Stitz-Zeager_College_Algebra_e-book

X 50 0 50 0 interpreting this slope as we have in

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: domain of f is (−∞, 2) ∪ (2, ∞). To indicate this, we write f (x) = x + 2, x = 2. So, except at x = 2, we graph the line y = x + 2. The slope m = 1 and the y -intercept is (0, 2). A second point on the graph is (1, f (1)) = (1, 3). Since our function f is not defined at x = 2, we put an open circle at the point that would be on the line y = x + 2 when x = 2, namely (2, 4). y 4 y 3 2 2 1 1 −3 −2 −1 1 2 3 x −1 3 − 2x f (x) = 4 1 f (x) = 2 3 x x2 −4 x−2 The last two functions in the previous example showcase some of the difficulty in defining a linear function using the phrase ‘of the form’ as in Definition 2.1, since some algebraic manipulations may be needed to rewrite a given function to match ‘the form.’ Keep in mind that the domains of 2− linear and constant functions are all real numbers, (−∞, ∞), and so while f (x) = x −24 simplified x to a formula f (x) = x + 2, f is not considered a linear function since its domain excludes x = 2. However, we would consider 2x2 + 2...
View Full Document

Ask a homework question - tutors are online