Corresponding Points On A Graph

Graphs of Functions

Contents: This folio corresponds to § 1.4 (p. 116) of the text.

Defining the Graph of a Function

The graph of a function f is the set of all points in the airplane of the form (x, f(x)). We could also define the graph of f to be the graph of the equation y = f(x). So, the graph of a function if a special case of the graph of an equation.

Example 1.

Let f(x) = xⁱⁱ - three.

Recall that when we introduced graphs of equations nosotros noted that if we can solve the equation for y, then it is easy to find points that are on the graph. We just choose a number for 10, and then compute the corresponding value of y. Graphs of functions are graphs of equations that have been solved for y!

The graph of f(x) in this example is the graph of y = x² - 3. It is easy to generate points on the graph. Choose a value for the outset coordinate, then evaluate f at that number to find the 2nd coordinate. The following table shows several values for x and the function f evaluated at those numbers.

ten

-2

-1

0

1

2

f(10)

1

-two

-3

-2

one

Each cavalcade of numbers in the table holds the coordinates of a point on the graph of f.

Practise 1:

(a) Plot the five points on the graph of f from the tabular array above, and based on these points, sketch the graph of f.

(b) Verify that your sketch is correct by using the Coffee Grapher to graph f. Just enter the formula x^2 - 3 in the f text box and click graph.

Example two.

Let f be the piecewise-divers function

To express f in a single formula for the Java Grapher or Java Figurer nosotros write

(5 - x^ii)*(xLE2) + (ten - 1)*(2Lx).

The gene (xLE2) has the value 1 for x <= 2 and 0 for x > two. Similarly, (2Lx) is i for 2 < x and 0 otherwise. If we evaluate the sum in a higher place at x = 3, the start product is 0 because (xLE2) is 0 and the second product is (3 - ane)*ane=2. In other words, for x > two, the formula evaluates to 10 - one. If x <= ii, and so the formula above is equal to five - 10^2, which is exactly what we want!

The graph of f is shown below.

Exercise ii:

Graph the piecewise-defined part

Answer

We have seen that some equations in x and y do not describe y every bit a function of 10. The algebraic style see if an equation determines y as a function of x is to solve for y. If at that place is non a unique solution, then y is not a function of x.

Suppose that we are given the graph of the equation. In that location is an easy fashion to see if this equation describes y as a function of x.

Return to Contents

Vertical Line Test

A set of points in the airplane is the graph of a role if and but if no vertical line intersects the graph in more than one point.

Example iii.

The graph of the equation y² = ten + 5 is shown beneath.

By the vertical line test, this graph is not the graph of a role, because there are many vertical lines that hit information technology more than once.

Remember of the vertical line test this fashion. The points on the graph of a function f accept the form (x, f(ten)), so once you know the first coordinate, the second is adamant. Therefore, in that location cannot exist two points on the graph of a office with the same beginning coordinate.

All the points on a vertical line have the same first coordinate, so if a vertical line hits a graph twice, then there are two points on the graph with the aforementioned first coordinate. If that happens, the graph is non the graph of a role. Videos: Animated Gif, MS Avi File, or Existent Video File

Render to Contents

Characteristics of Graphs

Consider the function f(10) = 2 x + ane. We recognize the equation y = two 10 + 1 as the Slope-Intercept form of the equation of a line with gradient 2 and y-intercept (0,1).

Think of a point moving on the graph of f. Every bit the signal moves toward the right it rises. This is what information technology means for a part to be increasing. Your text has a more precise definition, but this is the bones idea.

The function f above is increasing everywhere. In general, there are intervals where a function is increasing and intervals where it is decreasing.

The function graphed above is decreasing for x betwixt -3 and ii. It is increasing for x less than -3 and for ten greater than ii.

Videos: 10<-three Animated Gif, MS Avi File, or Existent Video File
Videos: -three<x<2 Blithe Gif, MS Avi File, or Real Video File
Videos: 2<x Animated Gif, MS Avi File, or Real Video File

Using interval note, we say that the role is

decreasing on the interval (-three, two)

increasing on (-infinity, -3) and (2, infinity)

Exercise 3:

Graph the part f(10) = x² - 6x + seven and discover the intervals where it is increasing and where information technology is decreasing. Respond

Some of the almost characteristics of a part are its Relative Extreme Values. Points on the functions graph respective to relative farthermost values are turning points, or points where the function changes from decreasing to increasing or vice versa. Let f exist the role whose graph is drawn below.

f is decreasing on (-infinity, a) and increasing on (a, b), so the point (a, f(a)) is a turning indicate of the graph. f(a) is chosen a relative minimum of f. Note that f(a) is not the smallest function value, f(c) is. Withal, if we consider simply the portion of the graph in the circle above a, then f(a) is the smallest second coordinate. Wait at the circle on the graph in a higher place b. While f(b) is not the largest part value (this function does not take a largest value), if nosotros look but at the portion of the graph in the circumvolve, then the point (b, f(b)) is in a higher place all the other points. So, f(b) is a relative maximum of f. f(c) is another relative minimum of f. Indeed, f(c) is the accented minimum of f, merely it is too i of the relative minima.

Here again we are giving definitions that appeal to your geometric intuition. The precise definitions are given in your text.

Render to Contents

Approximating Relative Extrema

Finding the exact location of a role's relative extrema generally requires calculus. However, graphing utilities such every bit the Coffee Grapher may exist used to approximate these numbers.

Annotation on terminology:

Suppose a is a number such that f(a) is a relative minimum. In applications, it is oft more of import to know where the function attains its relative minimum than it is to know what the relative minimum is.

For example, f(x) = 10³ - 4x² + 4x has a relative minimum of 0. It attains this relative minimum at ten = ii, so (ii,0) is a turning signal of the graph of f. We will call the point (2,0) a relative minimum point. In full general, a relative farthermost point is a betoken on the graph of f whose second coordinate is a relative extreme value of f.

Instance 4.

Approximate the relative extremes points of f(10) = xⁱⁱⁱ - x² - 6x.

When yous display the graph of f in the default viewing rectangle y'all run across that f has one relative maximum signal near (-one,four) and i relative minimum indicate near (two,-8). The approximations (-ane,four) and (two,-8) are not very close to the existent relative farthermost points, and so we will employ the zoom and trace features to improve the approximations.

When you click the Trace push, a bespeak on the graph of f is indicated with a minor circumvolve. The coordinates of that bespeak are reported in the 2 text boxes most the Trace push button. As y'all click the box with the right arrow >, the trace point moves to the right, staying on the graph of f. If y'all select a larger Step Size from the pull down menu, then the trace point moves further with each click. Also find that one time you take clicked the > push, and so y'all can use the enter key to move further right. It is possible to motion faster with the enter fundamental than with the mouse.

Using the default view, the lowest point found while tracing well-nigh the minimum point is (1.8, -8.208). Note that this is not the exact location of the minimum indicate. We demand to look at the trace points on either side of this point to get an idea of how shut we are. Find this trace point, make sure that the Step Size is set to 1, so find the points on either side of this bespeak. The table below lists the coordinates of these points.

x

y

1.7333333 -8.196741

1.eight -8.208

1.8666667 -8.180148

Note that unlike Java implementations will compute and study unlike accuracies, so the values you find may may be slightly different from those above.

The points in the table show that the real minimum point has an x coordinate somewhere between ane.7333333 and 1.8666667. Note that nosotros do not all the same accept plenty information to report the 10 value with even one decimal identify accurateness, considering if the second decimal place were a iv, then the value would round to 1.7. If the 2nd decimal place were a 7, and so the value would circular to 1.8. So we demand to improve our approximate past zooming in on the minimum point. There are several ways to practise this with the Grapher: Zoom In, Zoom Box, or set the coordinates of the viewing rectangle. In this example we will fix the view coordinates.

Type in Xminorth = i.74, Xmax = ane.86, Ymin = -8.21, Ymax = -8.19, and set the viewing rectangle to these coordinates by clicking the Reset button. Irresolute the viewing rectangle removes the trace indicate. To get it dorsum, click the Trace button again. Now you need to move left to get to the minimum point. There are two unlike x values that correspond to the lowest y value.

x

y

1.786 -8.20882

1.7864 -8.20882

We still do non know the exact location of the minimum point, but nosotros know that its x coordinate is between 1.786 and 1.7864. That means the ten coordinate will be i.786 after rounding to three decimal places, . Since nosotros are only reporting iii decimal places for the x coordinate, nosotros will too round the y coordinate to 3 decimal places, so our approximation is (ane.786, -8.209).

Practice 4:

Find the relative maximum point to two decimal place accurateness. Answer

Render to Contents

Even and Odd Functions

A part f is even if its graph is symmetric with respect to the y-axis. This criterion can be stated algebraically as follows: f is even if f(-x) = f(x) for all x in the domain of f. For example, if you evaluate f at 3 and at -three, then you will become the same value if f is even.

This status is very easy to check with the Java Grapher.

Example 5.

Open up the Grapher and type (x - ii)^ii into the f text box and click the Graph button. This function is not even, so when we graph its reflection about the y-axis, we will go a new graph.

In the g text box blazon f(-x), and click the Graph button The graph of chiliad is the reflection about the y-axis of the graph of f. Since nosotros run across two distinct graphs, nosotros know that f is non even.

Now supplant the text in the f box with x^two - 3; clear the text from the g box and graph the function. This graph is symmetric with respect to the y-axis, and then when you enter f(-ten) in the g box and graph over again, you do not see annihilation new. This is because the graph of thousand is the same as the graph of f.

A function f is odd if its graph is symmetric with respect to the origin. This criterion can exist stated algebraically as follows: f is odd if f(-x) = -f(x) for all x in the domain of f. For example, if you evaluate f at 3, you get the negative of f(-3) when f is odd.

If you enter any office in the f box of the Grapher and enter -f(-x) in the g box, then the graph of 1000 is the reflection through the origin of the graph of f. And then, if f is non odd, then you see ii singled-out graphs. If f is odd, you come across only one graph.

Example 6.

Graph f(x) = (x - 2)^ii and 1000(x) = -f(-10). Considering you see 2 distinct graphs, f is non odd.

At present enter f(x) = x*(ten^2 - ane) and "turn off" the graph of g by unchecking the box to the right of the text box and click Graph once more. With this role f, the graph of m is the same as the graph of f. So, when you turn on the graph of 1000 and click Graph over again, you see cipher new.

Render to Contents