The given function in the image is:

$f(x) = \frac{2x + 4}{x + 2}$

You are asked to find the following:

1. Domain
2. Range
3. x-intercepts
4. y-intercepts

Let’s go step by step to analyze this.

1. Domain:

The domain of a rational function is the set of all real numbers except those that make the denominator zero. Here, the denominator is $x + 2$. Setting it equal to zero: $x + 2 = 0 \implies x = -2$ Thus, the domain is all real numbers except $x = -2$, i.e., $\text{Domain} = (-\infty, -2) \cup (-2, \infty)$

2. Range:

To find the range, let's first simplify the function: $f(x) = \frac{2x + 4}{x + 2} = \frac{2(x + 2)}{x + 2}$ For $x \neq -2$, the $x + 2$ terms cancel out, leaving: $f(x) = 2$ Thus, for all $x \neq -2$, $f(x) = 2$. Therefore, the range is: $\text{Range} = \{2\}$

3. x-intercepts:

The x-intercepts occur when $f(x) = 0$, i.e., when the numerator is equal to zero. Setting $2x + 4 = 0$: $2x + 4 = 0 \implies x = -2$ However, $x = -2$ is excluded from the domain (it makes the denominator zero). Hence, there are no x-intercepts.

4. y-intercepts:

The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the function: $f(0) = \frac{2(0) + 4}{0 + 2} = \frac{4}{2} = 2$ Thus, the y-intercept is at $y = 2$, or the point $(0, 2)$.

Summary:

• Domain: $(-\infty, -2) \cup (-2, \infty)$
• Range: $\{2\}$
• x-intercepts: None
• y-intercepts: $(0, 2)$

Would you like more details or have any questions?

Here are five related questions:

1. How would the domain change if the function had a different denominator?
2. What happens to the range if the function was not simplified?
3. Can rational functions have multiple y-intercepts? Why or why not?
4. What is the significance of vertical asymptotes in rational functions?
5. How would horizontal asymptotes change if the degrees of the numerator and denominator were different?

Tip: Always check for any restrictions on the domain when simplifying rational functions!