Y 2 3x 4 graph the function. Building charts online. Basic properties of a quadratic function

A function graph is a visual representation of the behavior of some function on coordinate plane. Plots help to understand various aspects of a function that cannot be determined from the function itself. You can build graphs of many functions, and each of them will be given by a specific formula. The graph of any function is built according to a certain algorithm (if you forgot the exact process of plotting a graph of a particular function).

Steps

Plotting a Linear Function

Determine if the function is linear. A linear function is given by a formula of the form F (x) = k x + b (\displaystyle F(x)=kx+b) or y = k x + b (\displaystyle y=kx+b)(for example, ), and its graph is a straight line. Thus, the formula includes one variable and one constant (constant) without any exponents, root signs, and the like. Given a function of a similar form, plotting such a function is quite simple. Here are other examples of linear functions:

Use a constant to mark a point on the y-axis. The constant (b) is the “y” coordinate of the intersection point of the graph with the Y axis. That is, it is a point whose “x” coordinate is 0. Thus, if x = 0 is substituted into the formula, then y = b (constant). In our example y = 2x + 5 (\displaystyle y=2x+5) the constant is 5, that is, the point of intersection with the Y-axis has coordinates (0,5). Plot this point on the coordinate plane.

Find the slope of the line. It is equal to the multiplier of the variable. In our example y = 2x + 5 (\displaystyle y=2x+5) with the variable "x" is a factor of 2; thus, the slope is 2. The slope determines the angle of inclination of the straight line to the X-axis, that is, the larger the slope, the faster the function increases or decreases.

Write the slope as a fraction. The slope is equal to the tangent of the angle of inclination, that is, the ratio of the vertical distance (between two points on a straight line) to the horizontal distance (between the same points). In our example, the slope is 2, so we can say that the vertical distance is 2 and the horizontal distance is 1. Write this as a fraction: 2 1 (\displaystyle (\frac (2)(1))).

• If the slope is negative, the function is decreasing.

1. From the point where the line intersects with the Y axis, draw a second point using the vertical and horizontal distances. Schedule linear function can be built from two points. In our example, the point of intersection with the Y-axis has coordinates (0.5); from this point move 2 spaces up and then 1 space to the right. Mark a point; it will have coordinates (1,7). Now you can draw a straight line.

   Use a ruler to draw a straight line through two points. To avoid mistakes, find the third point, but in most cases the graph can be built using two points. Thus, you have plotted a linear function.

   Drawing points on the coordinate plane

   1. Define a function. The function is denoted as f(x). All possible values ​​of the variable "y" are called the range of the function, and all possible values ​​of the variable "x" are called the domain of the function. For example, consider the function y = x+2, namely f(x) = x+2.

      Draw two intersecting perpendicular lines. The horizontal line is the X-axis. The vertical line is the Y-axis.

      Label the coordinate axes. Break each axis into equal segments and number them. The intersection point of the axes is 0. For the X axis: positive numbers are plotted on the right (from 0), and negative numbers on the left. For the Y-axis: positive numbers are plotted on top (from 0), and negative numbers on the bottom.

      Find the "y" values ​​from the "x" values. In our example f(x) = x+2. Substitute certain "x" values ​​into this formula to calculate the corresponding "y" values. If given a complex function, simplify it by isolating the "y" on one side of the equation.

      • -1: -1 + 2 = 1
      • 0: 0 +2 = 2
      • 1: 1 + 2 = 3

   2. Draw points on the coordinate plane. For each pair of coordinates, do the following: find the corresponding value on the x-axis and draw a vertical line (dotted line); find the corresponding value on the y-axis and draw a horizontal line (dotted line). Mark the point of intersection of the two dotted lines; thus, you have plotted a graph point.

      Erase the dotted lines. Do this after plotting all the graph points on the coordinate plane. Note: the graph of the function f(x) = x is a straight line passing through the center of coordinates [point with coordinates (0,0)]; the graph f(x) = x + 2 is a line parallel to the line f(x) = x, but shifted up by two units and therefore passing through the point with coordinates (0,2) (because the constant is 2).

   Plotting a Complex Function

   Find the zeros of the function. The zeros of a function are the values ​​of the variable "x" at which y = 0, that is, these are the points of intersection of the graph with the x-axis. Keep in mind that not all functions have zeros, but this is the first step in the process of plotting a graph of any function. To find the zeros of a function, set it equal to zero. For example:

   Find and label the horizontal asymptotes. An asymptote is a line that the graph of a function approaches but never crosses (that is, the function is not defined in this area, for example, when divided by 0). Mark the asymptote with a dotted line. If the variable "x" is in the denominator of a fraction (for example, y = 1 4 − x 2 (\displaystyle y=(\frac (1)(4-x^(2))))), set the denominator to zero and find "x". In the obtained values ​​of the variable "x", the function is not defined (in our example, draw dashed lines through x = 2 and x = -2), because you cannot divide by 0. But asymptotes exist not only in cases where the function contains fractional expression. Therefore, it is recommended to use common sense:

How to build a parabola? There are several ways to graph a quadratic function. Each of them has its pros and cons. Let's consider two ways.

Let's start by plotting a quadratic function like y=x²+bx+c and y= -x²+bx+c.

Example.

Plot the function y=x²+2x-3.

Solution:

y=x²+2x-3 is a quadratic function. The graph is a parabola with branches up. Parabola vertex coordinates

From the vertex (-1;-4) we build a graph of the parabola y=x² (as from the origin. Instead of (0;0) - the vertex (-1;-4). From (-1;-4) we go to the right by 1 unit and up by 1, then left by 1 and up by 1, then: 2 - right, 4 - up, 2 - left, 4 - up, 3 - right, 9 - up, 3 - left, 9 - up. these 7 points are not enough, then - 4 to the right, 16 - up, etc.).

The graph of the quadratic function y= -x²+bx+c is a parabola whose branches are directed downwards. To build a graph, we are looking for the coordinates of the vertex and from it we build a parabola y= -x².

Example.

Plot the function y= -x²+2x+8.

Solution:

y= -x²+2x+8 is a quadratic function. The graph is a parabola with branches down. Parabola vertex coordinates

From the top we build a parabola y = -x² (1 - right, 1 - down; 1 - left, 1 - down; 2 - right, 4 - down; 2 - left, 4 - down, etc.):

This method allows you to build a parabola quickly and does not cause difficulties if you know how to plot the functions y=x² and y= -x². Disadvantage: if the vertex coordinates are fractional numbers, plotting is not very convenient. If you want to know the exact values ​​of the points of intersection of the graph with the x-axis, you will have to additionally solve the equation x² + bx + c = 0 (or -x² + bx + c = 0), even if these points can be directly determined from the figure.

Another way to build a parabola is by points, that is, you can find several points on the graph and draw a parabola through them (taking into account the fact that the line x=xₒ is its axis of symmetry). Usually, for this, they take the top of the parabola, the intersection points of the graph with the coordinate axes, and 1-2 additional points.

Plot the function y=x²+5x+4.

Solution:

y=x²+5x+4 is a quadratic function. The graph is a parabola with branches up. Parabola vertex coordinates

that is, the top of the parabola is the point (-2.5; -2.25).

Are looking for . At the point of intersection with the Ox axis y=0: x²+5x+4=0. Roots quadratic equation x1=-1, x2=-4, that is, we got two points on the graph (-1; 0) and (-4; 0).

At the intersection point of the graph with the Oy axis x=0: y=0²+5∙0+4=4. Got a point (0; 4).

To refine the graph, you can find an additional point. Let's take x=1, then y=1²+5∙1+4=10, that is, one more point of the graph - (1; 10). We mark these points on the coordinate plane. Taking into account the symmetry of the parabola with respect to the straight line passing through its vertex, we mark two more points: (-5; 6) and (-6; 10) and draw a parabola through them:

Plot the function y= -x²-3x.

Solution:

y= -x²-3x is a quadratic function. The graph is a parabola with branches down. Parabola vertex coordinates

The top (-1.5; 2.25) is the first point of the parabola.

At the points of intersection of the graph with the x-axis y=0, that is, we solve the equation -x²-3x=0. Its roots are x=0 and x=-3, that is, (0; 0) and (-3; 0) are two more points on the graph. The point (o; 0) is also the point of intersection of the parabola with the y-axis.

At x=1 y=-1²-3∙1=-4, i.e. (1; -4) is an additional point for plotting.

Building a parabola from points is a more time-consuming method compared to the first one. If the parabola does not intersect the Ox axis, more additional points will be required.

Before continuing the construction of graphs of quadratic functions of the form y=ax²+bx+c, let's consider the construction of graphs of functions using geometric transformations. Graphs of functions of the form y=x²+c are also most convenient to build using one of these transformations - parallel translation.

Rubric: |

Let's figure out how to build a graph with the module.

Let us find the points at the transition of which the sign of the moduli changes.
Each expression that under the modulus is equated 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;+∞). On each interval, you need to determine the sign of submodule 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 in each under the modular equation instead of the value of x.
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 module sign in the equation, and instead of the module sign we put brackets and we get the desired equation on the interval (-∞; -3).

y= — (x-3)-( — (x+3))=-x+3+x+3=6

On the interval (-∞;-3) we get a graph of a linear function (straight line) y \u003d 6

2. Consider the second interval (-3;3). Let's find how the equation of the graph will look like on this segment. Let's take any number from -3 to 3, for example, 0. Let's substitute the value of x for the value of 0.
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, we write a minus sign before the expression x-3, and a plus sign before the second expression.

y= — (x-3)-( + (x+3))=-x+3-x-3=-2x

On the interval (-3; 3) we get a graph of a linear function (straight line) y \u003d -2x

3. Consider the third interval (3;+∞). We take any value from this segment, for example 5, and substitute in each under the modular equation instead of the value x.

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 brackets and we get the desired equation on the interval (3;+∞).

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

On the interval (3; + ∞), we get a graph of a linear function (straight line) y \u003d -6

4. Now let's summarize. Let's plot y=|x-3|-|x+3|.
On the interval (-∞;-3) we build a graph of a linear function (straight line) y \u003d 6.
On the interval (-3; 3) we build a graph of a linear function (straight line) y \u003d -2x.
To build a graph y \u003d -2x, we select several points.
x=-3 y=-2*(-3)=6 got a point (-3;6)
x=0 y=-2*0=0 got a point (0;0)
x=3 y=-2*(3)=-6 got point (3;-6)
On the interval (3; + ∞) we build a graph of a linear function (straight line) y \u003d -6.

5. Now let's analyze the result and answer the question of the assignment, find the value of k for which the line y=kx has with the graph y=|x-3|-|x+3| this 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 straight 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 line y=kx with the graph y=|x-3|-|x+3|. This option suits us.

If k takes the value (-2;0), then the intersections of the 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 a set of solutions [-2;2], because the line y=kx will coincide with the graph y=|x-3|-|x+3| on this area. This option does not suit us.

If k is less than -2, then the line y=kx with the graph y=|x-3|-|x+3| will have one intersection. This option suits us.

If k=0, then the intersections of the line y=kx with the graph y=|x-3|-|x+3| there will also be one. This option suits us.

Answer: for k belonging to the interval (-∞;-2)U)