Question

(04.02r mc)
what are the domain and range of the function $f(x) = \frac{x^2 + 6x + 8}{x + 4}$?
\bigcirc \quad d: \\{x \in \mathbb{r} \mid x \
eq 4\\}; r: \\{y \in \mathbb{r} \mid y \
eq 0\\}
\bigcirc \quad d: \\{x \in \mathbb{r} \mid x \
eq -4\\}; r: \\{y \in \mathbb{r} \mid y \
eq -2\\}
\bigcirc \quad d: \\{x \in \mathbb{r} \mid x \
eq 4\\}; r: \\{y \in \mathbb{r} \mid y \
eq 2\\}
\bigcirc \quad d: \\{x \in \mathbb{r} \mid x \
eq -2\\}; r: \\{y \in \mathbb{r} \mid y \
eq 0\\}
question 4 \quad (multiple choice worth 5 points)
(04.01r lc)

Explanation:

Step1: Simplify the function

First, factor the numerator: \(x^2 + 6x + 8=(x + 2)(x + 4)\). So the function becomes \(f(x)=\frac{(x + 2)(x + 4)}{x + 4}\), where \(x
eq - 4\) (since the denominator cannot be zero). After canceling \(x + 4\) (for \(x
eq - 4\)), we get \(f(x)=x + 2\), \(x
eq - 4\).

Step2: Find the domain

The domain is all real numbers except the value that makes the original denominator zero. The denominator \(x + 4 = 0\) when \(x=-4\), so the domain \(D=\{x\in\mathbb{R}|x
eq - 4\}\).

Step3: Find the range

The simplified function is \(y=x + 2\) with a hole at \(x = - 4\). When \(x=-4\), the original function is undefined, but for the simplified linear function \(y=x + 2\), if we consider the hole, when \(x=-4\), \(y=-4 + 2=-2\). So the range is all real numbers except \(y = - 2\), i.e., \(R=\{y\in\mathbb{R}|y
eq - 2\}\).

Answer:

B. \(D:\{x\in\mathbb{R}|x
eq - 4\};R:\{y\in\mathbb{R}|y
eq - 2\}\)