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Unformatted text preview: x 29 3. f ( x ) = x 2 + 3 x8 x3 4. f ( x ) = x 2 + 3 x + 8 x 29 5. f ( x ) = x 2 + 3 x8 x 2 + 9 6. f ( x ) = x 23 x + 8 9x 2 Explanation: The graph has vertical asymptotes at x = 3, so a term x 29 must appear in the denominator of f ( x ); in particular, therefore, f ( x ) 6 = x 2 + 3 x8 x 2 + 9 and f ( x ) 6 = x 2 + 3 x8 x3 , eliminating two of the possible answers. The graph also has a horizontal asymptote y = 1, so the ratio of coecients of degree 2 terms in the numerator and denominator of f ( x ) must be 1. Thus f ( x ) 6 = x 23 x + 8 9x 2 ,...
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This note was uploaded on 02/14/2012 for the course MATH 408 K taught by Professor Clark,c.w./hoy,r.r during the Spring '08 term at University of Texas at Austin.
 Spring '08
 CLARK,C.W./HOY,R.R

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