This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For p(x) = (x² + 2x - 3)/(x² - 9), we factor both parts: numerator = (x + 3)(x - 1) and denominator = (x + 3)(x - 3). Choice B is correct because after canceling the common factor (x + 3), we get (x - 1)/(x - 3), which has a vertical asymptote only at x = 3. Choice C incorrectly identifies both x = -3 and x = 3 as asymptotes, missing that x = -3 creates a removable discontinuity due to the cancellation. To help students: Factor completely before identifying asymptotes, cancel common factors first, and remember that canceled factors create holes, not vertical asymptotes.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For t(x) = (2x+1)/(x+4), setting the denominator equal to zero gives x+4=0, so x=-4 is a vertical asymptote. As x approaches -4, the denominator approaches 0 while the numerator approaches 2(-4)+1 = -7, causing the function to approach ±∞ depending on the direction of approach. Choice C is correct because t(x)→±∞ as x→-4, which is the characteristic behavior at a vertical asymptote. Choice D incorrectly suggests the function approaches a finite value. To help students: At vertical asymptotes, rational functions always approach ±∞, never finite values, and the sign depends on the direction of approach.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For s(x) = (x²-5x+6)/(x²-9), we factor the numerator as (x-2)(x-3) and the denominator as (x+3)(x-3). The factor (x-3) cancels from both, leaving s(x) = (x-2)/(x+3) for x≠3, which has a vertical asymptote only at x=-3. Choice C is correct because x=-3 is where the remaining denominator equals zero after simplification. Choices A and B are incorrect because they identify zeros of the numerator or canceled factors. To help students: Always simplify by canceling common factors first, then identify vertical asymptotes from the remaining denominator factors.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For the function f(x) = (x²-4)/(x-2), we must first factor the numerator as (x+2)(x-2), giving us f(x) = (x+2)(x-2)/(x-2). Since the factor (x-2) cancels from both numerator and denominator, we get f(x) = x+2 for x≠2, which means there's a removable discontinuity (hole) at x=2, not a vertical asymptote. Choice D is correct because after simplification, no factors remain in the denominator to create vertical asymptotes. Choice A is incorrect because it identifies the location of the hole, not an asymptote. To help students: Emphasize the difference between vertical asymptotes (non-canceling factors in denominator) and holes (canceling factors), and always factor completely before identifying asymptotes.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For h(x) = (x+2)/[(x-1)(x+3)], the denominator is already factored, so we set each factor equal to zero: x-1=0 gives x=1, and x+3=0 gives x=-3. Choice B is correct because it identifies both vertical asymptotes at x=1 and x=-3 where the denominator equals zero. Choice A is incorrect because it confuses the numerator's zero with an asymptote location. To help students: Remember that vertical asymptotes come from denominator zeros only, not numerator zeros, and when the denominator is already factored, simply set each factor to zero.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For h(x) = (x + 4)/(x - 4), the denominator equals zero when x = 4, creating a vertical asymptote at this point. Choice C is correct because as x approaches 4, the denominator approaches 0 while the numerator approaches 8, causing the function to approach ±∞ (positive infinity from the right, negative infinity from the left). Choice A incorrectly suggests the function approaches 0, which would be horizontal asymptotic behavior. To help students: Analyze the sign of numerator and denominator near the asymptote, use test values slightly less than and greater than 4, and visualize how division by numbers approaching zero creates unbounded behavior.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For g(x) = (3x+1)/(x²-9), we need to factor the denominator as (x+3)(x-3) and find where it equals zero. Setting each factor to zero gives x = -3 and x = 3, which are the locations of the vertical asymptotes. Choice A is correct because x = ±3 represents both x = 3 and x = -3, where the function has vertical asymptotes. Choice B and C are incorrect because they only identify one of the two asymptotes. To help students: Practice factoring difference of squares (a²-b² = (a+b)(a-b)), and remember that each distinct factor in the denominator creates a separate vertical asymptote.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For f(x) = (x²+1)/(x-1), the numerator x²+1 cannot be factored over real numbers, and the denominator has only the factor (x-1). Since no factors cancel, setting x-1=0 gives the vertical asymptote at x=1. Choice B is correct because x=1 is the only value that makes the denominator zero. Choice D is incorrect because there is clearly a vertical asymptote where the denominator equals zero. To help students: Remember that not all polynomials factor nicely, and when performing polynomial division, the vertical asymptotes remain at the zeros of the original denominator.

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For g(x) = (3x + 1)/(x² - x - 6), we need to factor the denominator: x² - x - 6 = (x - 3)(x + 2). Choice A is correct because setting each factor equal to zero gives x - 3 = 0 → x = 3 and x + 2 = 0 → x = -2, identifying vertical asymptotes at x = -2 and x = 3. Choice B incorrectly identifies x = 2 and x = -3, which would come from factoring as (x - 2)(x + 3). To help students: Practice factoring quadratic expressions carefully, verify factors by expanding, and remember that vertical asymptotes occur at all values where the denominator equals zero (unless canceled by the numerator).

This question tests understanding of rational functions and vertical asymptotes in AP Precalculus. Vertical asymptotes occur where the denominator of a rational function equals zero and the function is undefined, signaling a potential infinite discontinuity. For the limit of 5/(x + 2) as x approaches -2 from the right (x → -2⁺), we analyze the sign of the denominator: when x is slightly greater than -2, x + 2 is positive and very small. Choice A is correct because 5 divided by a positive number approaching 0 gives +∞. Choice B incorrectly identifies -∞, which would occur when approaching from the left (x → -2⁻). To help students: Carefully note the direction of approach (+ means from the right), determine the sign of numerator and denominator near the asymptote, and remember that positive divided by positive approaching zero gives positive infinity.