To solve the problem, we will need to show that $a = 6$ and $b = -13$ for the function $f(x) = \frac{ax^2 + bx + 3}{x^2 - 4}$ given the conditions:

1. The graph of $f$ has a horizontal asymptote at $y = 3$.
2. $f$ has a removable discontinuity at $x = 2$.

Let's break down the steps:

1. Horizontal Asymptote at $y = 3$:

The horizontal asymptote for a rational function $\frac{p(x)}{q(x)}$ depends on the degrees of the numerator $p(x)$ and denominator $q(x)$:

• If the degree of $p(x)$ is equal to the degree of $q(x)$, the horizontal asymptote is given by $y = \frac{\text{leading coefficient of } p(x)}{\text{leading coefficient of } q(x)}$.

In this case, both the numerator and the denominator are quadratic polynomials:

$f(x) = \frac{ax^2 + bx + 3}{x^2 - 4}$

The leading coefficient of the numerator is $a$, and the leading coefficient of the denominator is 1. Since the horizontal asymptote is given as $y = 3$, we have:

$\frac{a}{1} = 3$

Thus, $a = 3$.

2. Removable Discontinuity at $x = 2$:

A removable discontinuity at $x = 2$ implies that $x = 2$ makes both the numerator and the denominator equal to zero. This is because for a discontinuity to be removable, the function must have the form $\frac{0}{0}$ at $x = 2$.

Substitute $x = 2$ into the denominator:

$x^2 - 4 = 2^2 - 4 = 0$

So, the denominator is zero at $x = 2$.

Next, substitute $x = 2$ into the numerator:

$ax^2 + bx + 3 = a(2)^2 + b(2) + 3 = 4a + 2b + 3$

Since the numerator must also equal zero at $x = 2$:

$4a + 2b + 3 = 0$

We already know $a = 3$, so substitute that into the equation:

$4(3) + 2b + 3 = 0$

Simplify and solve for $b$:

$12 + 2b + 3 = 0$

$15 + 2b = 0$

$2b = -15$

$b = -\frac{15}{2} = -7.5$

Given that the answer needed to show $b = -13$ may imply an issue. Let's correct that by checking the correct steps to solve this algebraically again:

Corrected Steps:

• Set $a = 3$
• Substitute back into the equation and solve for $b$.

Since there seems to be a mismatch, you might want to confirm that the problem intended this approach. The rationale used was correct for the conditions stated, but a mismatch in the horizontal asymptote may cause reconsideration or recalculation.

Do you want further elaboration on any of these steps or need a more comprehensive check? Here are five possible follow-up questions:

1. How do I derive horizontal asymptotes from any rational function?
2. What defines a removable discontinuity, and how can it be removed?
3. How would this change if the discontinuity wasn't removable?
4. Can we generalize the approach for any quadratic in the numerator?
5. What if the horizontal asymptote or discontinuity was at a different $x$-value?

Tip: When solving algebraic problems, always ensure that each variable aligns with all given conditions. Double-check each step to prevent computational errors.