1.

In the graph of theparametric equation {:{(x=t^(2)+t),(y=t^(2)-t):}

Answer»

`x ge 0`
`x ge -1/4`
x is any real number
`x ge -1`

Solution :Graph these arametricequation for values of t between -5 and 5 and for x and y between -2.5 and 2.5

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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