Stitz-Zeager_College_Algebra_e-book

30 centimeters 431 variation in many instances in the

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Unformatted text preview: To make this exclusion specific, we write h(x) = (2x+1)(x+1) , x = −1. x+2 3. To find the x-intercepts, as usual, we set h(x) = 0 and solve. Solving (2x+1)(x+1) = 0 yields x+2 x = − 1 and x = −1. The latter isn’t in the domain of h, so we exclude it. Our only x2 1 intercept is − 2 , 0 . To find the y -intercept, we set x = 0. Since 0 = −1, we can use the reduced formula for h(x) and we get h(0) = 1 for a y -intercept of 0, 1 . 2 2 4. From Theorem 4.1, we know that since x = −2 still poses a threat in the denominator of the reduced function, we have a vertical asymptote there. As for x = −1, we note the factor (x + 1) was canceled from the denominator when we reduced h(x), and so it no longer causes trouble there. This means we get a hole when x = −1. To find the y -coordinate of the hole, we substitute x = −1 into (2x+1)(x+1) , per Theorem 4.1 and get 0. Hence, we have a hole on x+2 12 Bet you never thought you’d never see that stuff again before the Final Exam! 4.2 Graphs of Rational Functio...
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This note was uploaded on 05/03/2013 for the course MATH Algebra taught by Professor Wong during the Fall '13 term at Chicago Academy High School.

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