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In the graph of theparametric equation {:{(x=t^(2)+t),(y=t^(2)-t):} |
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Answer» `x ge 0` apparently the x values are always greater than some value. Usethe TRACE function to MOVE the cursoras FAR left on the gaph as it will go this leads to (correct) guessof `x ge-1/4` this can be verified by completing the squre on the x equation `x=(t^(2)+t+1/4)-1/4=(t+1/2)^(2)-1/4` This represent a PARABOLA that opens up with vertex at `(-1/2,-1/4)` Therefore `x ge -1/4` |
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