To analyze the function $f(x) = \frac{1}{x + 4} - 3$, we will find the following:

1. Zeros (where $f(x) = 0$):

Set $f(x) = 0$ to find the zeros. $\frac{1}{x + 4} - 3 = 0$ Add 3 to both sides: $\frac{1}{x + 4} = 3$ Multiply both sides by $x + 4$: $1 = 3(x + 4)$ Solve for $x$: $1 = 3x + 12$ $3x = 1 - 12$ $3x = -11$ $x = -\frac{11}{3}$ Thus, the zero is at $x = -\frac{11}{3}$.

2. Vertical Asymptote:

The vertical asymptote occurs where the denominator is zero. The denominator of $\frac{1}{x + 4}$ is zero when: $x + 4 = 0$ $x = -4$ So, the vertical asymptote is at $x = -4$.

3. Holes:

Holes occur where there are common factors between the numerator and denominator, which cancel out. The function $f(x) = \frac{1}{x+4} - 3$ has no common factors to cancel, so there are no holes in this function.

4. Y-Intercept (where $x = 0$):

To find the y-intercept, set $x = 0$: $f(0) = \frac{1}{0 + 4} - 3 = \frac{1}{4} - 3 = \frac{1}{4} - \frac{12}{4} = \frac{-11}{4}$ Thus, the y-intercept is $\left( 0, \frac{-11}{4} \right)$.

Summary:

• Zero: $x = -\frac{11}{3}$
• Vertical asymptote: $x = -4$
• Holes: None
• Y-intercept: $\left( 0, \frac{-11}{4} \right)$

Would you like any further details or clarification?

Here are 5 related questions you might want to explore:

1. How do you graph a rational function like this one?
2. What is the behavior of $f(x)$ near the vertical asymptote?
3. How do you determine horizontal asymptotes for rational functions?
4. Can rational functions have slant asymptotes?
5. How do transformations affect rational functions like $\frac{1}{x+4} - 3$?

Tip: When analyzing rational functions, always check for holes by factoring both the numerator and denominator, as holes occur when terms cancel out.