# For f(t)=abs[t]+1; how do you find f(-5),f(0) and f(-9)?

Jul 7, 2015

$f \left(- 5\right) = 6$
$f \left(0\right) = 1$
$f \left(- 9\right) = 10$

#### Explanation:

Given $f \left(t\right) = \left\mid t \right\mid + 1$

For specific values of $t$:
$f \left(- 5\right) = \left\mid - 5 \right\mid + 1 = 5 + 1 = 6$

$f \left(0\right) = \left\mid 0 \right\mid + 1 = 0 + 1 = 1$

$f \left(- 9\right) = \left\mid - 9 \right\mid + 1 = 9 + 1 = 10$