Section 3: Interpreting the Graph of a Function

# Elementary and Intermediate Algebra: Graphs & Models (3rd Edition)

This preview shows pages 1–3. Sign up to view the full content.

Section 2.3 Interpreting the Graph of a Function 115 Version: Fall 2007 2.3 Interpreting the Graph of a Function In the previous section, we began with a function and then drew the graph of the given function. In this section, we will start with the graph of a function, then make a number of interpretations based on the given graph: function evaluations, the domain and range of the function, and solving equations and inequalities. The Vertical Line Test Consider the graph of the relation R shown in Figure 1 (a). Recall that we earlier defined a relation as a set of ordered pairs. Surely, the graph shown in Figure 1 (a) is a set of ordered pairs. Indeed, it is an infinite set of ordered pairs, so many that the graph is a solid curve. In Figure 1 (b), note that we can draw a vertical line that cuts the graph more than once. In Figure 1 (b), we’ve drawn a vertical line that cuts the graph in two places, once at ( x, y 1 ) , then again at ( x, y 2 ) , as shown in Figure 1 (c). This means that the domain object x is paired with two different range objects, namely y 1 and y 2 , so relation R is not a function. x y R x y R x y R y 2 y 1 ( x,y 2 ) ( x,y 1 ) (a) (b) (c) Figure 1. Explaining the vertical line test for functions. Recall the definition of a function. Definition 1. A relation is a function if and only if each object in its domain is paired with one and only one object in its range. Consider the mapping diagram in Figure 2 , where we’ve used arrows to indicate the ordered pairs ( x, y 1 ) and ( x, y 2 ) in Figure 1 (c). Note that x , an object in the domain of R , is mapped to two objects in the range of R , namely y 1 and y 2 . Hence, the relation R is not a function. Copyrighted material. See: 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
116 Chapter 2 Functions Version: Fall 2007 x y 1 y 2 R Figure 2. A mapping diagram repre- senting the points ( x, y 1 ) and ( x, y 2 ) in Figure 1 (c). This discussion leads to the following result, called the vertical line test for functions. The Vertical Line Test . If any vertical line cuts the graph of a relation more than once, then the relation is NOT a function. Hence, the circle pictured in Figure 3 (a) is a relation, but it is not the graph of a function. It is possible to cut the graph of the circle more than once with a vertical line, as shown in Figure 3 (a). On the other hand, the parabola shown in Figure 3 (b) is the graph of a function, because no vertical line will cut the graph more than once. x y x y (a) (b) Figure 3. Use the vertical line test to determine if the graph is the graph of a function. Reading the Graph for Function Values We know that the graph of f pictured in Figure 4 is the graph of a function. We know this because no vertical line will cut the graph of f more than once. We earlier defined the graph of f as the set of all ordered pairs ( x, f ( x )) , so that x is in the domain of f . Consequently, if we select a point P on the graph of f , as in Figure 4 (a), we label the point P ( x, f ( x )) . However, we can also label this point as P ( x, y ) , as shown in Figure 4 (b). This leads to a new interpretation of f ( x ) as the y -value of the point P . That is, f ( x ) is the
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern