Question

simplify and determine the domain of the function:
$f(x) = \frac{x^2 + 5x + 6}{x^2 - 3x - 10}$
a $\frac{x + 3}{x - 5}, x \
eq 5$
b $\frac{x + 3}{x - 5}, x \
eq -2, 5$
c $\frac{x^2 + 5x + 6}{x - 5}, x \
eq 5$
d $\frac{x^2 + 5x + 6}{x - 5}, x \
eq -2, 5$
e (partially visible) $\frac{x + 3}{x - 5}, x$ (rest unclear)

Explanation:

Step1: Factor numerator and denominator

Factor \(x^2 + 5x + 6\): \(x^2 + 5x + 6=(x + 2)(x + 3)\)
Factor \(x^2-3x - 10\): \(x^2-3x - 10=(x - 5)(x + 2)\)
So \(f(x)=\frac{(x + 2)(x + 3)}{(x - 5)(x + 2)}\)

Step2: Simplify the function

Cancel out the common factor \((x + 2)\) (note \(x
eq - 2\) to avoid division by zero), we get \(f(x)=\frac{x + 3}{x - 5}\)

Step3: Determine the domain

The original function's denominator \(x^2-3x - 10=(x - 5)(x + 2)\) cannot be zero. So \(x-5
eq0\) and \(x + 2
eq0\), which means \(x
eq5\) and \(x
eq - 2\)

Answer:

B. \(\frac{x + 3}{x - 5}, x
eq - 2, 5\)