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Stitz-Zeager_College_Algebra_e-book

# The base is a square of dimensions x by x and each

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Unformatted text preview: s, end behavior is determined by the leading term, so in the denominator, the x2 term wins out over the x term. 254 Rational Functions y 8 7 6 5 (+) (−) 0 (+) 0 (−) −2 −1 5 2 4 (+) 3 3 1 −9 −8 −7 −6 −5 −4 −3 −1 −1 1 2 4 5 6 7 8 9 x −2 −3 −4 Our next example gives us not only a hole in the graph, but also some slightly diﬀerent end behavior. Example 4.2.3. Sketch a detailed graph of h(x) = 2x3 + 5 x2 + 4 x + 1 . x2 + 3 x + 2 Solution. 1. For domain, you know the drill. Solving x2 + 3x + 2 = 0 gives x = −2 and x = −1. Our answer is (−∞, −2) ∪ (−2, −1) ∪ (−1, ∞). 2. To reduce h(x), we need to factor the numerator and denominator. To factor the numerator, we use the techniques12 set forth in Section 3.3 and we get 1 ! ¡ (2x + 1)(x + 1) 2x3 + 5x2 + 4x + 1 (2x + 1)(x + 1)2 (2x + 1)(x + 1)2 h(x) = = = \$ \$= 2 + 3x + 2 (x + x (x + 2)(x + 1) (x + 2)\$\$ 1) x+2 We will use this reduced formula for h(x) as long as we’re not substituting x = −1....
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