Plotting a linear function containing a module. How to Solve Equations with Modulus: Basic Rules

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Presentation for the lesson

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The purpose of the lesson:

• repeat the construction of graphs of functions containing the sign of the modulus;
• get acquainted with a new method of constructing a graph of a linear-piecewise function;
• to fix new method when solving problems.

Equipment:

• multimedia projector,
• posters.

During the classes

Knowledge update

On the screen slide 1 from the presentation.

What is the graph of the function y=|x| ? (slide 2).

(set of bisectors of 1 and 2 coordinate angles)

Find a correspondence between functions and graphs, explain your choice (slide 3).

Picture 1

Tell the algorithm for constructing graphs of functions of the form y=|f(x)| on the example of the function y=|x 2 -2x-3| (slide 4)

Student: to build a graph of this function, you need

Construct a parabola y=x 2 -2x-3

Figure 2

Figure 3

Tell the algorithm for constructing graphs of functions of the form y=f(|x|) using the example of the function y=x 2 -2|x|-3 (slide 6).

Build a parabola.

Part of the graph at x 0 is saved and displayed in symmetry with respect to the y-axis (slide 7)

Figure 4

Tell the algorithm for constructing graphs of functions of the form y=|f(|x|)| on the example of the function y=|x 2 -2|x|-3| (slide 8).

Student: To build a graph of this function, you need:

You need to build a parabola y \u003d x 2 -2x-3

We build y \u003d x 2 -2 | x | -3, save part of the graph and display it symmetrically with respect to the OS

We save the part above the OX, and display the lower part symmetrically with respect to the OX (slide 9)

Figure 5

The next task is written in notebooks.

1. Draw a graph of a linear-piecewise function y=|x+2|+|x-1|-|x-3|

Student on blackboard commenting:

We find the zeros of submodule expressions x 1 \u003d -2, x 2 \u003d 1, x 3 \u003d 3

Breaking the axis into intervals

For each interval, we write the function

at x< -2, у=-х-4

at -2 x<1, у=х

at 1 x<3, у = 3х-2

at x 3, y \u003d x + 4

We build a graph of a linear-piecewise function.

We have built a function graph using the module definition (slide 10).

Figure 6

I bring to your attention the “vertex method”, which allows you to plot a linear-piecewise function (slide 11). The children write down the construction algorithm in a notebook.

Vertex method

Algorithm:

1. Find the zeros of each submodule expression
2. Let's make a table in which, in addition to zeros, we write one value of the argument on the left and on the right
3. Let's put the points on the coordinate plane and connect them in series

2. Let's analyze this method on the same function y=|x+2|+|x-1|-|x-3|

The teacher is at the blackboard, the children are in their notebooks.

Vertex method:

Find the zeros of each submodule expression;

Let's make a table in which, in addition to zeros, we write one value of the argument on the left and on the right

Let's put the points on the coordinate plane and connect them in series.

The graph of a linear-piecewise function is a broken line with infinite extreme links (slide 12).

Figure 7

What method makes the graph faster and easier?

3. To fix this method, I propose to perform the following task:

For what values ​​of x does the function y=|x-2|-|x+1| takes on the largest value.

We follow the algorithm; student at the blackboard.

y=|x-2|-|x+1|

x 1 \u003d 2, x 2 \u003d -1

y(3)=1-4=3, connect the dots in series.

4. Additional task

For what values ​​of a does the equation ||4+x|-|x-2||=a have two roots.

5. Homework

a) For what values ​​of X is the function y =|2x+3|+3|x-1|-|x+2| takes the smallest value.

b) Plot the function y=||x-1|-2|-3| .

, Competition "Presentation for the lesson"

Presentation for the lesson

Back forward

Attention! The slide preview is for informational purposes only and may not represent the full extent of the presentation. If you are interested in this work, please download the full version.

The purpose of the lesson:

• repeat the construction of graphs of functions containing the sign of the modulus;
• get acquainted with a new method of constructing a graph of a linear-piecewise function;
• consolidate the new method in solving problems.

Equipment:

• multimedia projector,
• posters.

During the classes

Knowledge update

On the screen slide 1 from the presentation.

What is the graph of the function y=|x| ? (slide 2).

(set of bisectors of 1 and 2 coordinate angles)

Find a correspondence between functions and graphs, explain your choice (slide 3).

Picture 1

Tell the algorithm for constructing graphs of functions of the form y=|f(x)| on the example of the function y=|x 2 -2x-3| (slide 4)

Student: to build a graph of this function, you need

Construct a parabola y=x 2 -2x-3

Figure 2

Figure 3

Tell the algorithm for constructing graphs of functions of the form y=f(|x|) using the example of the function y=x 2 -2|x|-3 (slide 6).

Build a parabola.

Part of the graph at x 0 is saved and displayed in symmetry with respect to the y-axis (slide 7)

Figure 4

Tell the algorithm for constructing graphs of functions of the form y=|f(|x|)| on the example of the function y=|x 2 -2|x|-3| (slide 8).

Student: To build a graph of this function, you need:

You need to build a parabola y \u003d x 2 -2x-3

We build y \u003d x 2 -2 | x | -3, save part of the graph and display it symmetrically with respect to the OS

We save the part above the OX, and display the lower part symmetrically with respect to the OX (slide 9)

Figure 5

The next task is written in notebooks.

1. Draw a graph of a linear-piecewise function y=|x+2|+|x-1|-|x-3|

Student on blackboard commenting:

We find the zeros of submodule expressions x 1 \u003d -2, x 2 \u003d 1, x 3 \u003d 3

Breaking the axis into intervals

For each interval, we write the function

at x< -2, у=-х-4

at -2 x<1, у=х

at 1 x<3, у = 3х-2

at x 3, y \u003d x + 4

We build a graph of a linear-piecewise function.

We have built a function graph using the module definition (slide 10).

Figure 6

I bring to your attention the “vertex method”, which allows you to plot a linear-piecewise function (slide 11). The children write down the construction algorithm in a notebook.

Vertex method

Algorithm:

1. Find the zeros of each submodule expression
2. Let's make a table in which, in addition to zeros, we write one value of the argument on the left and on the right
3. Let's put the points on the coordinate plane and connect them in series

2. Let's analyze this method on the same function y=|x+2|+|x-1|-|x-3|

The teacher is at the blackboard, the children are in their notebooks.

Vertex method:

Find the zeros of each submodule expression;

Let's make a table in which, in addition to zeros, we write one value of the argument on the left and on the right

Let's put the points on the coordinate plane and connect them in series.

The graph of a linear-piecewise function is a broken line with infinite extreme links (slide 12).

Figure 7

What method makes the graph faster and easier?

3. To fix this method, I propose to perform the following task:

For what values ​​of x does the function y=|x-2|-|x+1| takes on the largest value.

We follow the algorithm; student at the blackboard.

y=|x-2|-|x+1|

x 1 \u003d 2, x 2 \u003d -1

y(3)=1-4=3, connect the dots in series.

4. Additional task

For what values ​​of a does the equation ||4+x|-|x-2||=a have two roots.

5. Homework

a) For what values ​​of X is the function y =|2x+3|+3|x-1|-|x+2| takes the smallest value.

b) Plot the function y=||x-1|-2|-3| .

Function of the form y=|x|.
The graph of the function on the interval - with the graph of the function y \u003d -x.

Consider first the simplest case - the function y=|x|. By definition of the module, we have:

Thus, for x≥0 the function y=|x| coincides with the function y \u003d x, and for x Using this explanation, it is easy to plot the function y \u003d | x | (Fig. 1).

It is easy to see that this graph is the union of that part of the graph of the function y \u003d x, which lies not below the OX axis, and the line obtained by mirror reflection about the OX axis, that part of it, which lies below the OX axis.
This method is also suitable for plotting the graph of the function y=|kx+b|.
If the graph of the function y=kx+b is shown in Figure 2, then the graph of the function y=|kx+b| is the line shown in Figure 3.

(!LANG:Example 1. Plot the function y=||1-x 2 |-3|.
Let's build a graph of the function y=1-x 2 and apply the "module" operation to it (the part of the graph located below the OX axis is reflected symmetrically relative to the OX axis).

Let's shift the chart down by 3.

Let's apply the "module" operation and get the final graph of the function y=||1-x 2 |-3|

Example 2 Plot the function y=||x 2 -2x|-3|.
As a result of the transformation, we get y=|x 2 -2x|=|(x-1) 2 -1|. Let's build a graph of the function y=(x-1) 2 -1: build a parabola y=x 2 and shift right by 1 and down by 1.

Let's apply the "module" operation to it (the part of the graph located below the OX axis is reflected symmetrically with respect to the OX axis).

Let's shift the graph down by 3 and apply the "module" operation, as a result we will get the final graph.

Example 3 Plot the function.
To expand a module, we need to consider two cases:
1)x>0, then the module will open with the sign "+" =
2) x =

Let's build a graph for the first case.

Let's discard the part of the graph, where x

Let's build a graph for the second case and similarly discard the part where x>0, as a result we get.

Let's combine the two graphs and get the final one.

Example 4 Plot the function.
First, let's build a graph of the function. For this, it is convenient to select the integer part, we get. Building on the table of values, we get a graph.

Let's apply the modulus operation (the part of the graph located below the OX axis is reflected symmetrically with respect to the OX axis). We get the final chart

Example 5 Plot the function y=|-x 2 +6x-8|. First, we simplify the function to y=1-(x-3) 2 and build its graph

Now we apply the “module” operation and reflect the part of the graph below the OX axis, relative to the OX axis

Example 6 Plot the function y=-x 2 +6|x|-8. We also simplify the function to y=1-(x-3) 2 and build its graph

Now we apply the “module” operation and reflect the part of the graph to the right of the oY axis, to the left side

Example 7 Plot a function . Let's plot the function

Let's plot the function

Let's perform a parallel transfer by 3 unit segments to the right and 2 up. The graph will look like:

Let's apply the "module" operation and reflect the part of the graph to the right of the straight line x=3 into the left half-plane.

The modulo sign is perhaps one of the most interesting phenomena in mathematics. In this regard, many schoolchildren have the question of how to build graphs of functions containing a module. Let's examine this issue in detail.

1. Plotting functions containing a module

Example 1

Plot the function y = x 2 – 8|x| + 12.

Solution.

Let us define the parity of the function. The value for y(-x) is the same as the value for y(x), so this function is even. Then its graph is symmetrical with respect to the Oy axis. We build a graph of the function y \u003d x 2 - 8x + 12 for x ≥ 0 and symmetrically display the graph relative to Oy for negative x (Fig. 1).

Example 2

The next graph is y = |x 2 – 8x + 12|.

– What is the range of the proposed function? (y ≥ 0).

- How is the chart? (Above or touching the x-axis).

This means that the graph of the function is obtained as follows: they plot the function y \u003d x 2 - 8x + 12, leave the part of the graph that lies above the Ox axis unchanged, and the part of the graph that lies under the abscissa axis is displayed symmetrically relative to the Ox axis (Fig. 2).

Example 3

To plot the function y = |x 2 – 8|x| + 12| carry out a combination of transformations:

y = x 2 - 8x + 12 → y = x 2 - 8|x| + 12 → y = |x 2 – 8|x| + 12|.

Answer: figure 3.

The considered transformations are valid for all kinds of functions. Let's make a table:

2. Plotting functions containing "nested modules" in the formula

We have already got acquainted with examples of a quadratic function containing a modulus, as well as with the general rules for constructing graphs of functions of the form y = f(|x|), y = |f(x)| and y = |f(|x|)|. These transformations will help us when considering the following example.

Example 4

Consider a function of the form y = |2 – |1 – |x|||. The expression that defines the function contains "nested modules".

Solution.

We use the method of geometric transformations.

Let's write down a chain of successive transformations and make the corresponding drawing (Fig. 4):

y = x → y = |x| → y = -|x| → y = -|x| + 1 → y = |-|x| + 1|→ y = -|-|x| + 1|→ y = -|-|x| + 1| + 2 → y = |2 –|1 – |x|||.

Let's consider the cases when symmetry and parallel translation transformations are not the main technique for plotting.

Example 5

Construct a graph of a function of the form y \u003d (x 2 - 4) / √ (x + 2) 2.

Solution.

Before building a graph, we transform the formula that defines the function and get another analytical definition of the function (Fig. 5).

y = (x 2 – 4)/√(x + 2) 2 = (x– 2)(x + 2)/|x + 2|.

Let's expand the module in the denominator:

For x > -2, y = x - 2, and for x< -2, y = -(x – 2).

Domain D(y) = (-∞; -2)ᴗ(-2; +∞).

Range E(y) = (-4; +∞).

Points at which the graph intersects with the coordinate axis: (0; -2) and (2; 0).

The function decreases for all x from the interval (-∞; -2), increases for x from -2 to +∞.

Here we had to reveal the sign of the modulus and plot the function for each case.

Example 6

Consider the function y = |x + 1| – |x – 2|.

Solution.

Expanding the sign of the module, it is necessary to consider all possible combinations of signs of submodule expressions.

There are four possible cases:

(x + 1 - x + 2 = 3, with x ≥ -1 and x ≥ 2;

(-x - 1 + x - 2 = -3, with x< -1 и x < 2;

(x + 1 + x - 2 = 2x - 1, for x ≥ -1 and x< 2;

(-x - 1 - x + 2 = -2x + 1, with x< -1 и x ≥ 2 – пустое множество.

Then the original function will look like:

(3, for x ≥ 2;

y = (-3, at x< -1;

(2x – 1, with -1 ≤ x< 2.

We got a piecewise given function, the graph of which is shown in Figure 6.

3. Algorithm for constructing graphs of functions of the form

y = a 1 | x – x 1 | + a 2 |x – x 2 | + … + a n |x – x n | + ax + b.

In the previous example, it was easy enough to expand the module signs. If there are more sums of modules, then it is problematic to consider all possible combinations of signs of submodule expressions. How can we graph the function in this case?

Note that the graph is a polyline, with vertices at points having abscissas -1 and 2. For x = -1 and x = 2, the submodule expressions are equal to zero. In a practical way, we approached the rule for constructing such graphs:

Graph of a function of the form y = a 1 |x – x 1 | + a 2 |x – x 2 | + … + a n |x – x n | + ax + b is a broken line with infinite end links. To construct such a polyline, it is enough to know all its vertices (vertex abscissas are zeros of submodule expressions) and one control point each on the left and right infinite links.

A task.

Plot the function y = |x| + |x – 1| + |x + 1| and find its smallest value.

Solution:

Zeros of submodule expressions: 0; -one; 1. Vertices of the polyline (0; 2); (-13); (13). Control point on the right (2; 6), on the left (-2; 6). We build a graph (Fig. 7). min f(x) = 2.

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