The problem asks to find the x-intercept and y-intercept of the function $f(x) = \frac{x-9}{x+3}$ and to explain how these are determined.

1. Finding the x-intercept:

The x-intercept occurs when $f(x) = 0$. This means the numerator of the fraction must equal $0$, as a fraction equals zero when the numerator is zero and the denominator is non-zero.

Set the numerator equal to zero: $x - 9 = 0$ $x = 9$

Thus, the x-intercept is: $(9, 0)$

2. Finding the y-intercept:

The y-intercept occurs when $x = 0$. Substitute $x = 0$ into the function $f(x) = \frac{x-9}{x+3}$.

Substitute: $f(0) = \frac{0 - 9}{0 + 3} = \frac{-9}{3} = -3$

Thus, the y-intercept is: $(0, -3)$

Explanation:

• For the x-intercept, set $f(x) = 0$ and solve for $x$, which involves finding when the numerator of the rational function equals zero.
• For the y-intercept, substitute $x = 0$ into the function and simplify.

Would you like me to clarify further or go into related concepts?

Related Questions:

1. How do vertical asymptotes relate to the denominator of the function $f(x) = \frac{x-9}{x+3}$?
2. What is the domain of $f(x)$ and why?
3. How can you find the horizontal asymptote of $f(x) = \frac{x-9}{x+3}$?
4. What happens near $x = -3$ (the vertical asymptote)?
5. How does the graph of $f(x) = \frac{x-9}{x+3}$ compare to a linear function?

Tip:

For rational functions, always check the numerator for x-intercepts, the denominator for vertical asymptotes, and the ratio of leading terms for horizontal asymptotes.