home

Graph of function y 3x 8. Quadratic and cubic functions

Constructing graphs of functions containing modules usually causes considerable difficulties for schoolchildren. However, everything is not so bad. It is enough to remember a few algorithms for solving such problems, and you can easily build a graph even for the most seemingly complex function. Let's figure out what kind of algorithms these are.

1. Plotting a graph of the function y = |f(x)|

Note that the set of function values y = |f(x)| : y ≥ 0. Thus, the graphs of such functions are always located entirely in the upper half-plane.

Plotting a graph of the function y = |f(x)| consists of the following simple four steps.

1) Carefully and carefully construct a graph of the function y = f(x).

2) Leave unchanged all points on the graph that are above or on the 0x axis.

3) Display the part of the graph that lies below the 0x axis symmetrically relative to the 0x axis.

Example 1. Draw a graph of the function y = |x 2 – 4x + 3|

1) We build a graph of the function y = x 2 – 4x + 3. Obviously, the graph of this function is a parabola. Let's find the coordinates of all points of intersection of the parabola with the coordinate axes and the coordinates of the vertex of the parabola.

x 2 – 4x + 3 = 0.

x 1 = 3, x 2 = 1.

Therefore, the parabola intersects the 0x axis at points (3, 0) and (1, 0).

y = 0 2 – 4 0 + 3 = 3.

Therefore, the parabola intersects the 0y axis at the point (0, 3).

Parabola vertex coordinates:

x in = -(-4/2) = 2, y in = 2 2 – 4 2 + 3 = -1.

Therefore, point (2, -1) is the vertex of this parabola.

Draw a parabola using the data obtained (Fig. 1)

2) The part of the graph lying below the 0x axis is displayed symmetrically relative to the 0x axis.

3) We get a graph of the original function ( rice. 2, shown in dotted line).

2. Plotting the function y = f(|x|)

Note that functions of the form y = f(|x|) are even:

y(-x) = f(|-x|) = f(|x|) = y(x). This means that the graphs of such functions are symmetrical about the 0y axis.

Plotting a graph of the function y = f(|x|) consists of the following simple chain of actions.

1) Graph the function y = f(x).

2) Leave that part of the graph for which x ≥ 0, that is, the part of the graph located in the right half-plane.

3) Display the part of the graph specified in point (2) symmetrically to the 0y axis.

4) As the final graph, select the union of the curves obtained in points (2) and (3).

Example 2. Draw a graph of the function y = x 2 – 4 · |x| + 3

Since x 2 = |x| 2, then the original function can be rewritten in the following form: y = |x| 2 – 4 · |x| + 3. Now we can apply the algorithm proposed above.

1) We carefully and carefully build a graph of the function y = x 2 – 4 x + 3 (see also rice. 1).

2) We leave that part of the graph for which x ≥ 0, that is, the part of the graph located in the right half-plane.

3) Display the right side of the graph symmetrically to the 0y axis.

(Fig. 3).

Example 3. Draw a graph of the function y = log 2 |x|

We apply the scheme given above.

1) Build a graph of the function y = log 2 x (Fig. 4).

3. Plotting the function y = |f(|x|)|

Note that functions of the form y = |f(|x|)| are also even. Indeed, y(-x) = y = |f(|-x|)| = y = |f(|x|)| = y(x), and therefore, their graphs are symmetrical about the 0y axis. The set of values of such functions: y ≥ 0. This means that the graphs of such functions are located entirely in the upper half-plane.

To plot the function y = |f(|x|)|, you need to:

1) Carefully construct a graph of the function y = f(|x|).

2) Leave unchanged the part of the graph that is above or on the 0x axis.

3) Display the part of the graph located below the 0x axis symmetrically relative to the 0x axis.

4) As the final graph, select the union of the curves obtained in points (2) and (3).

Example 4. Draw a graph of the function y = |-x 2 + 2|x| – 1|.

1) Note that x 2 = |x| 2. This means that instead of the original function y = -x 2 + 2|x| - 1

you can use the function y = -|x| 2 + 2|x| – 1, since their graphs coincide.

We build a graph y = -|x| 2 + 2|x| – 1. For this we use algorithm 2.

a) Graph the function y = -x 2 + 2x – 1 (Fig. 6).

b) We leave that part of the graph that is located in the right half-plane.

c) We display the resulting part of the graph symmetrically to the 0y axis.

d) The resulting graph is shown in the dotted line in the figure (Fig. 7).

2) There are no points above the 0x axis; we leave the points on the 0x axis unchanged.

3) The part of the graph located below the 0x axis is displayed symmetrically relative to 0x.

4) The resulting graph is shown in the figure with a dotted line (Fig. 8).

Example 5. Graph the function y = |(2|x| – 4) / (|x| + 3)|

1) First you need to plot the function y = (2|x| – 4) / (|x| + 3). To do this, we return to Algorithm 2.

a) Carefully plot the function y = (2x – 4) / (x + 3) (Fig. 9).

Note that this function is fractional linear and its graph is a hyperbola. To plot a curve, you first need to find the asymptotes of the graph. Horizontal – y = 2/1 (the ratio of the coefficients of x in the numerator and denominator of the fraction), vertical – x = -3.

2) We will leave that part of the graph that is above the 0x axis or on it unchanged.

3) The part of the graph located below the 0x axis will be displayed symmetrically relative to 0x.

4) The final graph is shown in the figure (Fig. 11).

website, when copying material in full or in part, a link to the source is required.

Let's look at how to build a graph with a module.

Let us find the points at the transition of which the sign of the modules changes.
We equate each expression under the modulus to 0. We have two of them x-3 and x+3.
x-3=0 and x+3=0
x=3 and x=-3

Our number line will be divided into three intervals (-∞;-3)U(-3;3)U(3;+∞). At each interval, you need to determine the sign of the modular expressions.

1. This is very easy to do, consider the first interval (-∞;-3). Let’s take any value from this segment, for example, -4, and substitute the value of x into each of the modular equations.
x=-4
x-3=-4-3=-7 and x+3=-4+3=-1

Both expressions have negative signs, which means we put a minus before the modulus sign in the equation, and instead of the modulus sign we put parentheses and we get the required equation on the interval (-∞;-3).

y= — (x-3)-( — (x+3))=-x+3+x+3=6

On the interval (-∞;-3) the graph was obtained linear function(direct) y=6

2. Consider the second interval (-3;3). Let's find what the graph equation will look like on this segment. Let's take any number from -3 to 3, for example, 0. Substitute 0 for the value x.
x=0
x-3=0-3=-3 and x+3=0+3=3

The first expression x-3 has a negative sign, and the second expression x+3 has a positive sign. Therefore, before the expression x-3 we write a minus sign, and before the second expression a plus sign.

y= — (x-3)-( + (x+3))=-x+3-x-3=-2x

On the interval (-3;3) we obtained a graph of a linear function (straight line) y=-2x

3. Consider the third interval (3;+∞). Let’s take any value from this segment, for example 5, and substitute the value x into each of the modular equations.

x=5
x-3=5-3=2 and x+3=5+3=8

For both expressions, the signs turned out to be positive, which means we put a plus in front of the modulus sign in the equation, and instead of the modulus sign we put parentheses and we get the required equation on the interval (3;+∞).

y= + (x-3)-( + (x+3))=x-3-x-3=-6

On the interval (3;+∞) we obtained a graph of a linear function (straight line) у=-6

4. Now let's summarize. Let's plot the graph y=|x-3|-|x+3|.
On the interval (-∞;-3) we build a graph of the linear function (straight line) y=6.
On the interval (-3;3) we build a graph of the linear function (straight line) y=-2x.
To construct a graph of y = -2x, we select several points.
x=-3 y=-2*(-3)=6 the result is a point (-3;6)
x=0 y=-2*0=0 the result is a point (0;0)
x=3 y=-2*(3)=-6 the result is point (3;-6)
On the interval (3;+∞) we build a graph of the linear function (straight line) у=-6.

5. Now let’s analyze the result and answer the question, find the value of k at which the straight line y=kx has with the graph y=|x-3|-|x+3| a given function has exactly one common point.

The straight line y=kx for any value of k will always pass through the point (0;0). Therefore, we can only change the slope of this line y=kx, and the coefficient k is responsible for the slope.

If k is any positive number, then there will be one intersection of the straight line y=kx with the graph y=|x-3|-|x+3|. This option suits us.

If k takes the value (-2;0), then the intersection of the straight line y=kx with the graph y=|x-3|-|x+3| there will be three. This option does not suit us.

If k=-2, there will be many solutions [-2;2], because the straight line y=kx will coincide with the graph y=|x-3|-|x+3| in this area. This option does not suit us.