Stitz-Zeager_College_Algebra_e-book

# The result is a fraction whose numerator is positive

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Unformatted text preview: cancellation possible, and we conclude that the lines x = − 3 and x = 3 are vertical asymptotes to the graph of y = f (x). The calculator veriﬁes this claim. − x+2 x 2. Solving x2 − 9 = 0 gives x = ±3. In lowest terms g (x) = x x−−−6 = (x−3)(x+2) = x+3 . Since 29 (x 3)(x+3) x = −3 continues to make trouble in the denominator, we know the line x = −3 is a vertical asymptote of the graph of y = g (x). Since x = 3 no longer produces a 0 in the denominator, we have a hole at x = 3. To ﬁnd the y -coordinate of the hole, we substitute x = 3 into x+2 x+3 2 4.1 Introduction to Rational Functions 237 5 and ﬁnd the hole is at 3, 6 . When we graph y = g (x) using a calculator, we clearly see the vertical asymptote at x = −3, but everything seems calm near x = 3. The graph of y = f (x) The graph of y = g (x) 3. The domain of h is all real numbers, since x2 + 9 = 0 has no real solutions. Accordingly, the graph of y = h(x) is devoid of both vertical asymptotes and holes. 4. Setting x2 + 4x + 4 = 0 gives us x = −2 as the only real number of concern. Simplifying, x2 − we see r(x) = x2 +4x−6 = (x−3)(x+2) = x−3...
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