## Precalculus: Mathematics for Calculus, 7th Edition

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