# Pset3Answers - Ec 115 - Problem Set 3 solutions. 1....

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Ec 115 - Problem Set 3 solutions. 1. Consider the function y = x 2 4 x + 5 : For what values of x is y = 0; i.e. solve for x where x 2 4 x + 5 = 0 : Draw the function y = x 2 4 x + 5 on a graph. Answer: rewrite this quadratic equation as x 2 + 4 x 5 = 0 and it is clear this factorises as ( x + 5)( x 1) = 0 : There are thus two solutions x = 1 ; 5 : Alternatively use the rule with a = 1 ; b = 4 ; c = 5 to get x = 4 ± p 16 + 20 2 = 4 ± 6 2 = 4 + 6 2 or 4 6 2 = 2 2 or - 10 2 = 1 or 5 : GTAs to draw graph with maximum at x = b= 2 a = 2 (where y = 4+8+5 = 9) : 2. Repeat question 1 but for the function y = 2 x 2 + 6 x + 5 : Answer: with the rule. As a = 2 ; b = 6 ; c = 5 we get: x = 6 ± p 36 40 4 = 6 ± p 4 4 which does not exist. As there are no solutions, this function does not cross the x-axis (it is always strictly positive). GTAs to draw with intercept y = 5 at x = 0 and function is

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## This note was uploaded on 12/31/2011 for the course ECON MR 102 taught by Professor Huyduong during the Winter '11 term at RMIT Vietnam.

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Pset3Answers - Ec 115 - Problem Set 3 solutions. 1....

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