Answer
Evaluate $g(t) = \frac{t+2}{t-2}$:
$g(-2) =0$
$g(2) = \frac{4}{0} = undefined$
$g(0) = -1$
$g(a) = \frac{a+2}{a-2}$
$g(a^2-2) = \frac{a^2}{a^2-4}$
$g(a+1) =\frac{a+3}{a-1}$
Work Step by Step
For t = -2
$g(-2) = \frac{(-2)+2}{(-2)-2}$... add top and subtract bottom
$=\frac{0}{-4} = 0$
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For t = 2
$g(2) = \frac{(2)+2}{(2)-2}$... add top and subtract bottom
$=\frac{4}{0} = undefined$
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For t = 0
$g(0) = \frac{(0)+2}{(0)-2}$... add top and subtract bottom
$=\frac{2}{-2} = -1$
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For t = a
$g(a) = \frac{a+2}{a-2}$
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For t = $a^2 - 2$
$g(a^2 - 2) = \frac{(a^2-2)+2}{(a^2-2)-2}$... add top and subtract bottom
$=\frac{a^2}{a^2-4}$
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For t = a + 1
$g(a+1) = \frac{(a+1)+2}{(a+1)-2}$... add top and subtract bottom
$=\frac{a+3}{a-1}$
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