1. Functions

a. …evaluate and simplify h(0)

h(x) =

-15-3x
______
12-6x

b. …what is the domain of h ? (Be sure to use set notation since domains are sets!)

2. Originally Posted by mjfaris
a. …evaluate and simplify h(0)

h(x) =

-15-3x
______
12-6x

b. …what is the domain of h ? (Be sure to use set notation since domains are sets!)

$\displaystyle h(0)=\frac{-15-3(0)}{12-6(0)}= \frac{-15}{12} = \frac{-5}{4}$

b) the deminator should not equal 0 so

$\displaystyle 12-6x = 0 \Rightarrow x=2$

so x all real numbers except 0 as an interval

$\displaystyle (-\infty, 0)\cup (0,\infty)$

3. Originally Posted by Amer
$\displaystyle h(0)=\frac{-15-3(0)}{12-6(0)}= \frac{-15}{12} = \frac{-5}{4}$

b) the deminator should not equal 0 so

$\displaystyle 12-6x = 0 \Rightarrow x=2$

so x all real numbers except 0 as an interval

$\displaystyle (-\infty, 0)\cup (0,\infty)$
You mean, surely,
"so x all real numbers except 2. As an interval

$\displaystyle (-\infty, 2)\cup (2, \infty)$