05_4258_Key - denominator equal zero. The horizontal...

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MAC 1147 (4258) - 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) (9 x 3 - 16 x - 18 x 2 + 32) ÷ ( x - 2) 9 x 2 - 16 x - 2 ) 9 x 3 - 18 x 2 - 16 x + 32 - 9 x 3 + 18 x 2 - 16 x + 32 16 x - 32 0 9 - 18 - 16 32 2 18 0 - 32 9 ±* · 2 0 ? + ± ±* · 2 - 16 ? + ± ±* · 2 0 ? + 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.) 6 - 7 i 1 - 2 i 6 - 7 i 1 - 2 i = 6 - 7 i 1 - 2 i ± 1 + 2 i 1 + 2 i ² = 6 + 5 i - 14 i 2 1 - 4 i 2 = 20 + 5 i 5 = 4 + i. 3 Find the domain of the function and identify any vertical or horizontal asymptotes. (2 points) f ( x ) = x 2 - 25 x 2 - 4 x - 5 f ( x ) = x 2 - 25 x 2 - 4 x - 5 = ( x + 5)( x - 5) ( x - 5)( x + 1) . So, the domain is all real numbers except 5 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 = 5 because the x-5 terms cancel, so there is just a hole in the function at x = 5. 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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