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math_1315_sum08_test_2_sol

# math_1315_sum08_test_2_sol - 1 2 3 4 5 6 7 8 Total MATH...

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1 2 3 4 5 6 7 8 Total MATH 1315 Exam 2 Instructor: C. Gunning Name: 1. 10 points Determine polynomial functions of the form: P ( x ) = a ( x - x 1 ) m 1 · ( x - x 2 ) m 2 · · · ( x - x k ) m k , to correspond to each of polynomials graphed below. (a) (i) x -intercept ( x - x k ) multiplicity m k -6 ( x + 6) odd (1) -3 ( x + 3) odd (1) +1 ( x - 1) odd (1) +4 ( x - 4) even (2) (ii) End Behavior (from graph): as x → ±∞ , P ( x ) + x n , where n is odd. (iii) Polynomial Function: P ( x ) = ( x + 6)( x + 3)( x - 1)( x - 4) 2 (b) (i) x -intercept ( x - x k ) multiplicity m k -2 ( x + 2) even (2) 0 x odd (1) +1 ( x - 1) even (2) +2 ( x - 2) odd (1) +4 ( x - 4) odd (1) (ii) End Behavior (from graph): as x → ±∞ , P ( x ) ≈ - x n , where n is odd. (iii) Polynomial Function: P ( x ) = - x ( x + 2) 2 ( x + 1) 2 ( x - 2)( x - 4) 1

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2. 10 points Graph the polynomial function: P ( x ) = - 2 x ( x + 3)( x - 2) 2 (1 - x ) , by hand. You must show complete work for credit. Label all intercepts, asymptotes, and other important features. First, let’s rearrange the polynomial to tra- ditional form: P ( x ) = - 2 x ( x + 3)( x - 2) 2 (1 - x ) P ( x ) = - 2 x ( x + 3)( x - 2) 2 ( - 1)( x - 1) P ( x ) = 2 x ( x + 3)( x - 2) 2 ( x - 1) P ( x ) = 2 x ( x + 3)( x - 2) 2 ( x - 1) This gives us the correct leading coefficient of a = 2, since a is positive, we know we have a positive coefficient on the end behavior.
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math_1315_sum08_test_2_sol - 1 2 3 4 5 6 7 8 Total MATH...

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