A vertical asymptote occurs where the denominator equals zero but the numerator does not. For choice B, when $$x = 3$$, the denominator $$x - 3 = 0$$ and the numerator $$x + 2 = 5 \neq 0$$, creating a vertical asymptote. Choice A has a hole at $$x = 3$$ since both numerator and denominator equal zero. Choice C has vertical asymptotes at $$x = \pm 2$$, not $$x = 3$$. Choice D has a vertical asymptote at $$x = -1$$, not $$x = 3$$.

Vertical asymptotes occur where the denominator equals zero but the numerator does not. The denominator $$(x + 1)(x - 4) = 0$$ when $$x = -1$$ or $$x = 4$$. At $$x = -1$$: numerator $$= 2(-1) - 5 = -7 \neq 0$$. At $$x = 4$$: numerator $$= 2(4) - 5 = 3 \neq 0$$. Both create vertical asymptotes. Choice B incorrectly includes $$x = \frac{5}{2}$$ (where numerator equals zero). Choice C incorrectly includes $$x = \frac{5}{2}$$. Choice D uses incorrect signs.

The denominator factors as $$x^2 - 5x + 6 = (x - 2)(x - 3)$$, giving potential vertical asymptotes at $$x = 2$$ and $$x = 3$$. For exactly two vertical asymptotes, the numerator must not equal zero at either $$x = 2$$ or $$x = 3$$. At $$x = 2$$: numerator $$= 2 + k$$, so $$k \neq -2$$. At $$x = 3$$: numerator $$= 3 + k$$, so $$k \neq -3$$. Therefore $$k$$ can be any real number except $$-2$$ and $$-3$$. Choice B uses wrong signs. Choice C would create holes, not asymptotes. Choice D uses incorrect values.

The denominator $$x^2 - 9 = (x - 3)(x + 3)$$ gives vertical asymptotes at $$x = 3$$ and $$x = -3$$. As $$x \to 3^-$$: numerator approaches $$9 > 0$$ and denominator $$(x-3)(x+3)$$ approaches $$0^- \cdot 6 = 0^-$$, so the limit is $$+\infty$$. As $$x \to -3^-$$: numerator approaches $$-3 < 0$$ and denominator $$(-6)(x+3)$$ approaches $$(-6)(0^-) = 0^+$$, so the limit is $$-\infty$$. Choices B, C, and D have incorrect sign combinations for the limit behavior.

The denominator $$(x + 3)^2(x - 7)$$ has zeros at $$x = -3$$ (multiplicity 2) and $$x = 7$$ (multiplicity 1). Since the numerator $$5x - 1$$ doesn't equal zero at either point ($$5(-3) - 1 = -16 \neq 0$$ and $$5(7) - 1 = 34 \neq 0$$), both zeros create vertical asymptotes with their respective multiplicities. The asymptote at $$x = -3$$ has multiplicity 2, and at $$x = 7$$ has multiplicity 1. Choices B and C have incorrect multiplicities. Choice D ignores the asymptote at $$x = 7$$.

For exactly one vertical asymptote at $$x = 4$$, the denominator must have $$x = 4$$ as its only zero where the numerator is non-zero. Choice B has denominator $$x - 4$$ with only one zero at $$x = 4$$, and numerator $$4 + 1 = 5 \neq 0$$ at this point. Choice A has two asymptotes at $$x = 4$$ and $$x = -1$$. Choice C has $$x^2 - 16 = (x-4)(x+4)$$, giving asymptotes at both $$x = 4$$ and $$x = -4$$. Choice D has $$x^2 - 8x + 16 = (x-4)^2$$, giving one asymptote at $$x = 4$$ with multiplicity 2.

For vertical asymptotes at $$x = 1$$ and $$x = -5$$, the denominator must have these as zeros where the numerator is non-zero. Choice B has denominator $$x^2 + 4x - 5 = (x - 1)(x + 5)$$, giving zeros at $$x = 1$$ and $$x = -5$$. Check numerator: at $$x = 1$$, $$2(1) - 3 = -1 \neq 0$$; at $$x = -5$$, $$2(-5) - 3 = -13 \neq 0$$. Both create asymptotes. Choice A has wrong signs. Choice C has asymptotes at $$x = -1$$ and $$x = 5$$. Choice D has $$x^2 - 6x + 5 = (x-1)(x-5)$$, giving asymptotes at $$x = 1$$ and $$x = 5$$.

Factor the denominator: $$x^3 - 2x^2 - 3x = x(x^2 - 2x - 3) = x(x - 3)(x + 1)$$. The denominator has zeros at $$x = 0, 3, -1$$. Check if the numerator $$2x + 1$$ equals zero at any of these points: At $$x = 0$$: $$2(0) + 1 = 1 \neq 0$$. At $$x = 3$$: $$2(3) + 1 = 7 \neq 0$$. At $$x = -1$$: $$2(-1) + 1 = -1 \neq 0$$. Since the numerator is non-zero at all three zeros of the denominator, all three create vertical asymptotes. Choices B, C, and D incorrectly exclude some asymptotes.

Factor both parts: numerator $$x^2 + 5x + 6 = (x + 2)(x + 3)$$ and denominator $$x^2 + 3x + 2 = (x + 1)(x + 2)$$. So $$h(x) = \frac{(x + 2)(x + 3)}{(x + 1)(x + 2)} = \frac{x + 3}{x + 1}$$ for $$x \neq -2$$. After canceling the common factor $$(x + 2)$$, there's a hole at $$x = -2$$ and one vertical asymptote at $$x = -1$$ where the simplified denominator equals zero but the numerator $$(-1) + 3 = 2 \neq 0$$. Choice A counts the hole as an asymptote. Choices C and D are incorrect counts.

For a vertical asymptote at $$x = -2$$, the denominator must equal zero at $$x = -2$$ while the numerator does not. Choice C has denominator $$x + 2$$ which equals zero when $$x = -2$$, and numerator $$2(-2) + 1 = -3 \neq 0$$, creating a vertical asymptote. Choice A has a hole at $$x = 3$$ and asymptote at $$x = -2$$. Choice B has asymptote at $$x = 3$$, not $$x = -2$$. Choice D factors to $$\frac{(x-3)(x+3)}{(x-3)(x+2)}$$ with asymptote at $$x = -2$$ after canceling.