+296 x(t)=t−3, y(t)=2t^2 −1, where t is on the interval [−2,2].
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y = ___,where x is on the interval [___,___].
+129852 x = t - 3 → t = x + 3 (1)
y = 2t^2 - 1 (2)
Sub (1) into (2) for t
y = 2 ( x + 3)^2 - 1 simplify
y =2 (x^2 + 6x + 9) - 1
y = 2x^2 + 12x + 18 -1
y = 2x^2 + 12x + 17
Note that in t is on [-2, 2]
Then x is on t -3 = -2 - 3 = -5 and 2 - 3 = -1
So x is on the interval [ -5, -1 ]
