This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions $f(x) = p(x)/q(x)$ have distinctive features: zeros where $p(x) = 0$ (numerator equals zero, these are x-intercepts), vertical asymptotes where $q(x) = 0$ (denominator equals zero, graph shoots to $±∞$), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have $(x - 2)$), that creates a hole (removable discontinuity) at $x = 2$, not a zero or asymptote—the common factor cancels! For $f(x) = \frac{(x-4)(x+1)}{(x-4)(x-2)}$, factor and cancel common $(x-4)$, leaving $\frac{(x+1)}{(x-2)}$ with hole at $x=4$ ($y=\frac{4+1}{4-2}=\frac{5}{2}$); VA at $x=2$ from remaining denominator zero. Choice C correctly identifies VA at $x=2$ and hole at $x=4$, capturing the removable discontinuity. A distractor like choice A swaps the hole and VA, but after canceling, the denominator zero at $x=4$ is removed, so no VA there—always simplify first! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → $y = 0$; degrees equal → $y =$ ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! For $$f(x) = \frac{(x+2)(x-1)}{x^2 - 4} = \frac{(x+2)(x-1)}{(x+2)(x-2)},$$ cancel $(x+2)$, leaving $$\frac{x-1}{x-2}$$ with hole at $x=-2$; zero at $x=1$ from simplified numerator, VA at $x=2$. Choice D correctly identifies zero at $x=1$, VA at $x=2$, and hole at $x=-2$. A distractor like choice A confuses the hole with a zero or VA, but after canceling, $x=-2$ is neither—simplify to see clearly! The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → y = 0; degrees equal → y = ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is $y = 0$ (graph flattens toward x-axis as $x \to \pm \infty$), (2) if degrees equal, HA is $y = \frac{\text{numerator leading coefficient}}{\text{denominator leading coefficient}}$ (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x) = (3x^2 - x + 7)/(2x^2 + 5x - 1), degrees equal (both 2), so HA $y = \frac{3}{2}$ from leading coefficients. Choice B correctly applies the equal-degree rule to find HA $y=\frac{3}{2}$. A distractor like choice A reverses the ratio to 2/3, but it's numerator over denominator leading coefficients—remember the order! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: $(x^2 - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2)$ has a hole at x = 2 (cancels), leaving simplified f(x) = x + 2 with a gap at x = 2.

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is y = 0 (graph flattens toward x-axis as x → ±∞), (2) if degrees equal, HA is y = (numerator leading coefficient)/(denominator leading coefficient) (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x) = (x+1)/((x-2)(x+3)), zero at x=-1 from numerator; VAs at x=2 and x=-3 from denominator; deg num 1 < deg den 2, so HA y=0. Choice A correctly identifies zero at x=-1, VAs at x=2 and -3, and HA y=0 for sketching. A distractor like choice B shifts values, perhaps misfactoring, but double-check roots—keep up the great work! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) has a hole at x = 2 (cancels), leaving simplified f(x) = x + 2 with a gap at x = 2.

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is $y = 0$ (graph flattens toward $x$-axis as $x \to \pm \infty$), (2) if degrees equal, HA is $y = \frac{\text{numerator leading coefficient}}{\text{denominator leading coefficient}}$ (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For $f(x) = \frac{2x^2 - 8}{x^2 + 1}$, factor numerator as $2(x^2 - 4) = 2(x-2)(x+2)$, so zeros at $x=\pm 2$; denominator $x^2 + 1 = 0$ has no real roots (no VAs); degrees equal (both 2), so HA $y=2/1=2$, and the graph crosses x-axis at $\pm 2$ while approaching $y=2$ horizontally. Choice A correctly identifies zeros at $x=\pm 2$ and HA $y=2$, allowing an accurate sketch without vertical asymptotes. A distractor like choice B uses $y=0$, which would apply if deg(num) < deg(den), but here degrees match, so use the leading coefficient ratio instead—keep practicing those rules! Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If $(x - 3)$ appears in both, it cancels, creating a hole at $x = 3$ (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting $x = 3$ into the simplified function. Holes are easy to miss—always check for common factors! Example: $\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2}$ has a hole at $x = 2$ (cancels), leaving simplified $f(x) = x + 2$ with a gap at $x = 2$.

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions f(x) = p(x)/q(x) have distinctive features: zeros where p(x) = 0 (numerator equals zero, these are x-intercepts), vertical asymptotes where q(x) = 0 (denominator equals zero, graph shoots to ±∞), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have (x - 2)), that creates a hole (removable discontinuity) at x = 2, not a zero or asymptote—the common factor cancels! Factoring gives numerator (x-3)(x+3) and denominator (x-3)(x+2), so cancel (x-3) for simplified (x+3)/(x+2) with hole at x=3 where y=6/5, zero at x=-3, and vertical asymptote at x=-2. Choice B correctly identifies the zero at x=-3, vertical asymptote at x=-2, and hole at x=3 with y=6/5. A distractor like Choice A ignores the common factor, listing x=3 as both zero and asymptote, but cancellation creates a hole instead. Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If (x - 3) appears in both, it cancels, creating a hole at x = 3 (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting x = 3 into the simplified function. Holes are easy to miss—always check for common factors! Example: (x² - 4)/(x - 2) = (x + 2)(x - 2)/(x - 2) has a hole at x = 2 (cancels), leaving simplified f(x) = x + 2 with a gap at x = 2.

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is $y = 0$ (graph flattens toward $x \to \pm \infty$), (2) if degrees equal, HA is $y = \frac{\text{numerator leading coefficient}}{\text{denominator leading coefficient}}$ (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For f(x) = (3x^2 -7)/(2x^2 +5x -1), both numerator and denominator are degree 2, so the horizontal asymptote is $y = \frac{3}{2}$ from the ratio of leading coefficients $\frac{3}{2}$. Choice C correctly applies the equal degrees rule to find $y = \frac{3}{2}$. A distractor like Choice A might mistakenly use non-leading coefficients, such as $\frac{2}{3}$ from constants, but always use the leading coefficients for equal degrees. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → $y = 0$; degrees equal → $y = $ ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions $f(x) = \frac{p(x)}{q(x)}$ have distinctive features: zeros where $p(x) = 0$ (numerator equals zero, these are x-intercepts), vertical asymptotes where $q(x) = 0$ (denominator equals zero, graph shoots to $\pm\infty$), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have $(x - 2)$), that creates a hole (removable discontinuity) at $x = 2$, not a zero or asymptote—the common factor cancels! For $$f(x) = \frac{(x-5)(x+2)}{(x-5)(x-1)}$$, factor and cancel the common $(x-5)$ to get simplified $\frac{x+2}{x-1}$ with a hole at $x=5$ where $y=\frac{5+2}{5-1}=\frac{7}{4}$, zero at $x=-2$, vertical asymptote at $x=1$, and horizontal asymptote $y=1$ since degrees match. Choice A correctly identifies the zero at $x=-2$, vertical asymptote at $x=1$, hole at $x=5$ with $y=\frac{7}{4}$, and horizontal asymptote $y=1$. Distractors like Choice B fail to recognize the common factor, treating $x=5$ as a zero instead of a hole, but always check for cancellations first. Hole detection: before finalizing zeros and asymptotes, factor both numerator and denominator completely and check for common factors. If $(x - 3)$ appears in both, it cancels, creating a hole at $x = 3$ (point missing from graph) rather than a zero or asymptote. Calculate the y-coordinate of the hole by substituting $x = 3$ into the simplified function. Holes are easy to miss—always check for common factors! Example: $$\frac{x^2 - 4}{x - 2} = \frac{(x + 2)(x - 2)}{x - 2}$$ has a hole at $x = 2$ (cancels), leaving simplified $f(x) = x + 2$ with a gap at $x = 2$.

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Horizontal asymptotes depend on degree comparison: (1) if numerator degree less than denominator degree, HA is $y = 0$ (graph flattens toward x-axis as $x \to \pm \infty$), (2) if degrees equal, HA is $y = ($numerator leading coefficient$)/($denominator leading coefficient$)$ (graph approaches this horizontal line), (3) if numerator degree exceeds denominator by exactly 1, there's an oblique (slant) asymptote found by polynomial division, (4) if numerator degree exceeds by 2+, no horizontal or oblique asymptote—end behavior is more like a polynomial. These degree rules determine long-term graph behavior! For $f(x) = \frac{2x-5}{x^2+1}$, numerator degree 1 is less than denominator degree 2, so horizontal asymptote is $y=0$. Choice C correctly applies the degree comparison rule for numerator degree < denominator degree to find $y=0$. A distractor like Choice D confuses it with oblique asymptote criteria, but oblique requires numerator degree exactly one higher, not lower. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: deg(num) < deg(den) → $y = 0$; degrees equal → $y =$ ratio of leading coefficients; deg(num) > deg(den) → no HA. (4) OBLIQUE ASYMPTOTE: if deg(num) = deg(den) + 1, divide to find it. Follow these steps systematically!

This question tests your ability to graph rational functions by finding zeros (from the numerator), vertical asymptotes (from the denominator), and horizontal or oblique asymptotes (from comparing degrees and using end behavior). Rational functions $f(x) = \frac{p(x)}{q(x)}$ have distinctive features: zeros where $p(x) = 0$ (numerator equals zero, these are x-intercepts), vertical asymptotes where $q(x) = 0$ (denominator equals zero, graph shoots to $\pm\infty$), and horizontal or oblique asymptotes describing end behavior. The key is: numerator gives zeros, denominator gives vertical asymptotes. IMPORTANT: if numerator and denominator share a factor (like both have $(x - 2)$), that creates a hole (removable discontinuity) at $x = 2$, not a zero or asymptote—the common factor cancels! Factoring gives numerator $(x+3)(x-2)$ and denominator $(x+4)(x-4)$, with no common factors, so zeros at $x=-3$ and $x=2$, vertical asymptotes at $x=4$ and $x=-4$. Choice B correctly identifies zeros at $x=2,-3$ and vertical asymptotes at $x=4,-4$. A distractor like Choice A swaps them, perhaps confusing numerator and denominator roles, but zeros are always from numerator roots. The rational function feature-finding roadmap: (1) ZEROS: set numerator = 0, solve (these are x-intercepts), (2) VERTICAL ASYMPTOTES: set denominator = 0, solve (but check for common factors with numerator—if common, it's a hole, not VA!), (3) HORIZONTAL ASYMPTOTE: compare degrees: $\deg(\text{num}) < \deg(\text{den}) \rightarrow y = 0$; degrees equal $\rightarrow y =$ ratio of leading coefficients; $\deg(\text{num}) > \deg(\text{den}) \rightarrow$ no HA. (4) OBLIQUE ASYMPTOTE: if $\deg(\text{num}) = \deg(\text{den}) + 1$, divide to find it. Follow these steps systematically!