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Math SAT question

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As far as I can tell, you can't rule 1 or 2 out because of the negative x, because the 'squared' would still leave y as a positive number. You can, however, rule them out because the lowest possible value for y would be 1 (because of the + 1), and you can see from the chart that twice, the value of y reaches 0. The same would hold true for answer #3. I believe the correct answer is #4, because at x = 0 then y = 0^2  1 which means y = 1 leaving y = 1. That is correct for the graph. But as you move away from zero, say, x = .5, you get: y = .5^2  1 which is y = .25  1 or y = .75 or y = .75 So as x starts to move away from zero, y initially continues to drop. At x = 1, you're left with y = 0. Which matches the graph. Then as you continue past x = 1 ( x = 2, 3, etc), the value of y starts to raise again. Last edited by Fazor; August 07, 2009 at 10:23 AM. 
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#5




Thanks guys! ^_^

#6




Since I more or less just worked through the problem in my own, rambling style (my personal approach); let me add one tip:
With these kinds of problems, it often helps to start by solving for x = 0 and then y = 0. 
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