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How do you find the range of #y = t^2 + t - 2#; given Domain = {-2, -1, 0, 1}?

Answer 1

Eva Abramson

The y coordinates of the points inside the domain provide the accompanying range.

Replacing:

#(-2)^2 - 2 - 2 = 0#

#(-1)^2 - 1 - 2 = -2#

#0^2 - 0 - 2 = -2#

#(1)^2 - 1 - 2 = -2#

Hence, the range is #{0, -2, -2, -2}#, or #{y| 0, -2, -2, -2}#.

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Answer 2

Jameson Allen

Substitute each value from the domain into the equation and determine the corresponding y-values.

[ (-2): \quad y = (-2)^2 + (-2) - 2 ]

[ (-1): \quad y = (-1)^2 + (-1) - 2 ]

[ (0): \quad y = (0)^2 + (0) - 2 ]

[ (1): \quad y = (1)^2 + (1) - 2 ]

The resulting set of y-values represents the range.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.