05_3067_Key - denominator equal zero. The horizontal...

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MAC 1147 (3067) - Quiz #5 Name: Key For full credit, you must show all work and circle your final answer. 1 Use long division OR synthetic division to divide. (1.5 points) (2 x 3 + 14 x 2 - 20 x + 7) ÷ ( x + 6) 2 x 2 + 2 x - 32 x + 6 ) 2 x 3 + 14 x 2 - 20 x + 7 - 2 x 3 - 12 x 2 2 x 2 - 20 x - 2 x 2 - 12 x - 32 x + 7 32 x + 192 199 2 14 - 20 7 - 6 - 12 - 12 192 2 ±* · ( - 6) 2 ? + ± ±* · ( - 6) - 32 ? + ± ± * · ( - 6) 199 ? + 2 Write the quotient in standard form. (1.5 points) ( Hint: The standard form of a complex number is a + bi , where a and b are real numbers.) 5 + i 5 - i 5 + i 5 - i = 5 + i 5 - i ± 5 + i 5 + i ² = 25 + 10 i + i 2 25 - i 2 = 24 + 10 i 26 = 12 13 + 5 13 i. 3 Find the domain of the function and identify any vertical or horizontal asymptotes. (2 points) f ( x ) = x 2 - 4 x 2 - 3 x + 2 f ( x ) = x 2 - 4 x 2 - 3 x + 2 = ( x + 2)( x - 2) ( x - 2)( x - 1) . So, the domain is all real numbers except 2 and 1, since those are the two numbers which make the
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Unformatted text preview: denominator equal zero. The horizontal asymptote is at y = 1 because the highest power of x in both the numerator and the demoninator is x 2 , and each has coecient 1, so the horizontal asymptote is 1 / 1 = 1. The vertical asymptote is at x = 1. Note that there is not a vertical asymptote at x = 2 because the x-2 terms cancel, so there is just a hole in the function at x = 2. University of Florida Honor Code: On my honor, I have neither given nor received unauthorized aid in doing this assignment. Signature...
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This note was uploaded on 12/27/2011 for the course MAC 1147 taught by Professor German during the Summer '08 term at University of Florida.

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