Question

part a
identify the rule for $\frac{f}{g}$ when $f(x) = -3x - 6$ and $g(x) = x^2 - x - 6$.
a. $\frac{f(x)}{g(x)} = \frac{-2}{(x + 2)}$
b. $\frac{f(x)}{g(x)} = \frac{-3(x - 2)}{(x - 6)(x + 1)}$
c. $\frac{f(x)}{g(x)} = \frac{3}{(x + 2)}$
d. $\frac{f(x)}{g(x)} = \frac{-3}{x - 3}$
part b

Explanation:

Step1: Factor numerator $f(x)$

$f(x) = -3x - 6 = -3(x + 2)$

Step2: Factor denominator $g(x)$

$g(x) = x^2 - x - 6 = (x - 3)(x + 2)$

Step3: Form and simplify $\frac{f(x)}{g(x)}$

Cancel common factor $(x+2)$:
$\frac{f(x)}{g(x)} = \frac{-3(x + 2)}{(x - 3)(x + 2)} = \frac{-3}{x - 3}$

Answer:

D. $\frac{f(x)}{g(x)} = \frac{-3}{x - 3}$