10Aquiz2

# 10Aquiz2 - f x is zero So the vertical asymptotes are x =-4...

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MATH 10A Quiz 2 Name (Section B0 ) Jan 21 2011 Problem 1 (4 points): Find the period , amplitude , and range of the function y = 7 sin (2 t ) + 1 . Solution: The amplitude is 7, and the period is 2 π | 2 | = π . This function oscillates about the line y = 1 with an amplitude of 7, so its range is [ - 6 , 8]. Problem 2 (6 points): Find the vertical and horizontal asymptotes of the function f ( x ) = - 2 x 3 + 18 x 2 - 22 x - 42 x 3 - 6 x 2 - 15 x + 100 = - 2( x + 1)( x - 3)( x - 7) ( x + 4)( x - 5) 2 . Solution: The vertical asymptotes occur where the denominator of
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Unformatted text preview: f ( x ) is zero. So the vertical asymptotes are x =-4 and x = 5 . To ﬁnd the horizontal asymptotes, determine the end behviour of f ( x ). As x → ∞ or x → -∞ , the highest powers of x dominate, and the other terms are insigniﬁcant. So f ( x ) =-2 x 3 + 18 x 2-22 x-42 x 3-6 x 2-15 x + 100 behaves like-2 x 3 x 3 =-2 . So the function has a horizontal asymptote y =-2 ....
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## This note was uploaded on 03/28/2011 for the course MATH 10A taught by Professor Arnold during the Winter '07 term at UCSD.

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