To solve this problem, we need to determine the constants $A$ and $B$ such that the function

$f(x) = \frac{Ax + 9}{33 + Bx}$

has $x = 3$ as a vertical asymptote and $y = -3$ as a horizontal asymptote.

1. Vertical Asymptote Condition

A vertical asymptote occurs where the denominator of the function equals zero, because the function approaches infinity. Thus, to have a vertical asymptote at $x = 3$:

$33 + B(3) = 0$

Solving for $B$:

$33 + 3B = 0 \implies B = -11$

2. Horizontal Asymptote Condition

For rational functions, the horizontal asymptote is determined by the degrees of the numerator and the denominator:

• If the degree of the numerator equals the degree of the denominator, the horizontal asymptote is the ratio of the leading coefficients.

The degrees are the same here (both are linear, degree 1), so the horizontal asymptote is determined by:

$\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \frac{Ax + 9}{33 + Bx} = \frac{A}{B}$

We are given that the horizontal asymptote is $y = -3$, so:

$\frac{A}{B} = -3$

Given that $B = -11$ from the vertical asymptote condition, we substitute it into the equation:

$\frac{A}{-11} = -3 \implies A = 33$

Final Values

Thus, the constants are:

$A = 33, \quad B = -11$

Would you like further details or have any questions?

Related Questions:

1. How would the function change if the vertical asymptote was at $x = 5$?
2. What if the horizontal asymptote was $y = 2$?
3. How does the function behave near its vertical asymptote?
4. Can you explain why the degrees of the numerator and denominator matter for horizontal asymptotes?
5. How would the function be altered if the numerator was quadratic?

Tip:

For horizontal asymptotes, always check the relative degrees of the numerator and denominator—it helps determine the end behavior of the function.