This problem tests linking infinite limits to vertical asymptotes in functions with added constants. For q(x) = 2/(x-4) + 1, the denominator zero at x=4 with nonzero numerator term drives the behavior. Approaching from the left yields negative infinity, and from the right positive infinity, despite the +1 not affecting the infinite trend. These opposing infinite limits confirm a vertical asymptote at x=4. A common distractor confuses this with a horizontal asymptote at y=1 due to the added constant, but horizontal asymptotes relate to limits at infinity, not vertical ones. A general strategy is to isolate terms causing division by zero and evaluate one-sided limits for infinity.

This question tests your ability to connect infinite limits with vertical asymptotes in transformed reciprocal functions. For u(x) = 5/x - 2, the 5/x term causes infinite behavior at x=0: positive infinity from the right and negative infinity from the left. Subtracting 2 shifts the graph vertically but doesn't prevent the limits from being infinite. Thus, there's a vertical asymptote at x=0 with opposite-signed infinite limits. Choice D is tempting but fails because the shift affects horizontal asymptotes, not the vertical one from the 1/x core. Always isolate the term causing division by zero and assess its infinite limits, regardless of added constants.

This question tests your ability to identify vertical asymptotes in transformed rational functions. For $q(x)=\frac{2}{x-4}+1$, the rational part $\frac{2}{x-4}$ has a denominator that equals zero when $x=4$. As $x\to 4^-$, we have $(x-4)\to 0^-$, so $\frac{2}{x-4}\to -\infty$, making $q(x)\to -\infty$. As $x\to 4^+$, we have $(x-4)\to 0^+$, so $\frac{2}{x-4}\to +\infty$, making $q(x)\to +\infty$. This confirms a vertical asymptote at $x=4$. Choice D incorrectly calls this a horizontal asymptote, confusing the direction of the asymptote line. Remember that vertical asymptotes are vertical lines $x=a$ where the function has infinite limits, while horizontal asymptotes are horizontal lines $y=b$ describing end behavior.

This question tests your ability to identify vertical asymptotes by analyzing when denominators equal zero and limits become infinite. For $f(x)=\frac{3x+1}{x^2-4}$, we need to find where the denominator $x^2-4=0$, which factors as $(x-2)(x+2)=0$, giving us $x=2$ and $x=-2$. At both these values, the numerator $3x+1$ is non-zero (it equals 7 and -5 respectively), so the function has infinite limits at these points. Since the denominator changes sign as we cross each zero, we get vertical asymptotes at both $x=-2$ and $x=2$. Choice A incorrectly suggests an asymptote at $x=0$ where the function is actually well-defined. When finding vertical asymptotes, always check that the numerator doesn't also equal zero at the same point, as that could indicate a removable discontinuity instead.

This problem tests simplifying algebraic expressions before taking limits. For $u(x)=\frac{x-1}{\sqrt{x-1}}$ with $x>1$, we can rewrite this as $u(x)=\frac{x-1}{(x-1)^{1/2}}=(x-1)^{1-1/2}=(x-1)^{1/2}=\sqrt{x-1}$. As $x\to1^+$, we have $\sqrt{x-1}\to\sqrt{0^+}=0$. Since the limit is finite (zero), there is no vertical asymptote at $x=1$. Choice A incorrectly assumes an infinite limit without simplifying. Always simplify expressions algebraically before evaluating limits to avoid missing cancellations.

This problem requires recognizing vertical asymptotes in transformed rational functions through infinite limit behavior. The function t(x) = 2/(x-7) + 5 is a transformation of the basic rational function 1/x. The term 2/(x-7) has a vertical asymptote at x = 7, where the denominator equals zero. As x approaches 7 from the right (x → 7⁺), (x-7) approaches 0 through positive values, making 2/(x-7) approach +∞, and thus t(x) approaches +∞. The constant term +5 shifts the graph vertically but doesn't affect the location of the vertical asymptote. Choice C incorrectly places the asymptote at y = 7, confusing vertical and horizontal asymptotes. For functions of the form a/(x-h) + k, vertical asymptotes always occur at x = h.

This question tests your ability to connect infinite limits with vertical asymptote behavior. For $f(x)=\frac{3x+1}{(x-2)(x+4)}$, as $x$ approaches 2, the denominator approaches 0 while the numerator approaches $3(2)+1=7$. When $x$ approaches 2 from the left ($x<2$), the factor $(x-2)$ is negative and $(x+4)$ is positive, making the denominator negative overall, so $f(x)\to-\infty$. When $x$ approaches 2 from the right ($x>2$), the factor $(x-2)$ is positive and $(x+4)$ is positive, making the denominator positive, so $f(x)\to+\infty$. Choice A incorrectly claims both one-sided limits approach $+\infty$, failing to account for the sign change. The key strategy is to analyze the sign of each factor near the asymptote to determine whether the function approaches $+\infty$ or $-\infty$ from each side.

This problem requires analyzing infinite limits for a rational function with a factorable denominator. For $r(x)=\frac{2}{x^2-4}=\frac{2}{(x-2)(x+2)}$, near $x=2$, the factor $(x+2)$ is approximately 4 (positive). When approaching from the left ($x<2$), the factor $(x-2)$ is negative, making the denominator negative overall, so $r(x)\to-\infty$. When approaching from the right ($x>2$), the factor $(x-2)$ is positive, making the denominator positive, so $r(x)\to+\infty$. This matches choice C, confirming that $x=2$ is a vertical asymptote with different one-sided infinite limits. Choice E incorrectly claims both limits are $+\infty$, missing the sign change. The strategy is to determine the sign of each factor in the denominator on each side of the potential asymptote.

This question tests your ability to connect infinite limits with vertical asymptotes when one-sided limits differ. If the left-hand limit at x=-3 is negative infinity and the right-hand is finite (5), the infinite side indicates unbounded behavior approaching x=-3. This supports a vertical asymptote, even if only one side is infinite. The finite side means the graph approaches 5 from the right, but the infinite side dominates for asymptote classification. Choice E is tempting but fails because vertical asymptotes require at least one infinite one-sided limit, not necessarily both. Always evaluate both one-sided limits separately; an infinite result from either confirms a vertical asymptote.

This question tests your ability to identify vertical asymptotes by analyzing infinite limits of rational functions. The function f(x) = (3x-1)/[(x+2)(x-5)] has its denominator equal to zero when x = -2 and x = 5, while the numerator is non-zero at these values. As x approaches -2 or 5, the denominator approaches 0 while the numerator approaches finite non-zero values, causing the function to approach ±∞. This creates vertical asymptotes at x = -2 and x = 5. Choice B incorrectly identifies these as horizontal asymptotes, which would be lines of the form y = k, not x = k. When a rational function's denominator has zeros that don't cancel with the numerator, vertical asymptotes occur at those x-values.