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) = (x²-4)/(x-3), factor the numerator: x²-4 = (x-2)(x+2). No common factors exist with denominator (x-3). Zeros occur when (x-2)(x+2) = 0, giving x = 2 and x = -2. Vertical asymptote occurs when x-3 = 0, giving x = 3. Since numerator has degree 2 and denominator has degree 1 (degree difference = 1), there's an oblique asymptote. Perform polynomial division: (x²-4)÷(x-3) = x+3 with remainder 5, so f(x) = x+3 + 5/(x-3). The oblique asymptote is y = x+3. Choice A correctly identifies zeros at x = ±2, VA at x = 3, and oblique asymptote y = x+3. Choice D incorrectly claims no VA and gives the wrong oblique asymptote equation. 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) = [(x-3)(x+1)] / [(x-3)(x-2)], common (x-3) cancels, leaving (x+1)/(x-2) with hole at x=3; zero at x=-1, VA at x=2. Choice A correctly lists zero at x=-1, VA at x=2, hole at x=3. Distractors like choice B ignore the hole, listing zero at x=3 incorrectly—remember, holes mean undefined, not zero! 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! 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. You're a pro at holes now!

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) = (3x-1)/(x²+4), the numerator has degree 1 and denominator has degree 2. Since degree(numerator) < degree(denominator), the horizontal asymptote is y = 0. As x → ±∞, the denominator grows much faster than the numerator, causing f(x) → 0. Think of it as 3x/x² ≈ 3/x → 0 for large |x|. Choice B correctly states that as x → ±∞, f(x) → 0 with horizontal asymptote y = 0. Choice A incorrectly claims y = 3, likely confusing this with the case of equal degrees, while Choice D incorrectly suggests a slant asymptote when the numerator degree is actually less than the denominator degree. 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) = (x²-9)/(x²-4x-5), first factor: numerator x²-9 = (x-3)(x+3), and denominator x²-4x-5 = (x-5)(x+1). Zeros occur when (x-3)(x+3) = 0, giving x = 3 and x = -3. Vertical asymptotes occur when (x-5)(x+1) = 0, giving x = 5 and x = -1. No common factors exist, so no holes. Choice A correctly identifies zeros at x = 3, -3 and vertical asymptotes at x = 5, -1. Choice B incorrectly swaps zeros and VAs—a common mistake when not carefully tracking which features come from numerator vs denominator. 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 $f(x) = \frac{2x^2 - 8}{x^2 + 1}$, 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! For f(x) = (x² - 9)/(x + 2), first factor the numerator: x² - 9 = (x + 3)(x - 3). So f(x) = (x + 3)(x - 3)/(x + 2). Zeros occur when numerator = 0: (x + 3)(x - 3) = 0 gives x = -3 and x = 3. The vertical asymptote occurs when denominator = 0: x + 2 = 0 gives x = -2. Since the numerator has degree 2 and denominator has degree 1, and deg(num) = deg(den) + 1, there's an oblique asymptote. Dividing x² - 9 by x + 2: First term: x² ÷ x = x. Multiply: x(x + 2) = x² + 2x. Subtract: (x² - 9) - (x² + 2x) = -2x - 9. Second term: -2x ÷ x = -2. Multiply: -2(x + 2) = -2x - 4. Subtract: (-2x - 9) - (-2x - 4) = -5. So f(x) = x - 2 + (-5)/(x + 2), giving oblique asymptote y = x - 2. Choice A correctly identifies zeros at x = ±3, VA at x = -2, and oblique asymptote y = x - 2. Choice D incorrectly has VA at x = 2 instead of x = -2—remember to solve x + 2 = 0, not confuse the sign! 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! Always factor completely before identifying features—this helps spot patterns and makes calculations easier.

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) = (x²+3x+2)/(x+1), the numerator has degree 2 and denominator has degree 1, so degree(num) = degree(den) + 1, indicating an oblique asymptote. To find it, perform polynomial division: (x²+3x+2)÷(x+1). Using long division or factoring first: x²+3x+2 = (x+1)(x+2), so f(x) = (x+1)(x+2)/(x+1) = x+2 (except at x = -1 where there's a hole). The oblique asymptote is y = x+2. Choice A correctly identifies the oblique asymptote as y = x+2. Choice B incorrectly gives y = x+1, likely from a division error, while Choice D incorrectly claims a horizontal asymptote when the degree difference clearly indicates an oblique asymptote. 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) = (x-2)(x+1)/(x-4)(x+3), zeros occur when (x-2)(x+1) = 0, giving x = 2 and x = -1. Vertical asymptotes occur when (x-4)(x+3) = 0, giving x = 4 and x = -3. Since both numerator and denominator have degree 2, the horizontal asymptote is y = 1/1 = 1 (ratio of leading coefficients). Choice B correctly identifies zeros at x = 2, -1, vertical asymptotes at x = 4, -3, and horizontal asymptote y = 1. Choice A incorrectly swaps the zeros and vertical asymptotes—remember, zeros come from numerator, VAs from denominator! 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 → ±∞), (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 → ±∞), (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) = (x²+x-6)/(x-2), first factor the numerator: x²+x-6 = (x+3)(x-2). Notice (x-2) appears in both numerator and denominator! This common factor cancels, creating a hole at x = 2, not a vertical asymptote. After canceling, f(x) = x+3 (except at x = 2). Since the original function has degree 2 numerator and degree 1 denominator (degree difference = 1), we perform polynomial division: (x²+x-6)÷(x-2) = x+3 with remainder 0, confirming the oblique asymptote y = x+3. Choice C correctly identifies a hole at x = 2 and oblique asymptote y = x+3. Choice A incorrectly claims a vertical asymptote at x = 2—missing that the common factor 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.