Question

what are the domain and range of the function $f(x)=\frac{x^2 - 4x - 12}{x + 2}$?\
d: $\\{x \in \mathbb{r}\mid x \
eq -2\\}$, r: $\\{y \in \mathbb{r}\mid y \
eq -8\\}$\
d: $\\{x \in \mathbb{r}\mid x \
eq 4\\}$, r: $\\{y \in \mathbb{r}\mid y \
eq -2\\}$\
d: $\\{x \in \mathbb{r}\mid x \
eq 2\\}$, r: $\\{y \in \mathbb{r}\mid y \
eq 8\\}$\
d: $\\{x \in \mathbb{r}\mid x \
eq -4\\}$, r: $\\{y \in \mathbb{r}\mid y \
eq 2\\}$

Explanation:

Step1: Simplify the function

First, factor the numerator \(x^{2}-4x - 12\). We know that \(x^{2}-4x - 12=(x - 6)(x+ 2)\)? Wait, no, let's do it correctly. \(x^{2}-4x-12=(x - 6)(x + 2)\)? Wait, \((x-6)(x + 2)=x^{2}+2x-6x - 12=x^{2}-4x - 12\), yes. But the denominator is \(x + 2\). So the function \(f(x)=\frac{x^{2}-4x - 12}{x + 2}=\frac{(x - 6)(x + 2)}{x + 2}\), for \(x
eq - 2\) (since denominator cannot be zero), we can cancel out \(x + 2\) (when \(x
eq - 2\)), so \(f(x)=x-6\) for \(x
eq - 2\).

Step2: Find the domain

The domain of a function is all real numbers except where the denominator is zero. The denominator is \(x + 2\), so \(x+2
eq0\implies x
eq - 2\). So the domain \(D=\{x\in\mathbb{R}|x
eq - 2\}\).

Step3: Find the range

The simplified function is \(y=x - 6\) with the restriction that \(x
eq - 2\). When \(x=-2\), \(y=-2 - 6=-8\). But since \(x
eq - 2\), \(y
eq - 8\). So the range \(R=\{y\in\mathbb{R}|y
eq - 8\}\).

Answer:

D: \(\{x\in\mathbb{R}|x
eq - 2\}\), R: \(\{y\in\mathbb{R}|y
eq - 8\}\) (which corresponds to the first option D among the given choices: D: \(\{x\in\mathbb{R}|x
eq - 2\}\), R: \(\{y\in\mathbb{R}|y
eq - 8\}\))