Stitz-Zeager_College_Algebra_e-book

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

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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...
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