MATH133_U2_answer

MATH133_U2_answer - Student Answer Form Unit 2 March 25,...

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March 25, 2011 1. a. x^2- 6x -16 = 0 (x + 2) (x - 8 ) = 0 x + 2 = 0 x – 8 = 0 x = 8 x = -2, 8 b. 6x^2 + 3x -18 = 0 6x^3 = 18 6x^3 = 18 (6x^3)/(6)=(18)/(6) 6x^3/6=18/6 X^3= 18/6 X^3=3 X≈ 1.44 c. 2x^2 – 3x – 5 = 0 D= b^2 – 4ac D= (-3)^2-4 * 2* (-5) D= 9- (40) D= 9 + 40 D = 49 2. a. The solution I determined by using only the graph is x = (-5, 1). X^2 + 4x – 5 = 0 I obtained my answer by locating the point for the solutions by finding the point to where the line meets the x- axis. The first point is -5 because the line meets the x- axis on the -5 point which determines that one solution is -5. The second point is 1 because the line meets at the x- axis on the 1 point. b. The function has a maximum and a minimum. The maximum would be -10 and the minimum is 10. The trend line puts the max and min at the numbers listed above. I used a ruler to determine the max and min. c.
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MATH133_U2_answer - Student Answer Form Unit 2 March 25,...

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