The function provided is:

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

We are tasked with finding the domain, the x- and y-intercepts, graphing the function, and labeling all asymptotes.

1. Domain

The domain of a rational function is determined by the values of $x$ for which the denominator is not zero. In this case:

$x + 4 \neq 0 \implies x \neq -4$

So, the domain is:

$(-\infty, -4) \cup (-4, \infty)$

2. X-Intercept

To find the x-intercept, set $f(x) = 0$. This occurs when the numerator equals zero:

$x - 2 = 0 \implies x = 2$

Thus, the x-intercept is at $x = 2$, or the point $(2, 0)$.

3. Y-Intercept

The y-intercept occurs when $x = 0$. Substituting $x = 0$ into the function:

$f(0) = \frac{0 - 2}{0 + 4} = \frac{-2}{4} = -\frac{1}{2}$

Thus, the y-intercept is $(0, -\frac{1}{2})$.

4. Asymptotes

• Vertical Asymptote: Occurs where the denominator equals zero, i.e., $x = -4$.
• Horizontal Asymptote: For rational functions where the degrees of the numerator and denominator are equal (both are degree 1 in this case), the horizontal asymptote is given by the ratio of the leading coefficients. Here, both leading coefficients are 1, so the horizontal asymptote is:

$y = 1$

5. Graphing the Function

• X-Intercept: $(2, 0)$
• Y-Intercept: $(0, -\frac{1}{2})$
• Vertical Asymptote: $x = -4$
• Horizontal Asymptote: $y = 1$

Would you like me to generate a graph of this function?

Further Exploration:

1. What happens to the graph as $x$ approaches the vertical asymptote $x = -4$?
2. How do the intercepts change if the numerator or denominator is altered?
3. Can this function have any slant asymptotes?
4. What is the significance of the horizontal asymptote for large values of $x$?
5. How does changing the degree of the numerator or denominator affect the graph?

Tip:

When analyzing rational functions, always check for vertical asymptotes by finding the values that make the denominator zero. This can provide useful insights for sketching the graph.