0, то прямая y = ax + b obtained by a parallel shift of the straight line y=ah on the - b units down. As an example, Figure 8 shows the construction of a straight line at = x / 2 - 3
direct y = ax + b can be built in another way.
Any line is completely determined by its two points. Therefore, to plot the function y = ax + b it is enough to find any two of its points, and then draw a straight line through them. Let's explain this with the example of the function at = - 2X + 3.
At X = 0 at = 3, while X = 1 at = 1. Therefore, two points: M with coordinates (0; 3) and N with coordinates (1; 1) - lie on our line. Marking these points on the coordinate plane and connecting them with a straight line (Fig. 9), we obtain a graph of the function at = - 2X + 3.
Instead of the points M and N, one could, of course, take the other two points. For example, as values X we could choose not 0 and 1, as above, but 1 and 2.5. Then for at we would get the values 5 and - 2, respectively. Instead of points M and N, we would have points P with coordinates (- 1; 5) and Q with coordinates (2.5; - 2). These two points, as well as the points M and N, completely determine the desired line at = - 2X + 3.
Exercises
15. On the same figure, build graphs of functions:
a) at = - 4; b) at = -2; in) at = 0; G) at = 2; e) at = 4.
Do these graphs intersect with the coordinate axes? If they intersect, then specify the coordinates of the intersection points.
16. On the same figure, plot function graphs:
a) at = x / four ; b) at = x / 2; in) at =X ; G) at = 2X ; e) at = 4X .
17. On the same figure, build graphs of functions:
a) at = - x / four ; b) at = - x / 2; in) at = - X ; G) at = - 2X ; e) at = - 4X .
Build graphs of these functions (No. 18-21) and determine the coordinates of the points of intersection of these graphs with the coordinate axes.
18. at = 3+ X . 20. at = - 4 - X .
19. at = 2X - 2. 21. at = 0,5(1 - 3X ).
22. Graph a function
at = 2x - 4;
using this graph, find out: a) for what values x y = 0;
b) at what values X values at negative and at what - positive;
c) at what values X quantities X and at have the same signs;
d) at what values X quantities X and at have different signs.
23. Write the equations of the lines shown in figures 10 and 11.
24. Which of the physical laws known to you are described using linear functions?
25. How to graph a function at = - (ax + b ) if the graph of the function is given y = ax + b ?
Tasks on the properties and graphs of a quadratic function, as practice shows, cause serious difficulties. This is rather strange, because the quadratic function is passed in the 8th grade, and then the entire first quarter of the 9th grade is "extorted" by the properties of the parabola and its graphs are built for various parameters.
This is due to the fact that forcing students to build parabolas, they practically do not devote time to "reading" graphs, that is, they do not practice comprehending the information received from the picture. Apparently, it is assumed that, having built two dozen graphs, a smart student himself will discover and formulate the relationship between the coefficients in the formula and the appearance of the graph. In practice, this does not work. For such a generalization, serious experience in mathematical mini-research is required, which, of course, most ninth-graders do not have. Meanwhile, in the GIA they propose to determine the signs of the coefficients precisely according to the schedule.
We will not demand the impossible from schoolchildren and simply offer one of the algorithms for solving such problems.
So, a function of the form y=ax2+bx+c is called quadratic, its graph is a parabola. As the name implies, the main component is ax 2. That is a should not be equal to zero, the remaining coefficients ( b and With) can be equal to zero.
Let's see how the signs of its coefficients affect the appearance of the parabola.
The simplest dependence for the coefficient a. Most schoolchildren confidently answer: "if a> 0, then the branches of the parabola are directed upwards, and if a < 0, - то вниз". Совершенно верно. Ниже приведен график квадратичной функции, у которой a > 0.
y = 0.5x2 - 3x + 1
AT this case a = 0,5
And now for a < 0:
y = - 0.5x2 - 3x + 1
In this case a = - 0,5
Influence of coefficient With also easy enough to follow. Imagine that we want to find the value of a function at a point X= 0. Substitute zero into the formula:
y = a 0 2 + b 0 + c = c. It turns out that y=s. That is With is the ordinate of the point of intersection of the parabola with the y-axis. As a rule, this point is easy to find on the graph. And determine whether it lies above zero or below. That is With> 0 or With < 0.
With > 0:
y=x2+4x+3
With < 0
y = x 2 + 4x - 3
Accordingly, if With= 0, then the parabola will necessarily pass through the origin:
y=x2+4x
More difficult with the parameter b. The point by which we will find it depends not only on b but also from a. This is the top of the parabola. Its abscissa (axis coordinate X) is found by the formula x in \u003d - b / (2a). In this way, b = - 2ax in. That is, we act as follows: on the graph we find the top of the parabola, determine the sign of its abscissa, that is, we look to the right of zero ( x in> 0) or to the left ( x in < 0) она лежит.
However, this is not all. We must also pay attention to the sign of the coefficient a. That is, to see where the branches of the parabola are directed. And only after that, according to the formula b = - 2ax in determine sign b.
Consider an example:
Branches pointing upwards a> 0, the parabola crosses the axis at below zero means With < 0, вершина параболы лежит правее нуля. Следовательно, x in> 0. So b = - 2ax in = -++ = -. b < 0. Окончательно имеем: a > 0, b < 0, With < 0.
Linear function definition
Let us introduce the definition of a linear function
Definition
A function of the form $y=kx+b$, where $k$ is nonzero, is called a linear function.
The graph of a linear function is a straight line. The number $k$ is called the slope of the line.
For $b=0$ the linear function is called the direct proportionality function $y=kx$.
Consider Figure 1.
Rice. 1. The geometric meaning of the slope of the straight line
Consider triangle ABC. We see that $BC=kx_0+b$. Find the point of intersection of the line $y=kx+b$ with the axis $Ox$:
\ \
So $AC=x_0+\frac(b)(k)$. Let's find the ratio of these sides:
\[\frac(BC)(AC)=\frac(kx_0+b)(x_0+\frac(b)(k))=\frac(k(kx_0+b))((kx)_0+b)=k \]
On the other hand $\frac(BC)(AC)=tg\angle A$.
Thus, the following conclusion can be drawn:
Conclusion
Geometric meaning of the coefficient $k$. The slope of the straight line $k$ is equal to the tangent of the slope of this straight line to the axis $Ox$.
Study of the linear function $f\left(x\right)=kx+b$ and its graph
First, consider the function $f\left(x\right)=kx+b$, where $k > 0$.
- $f"\left(x\right)=(\left(kx+b\right))"=k>0$. Therefore, this function increases over the entire domain of definition. There are no extreme points.
- $(\mathop(lim)_(x\to -\infty ) kx\ )=-\infty $, $(\mathop(lim)_(x\to +\infty ) kx\ )=+\infty $
- Graph (Fig. 2).
Rice. 2. Graphs of the function $y=kx+b$, for $k > 0$.
Now consider the function $f\left(x\right)=kx$, where $k
- The scope is all numbers.
- The scope is all numbers.
- $f\left(-x\right)=-kx+b$. The function is neither even nor odd.
- For $x=0,f\left(0\right)=b$. For $y=0,0=kx+b,\ x=-\frac(b)(k)$.
Intersection points with coordinate axes: $\left(-\frac(b)(k),0\right)$ and $\left(0,\ b\right)$
- $f"\left(x\right)=(\left(kx\right))"=k
- $f^("")\left(x\right)=k"=0$. Therefore, the function has no inflection points.
- $(\mathop(lim)_(x\to -\infty ) kx\ )=+\infty $, $(\mathop(lim)_(x\to +\infty ) kx\ )=-\infty $
- Graph (Fig. 3).
