These skills are heavily tested in graph-matching form ("The graph in the xy-plane of which of the following equations..."). One calculator caution up front: Section 1's online graphing calculator will never show you a hole — a hole is a single missing point, invisible at pixel resolution — so the algebra below is the only reliable detector.
A rational function r(x) = p(x)/q(x) (polynomial over polynomial) is undefined wherever q(x) = 0. Every such x-value is out of the domain; the only question is what the graph does near it.
Factor top and bottom completely. For each x-value a that zeroes either one, compare multiplicities — m in the numerator, n in the denominator:
| Case | Behavior at x = a |
|---|---|
| m ≥ 1, n = 0 | zero of r; x-intercept (a, 0) |
| m < n | vertical asymptote x = a |
| m ≥ n ≥ 1 | hole at x = a (point missing from the graph) |
Slogan: denominator wins → asymptote; numerator ties or wins → hole.
To find a hole's y-coordinate, cancel the common factors and evaluate what's left:
r(x) = (x − 3)(x + 3) / [(x − 3)(x + 2)] → simplified: (x + 3)/(x + 2)
hole at x = 3: y = 6/5 → hole at (3, 6/5)
A hole at height 0 is still not an x-intercept — the function is undefined there, so the graph skips the point.
Near a vertical asymptote, outputs grow without bound — the graph shoots up or down along the dashed line. The two sides can disagree; a quick sign check (test a value just left and just right) tells which way each branch goes. When the asymptote factor's net power is even, as in 1/(x − 4)², both branches point the same way.
For large |x|, a rational function behaves like the ratio of its leading terms. With numerator degree n and denominator degree d:
Don't recall the rule from memory — recompute the leading-term ratio each time; it takes five seconds. Two facts the exam tests directly: a graph can cross its horizontal asymptote at moderate x-values (the asymptote only governs the far ends), and both far ends approach the same horizontal line in regimes 1 and 2.
① Factor top and bottom. ② List every x that zeroes either. ③ Classify each (zero / hole / vertical asymptote). ④ Horizontal asymptote by degree comparison. ⑤ y-intercept = r(0). Five steps, and every graph-matching question about that function is pre-answered.
[GRAPH: r(x) = 2x/(x − 1) in the xy-plane. Dashed vertical asymptote x = 1 and dashed horizontal asymptote y = 2. Left branch passes through the origin (0, 0), falling toward y = 2 on the far left and plunging downward as x approaches 1 from the left. Right branch descends from high above near x = 1 toward y = 2 on the far right. Annotation: "VA where denominator = 0; HA = ratio of leading coefficients 2/1".]
Power functions y = xⁿ: even n gives a U-shaped graph symmetric about the y-axis (through (−1, 1), (0, 0), (1, 1)); odd n gives an S-through-the-origin shape symmetric about the origin (through (−1, −1), (0, 0), (1, 1)). Higher n flattens the graph near the origin and steepens it beyond |x| = 1.
Root functions: y = √x has domain x ≥ 0 and range y ≥ 0 — half a sideways parabola, rising ever more slowly. Shifts move the domain: √(x − 4) requires x − 4 ≥ 0, so x ≥ 4 (the endpoint is included — no asymptote here, the graph starts at the point (4, 0)). The cube root y = ∛x accepts every real number, positive or negative, and passes through the origin.
y = |x| is a V with vertex at the origin and slopes −1, +1. The transformed family
y = a|x − h| + k vertex (h, k), opens up if a > 0, down if a < 0
Read graphs the same way you read vertex form for parabolas: locate the vertex, check the opening direction, confirm with one more point. Inside the bars, (x − h) shifts right by h — the sign trap is identical to the parabola's.
A piecewise function applies different formulas on different pieces of the domain:
f(x) = { x² if x < 1
{ 3x − 2 if x ≥ 1
To evaluate, first ask which condition does the input satisfy, then use only that branch: f(−2) = (−2)² = 4 (since −2 < 1), f(3) = 3(3) − 2 = 7 (since 3 ≥ 1). When graphing, mark included endpoints with solid dots and excluded ones with open dots; the pieces need not connect.
| Graph signature | Family |
|---|---|
| straight line | linear |
| single U or ∩, one turning point | quadratic |
| S-curve or multiple turning points, no breaks | higher polynomial |
| breaks: vertical asymptotes or holes; levels off sideways | rational |
| sharp corner (V) | absolute value |
| starts at an endpoint, curves one way | square root |
| disconnected formulas, open/solid dots | piecewise |
Questions 1–3 refer to the function r defined by r(x) = (x² − 9)/(x² − x − 6).
Q1. C. Factored: r(x) = (x − 3)(x + 3)/[(x − 3)(x + 2)]. At x = −3 the numerator alone is zero → x-intercept. At x = 3 both are zero (a hole — not in the domain, so not a zero of r). A) counts the hole at 3 as a zero, but r(3) is undefined. B) picks the hole and discards the actual zero. D) throws in the asymptote value −2 as well. E) mistakes the denominator zero for a zero of r. Fix: a zero must survive in the domain — check the denominator before declaring any x-intercept.
Q2. D. x = −2 zeroes the denominator only (numerator multiplicity 0 < denominator multiplicity 1) → vertical asymptote. x = 3 cancels (multiplicities tie) → hole, not asymptote. A) treats every denominator zero of the original as an asymptote without checking cancellation. B) picks the cancelled value only. C) lists the numerator's zeros. E) sign error on the factor (x + 2). Fix: only denominator zeros that do NOT fully cancel produce vertical asymptotes — compare multiplicities first.
Q3. D. Cancel the common factor (x − 3), leaving (x + 3)/(x + 2); evaluating at x = 3 gives 6/5, so the hole is (3, 6/5). A) assumes holes sit on the x-axis. B) puts the hole at the asymptote's x-value. C) inverts the fraction, computing (x + 2)/(x + 3) at 3. E) reports the zero as a hole. Fix: hole height = simplified form evaluated at the cancelled x-value.
Q4. B. Degrees are equal (3 = 3), so the graph levels off at the ratio of leading coefficients: 2/5. A) applies the bottom-heavy rule (y = 0) though the degrees match. C) inverts the ratio. D) uses the numerator's leading coefficient alone. E) applies the top-heavy rule (no asymptote) though the degrees match. Fix: don't recall the horizontal-asymptote rule — recompute the leading-term ratio 2x³/5x³ = 2/5 each time.
Q5. E. (x + 1) appears top and bottom → hole at −1; (x − 5) bottom only → vertical asymptote at 5; (x − 2) top only → zero at 2. A) swaps the roles: hole at 5, asymptote at −1. B) puts the asymptote at 2 and the zero at 5. C) leaves (x + 1) in the denominator only, making x = −1 an asymptote instead of a hole. D) flips the signs, placing the hole at 1 and the asymptote at −5. Fix: each feature is a factor-placement instruction — zero: top only; asymptote: bottom only; hole: both.
Q6. B. Bottom-heavy (numerator degree 2 < denominator degree 4): outputs shrink toward 0, so y = 0. A) takes the ratio of leading coefficients, which applies only when degrees are equal. C) divides the constant terms. D) uses the larger degree as the asymptote's height. E) inverts the comparison, treating a bottom-heavy function as top-heavy. Fix: compare degrees first — smaller on top means the horizontal asymptote is y = 0, regardless of coefficients.
Q7. A. The denominator is x² + 1, which is at least 1 for every real x and never zero, so nothing is excluded: the domain is all real numbers. The student excluded the zeros of the numerator — those are x-intercepts, which belong in the domain. B) endorses exactly that numerator/denominator reversal. C) solves x² + 1 = 0 as if it were x² − 1 = 0. D) half-accepts the faulty exclusion. E) invents a single excluded value with no algebraic source. Fix: domain restrictions come from the denominator only — and a denominator with no real zeros restricts nothing.
Q8. C. The expression under a square root must be nonnegative: x − 4 ≥ 0, so x ≥ 4; the endpoint 4 is included, where f(4) = 0. A) wrongly excludes the endpoint — square-root domains are closed at the boundary (unlike log domains). B) solves x + 4 ≥ 0, a sign slip. D) reverses the inequality. E) confuses the square root with the cube root, which does accept all reals. Fix: set the radicand ≥ 0 (with equality allowed) and solve; only even roots restrict the domain.
Q9. C. −2 < 1, so f(−2) uses the first branch: (−2)² = 4; 3 ≥ 1, so f(3) uses the second: 3(3) − 2 = 7; the sum is 11. A) uses the wrong branch for both inputs (−8 + 9). B) applies 3x − 2 to −2, getting −8 + 7 = −1. D) squares the 3, using the wrong branch for the second input (4 + 9). E) reports f(3) alone and forgets the first term. Fix: for each input, check the condition FIRST, then substitute into that branch only.
Q10. A. Vertex (2, −3) with upward opening and unit slopes is y = |x − 2| − 3; check (0, −1): |−2| − 3 = −1 ✓ and (5, 0): |3| − 3 = 0 ✓. B) shifts left instead of right — inside the bars, (x − 2) moves the corner to +2. C) shifts up 3 instead of down, putting the vertex at (2, 3). D) opens downward. E) swaps the two shift amounts, moving the vertex to (3, −2). Fix: read absolute-value graphs like vertex form — corner at (h, k) from a|x − h| + k, with the inside sign reversed from the shift direction.
Q11. D. Required features: vertical asymptote x = 1 (denominator zero at 1), horizontal asymptote y = 2 (equal degrees, leading ratio 2/1), and passing through (0, 0). y = 2x/(x − 1) delivers all three. A) puts the vertical asymptote at x = 2 and levels off at y = 1. B) has the asymptote at x = −1, a sign error. C) has the right vertical asymptote but the 2 in the denominator makes the horizontal asymptote y = 1/2. E) swaps numerator and denominator roles: asymptote at x = 0, zero at x = 1. Fix: match graph to formula feature-by-feature — VA from the denominator's zero, HA from the leading ratio, then confirm one plotted point.
Q12. E. Since x² − 1 = (x − 1)(x + 1), the factor cancels: r agrees with y = x + 1 everywhere EXCEPT x = 1, where the original denominator is zero and r is undefined — a hole at (1, 2). The calculator can't display a one-point gap, which is precisely why the conclusion fails. A) forgets that cancelling a factor doesn't repair the domain. B) invents a sign condition; the hole is at x = 1 regardless. C) misclassifies the cancelled factor — matching multiplicities give a hole, not an asymptote. D) botches the simplification; (x² − 1)/(x − 1) = x + 1, not x − 1. Fix: record excluded x-values BEFORE cancelling — the cancelled factor leaves a hole the calculator will never show.
Q1. C. Factored: r(x) = (x − 3)(x + 3)/[(x − 3)(x + 2)]. At x = −3 the numerator alone is zero → x-intercept. At x = 3 both are zero (a hole — not in the domain, so not a zero of r). A) counts the hole at 3 as a zero, but r(3) is undefined. B) picks the hole and discards the actual zero. D) throws in the asymptote value −2 as well. E) mistakes the denominator zero for a zero of r. Fix: a zero must survive in the domain — check the denominator before declaring any x-intercept.
Q2. D. x = −2 zeroes the denominator only (numerator multiplicity 0 < denominator multiplicity 1) → vertical asymptote. x = 3 cancels (multiplicities tie) → hole, not asymptote. A) treats every denominator zero of the original as an asymptote without checking cancellation. B) picks the cancelled value only. C) lists the numerator's zeros. E) sign error on the factor (x + 2). Fix: only denominator zeros that do NOT fully cancel produce vertical asymptotes — compare multiplicities first.
Q3. D. Cancel the common factor (x − 3), leaving (x + 3)/(x + 2); evaluating at x = 3 gives 6/5, so the hole is (3, 6/5). A) assumes holes sit on the x-axis. B) puts the hole at the asymptote's x-value. C) inverts the fraction, computing (x + 2)/(x + 3) at 3. E) reports the zero as a hole. Fix: hole height = simplified form evaluated at the cancelled x-value.
Q4. B. Degrees are equal (3 = 3), so the graph levels off at the ratio of leading coefficients: 2/5. A) applies the bottom-heavy rule (y = 0) though the degrees match. C) inverts the ratio. D) uses the numerator's leading coefficient alone. E) applies the top-heavy rule (no asymptote) though the degrees match. Fix: don't recall the horizontal-asymptote rule — recompute the leading-term ratio 2x³/5x³ = 2/5 each time.
Q5. E. (x + 1) appears top and bottom → hole at −1; (x − 5) bottom only → vertical asymptote at 5; (x − 2) top only → zero at 2. A) swaps the roles: hole at 5, asymptote at −1. B) puts the asymptote at 2 and the zero at 5. C) leaves (x + 1) in the denominator only, making x = −1 an asymptote instead of a hole. D) flips the signs, placing the hole at 1 and the asymptote at −5. Fix: each feature is a factor-placement instruction — zero: top only; asymptote: bottom only; hole: both.
Q6. B. Bottom-heavy (numerator degree 2 < denominator degree 4): outputs shrink toward 0, so y = 0. A) takes the ratio of leading coefficients, which applies only when degrees are equal. C) divides the constant terms. D) uses the larger degree as the asymptote's height. E) inverts the comparison, treating a bottom-heavy function as top-heavy. Fix: compare degrees first — smaller on top means the horizontal asymptote is y = 0, regardless of coefficients.
Q7. A. The denominator is x² + 1, which is at least 1 for every real x and never zero, so nothing is excluded: the domain is all real numbers. The student excluded the zeros of the numerator — those are x-intercepts, which belong in the domain. B) endorses exactly that numerator/denominator reversal. C) solves x² + 1 = 0 as if it were x² − 1 = 0. D) half-accepts the faulty exclusion. E) invents a single excluded value with no algebraic source. Fix: domain restrictions come from the denominator only — and a denominator with no real zeros restricts nothing.
Q8. C. The expression under a square root must be nonnegative: x − 4 ≥ 0, so x ≥ 4; the endpoint 4 is included, where f(4) = 0. A) wrongly excludes the endpoint — square-root domains are closed at the boundary (unlike log domains). B) solves x + 4 ≥ 0, a sign slip. D) reverses the inequality. E) confuses the square root with the cube root, which does accept all reals. Fix: set the radicand ≥ 0 (with equality allowed) and solve; only even roots restrict the domain.
Q9. C. −2 < 1, so f(−2) uses the first branch: (−2)² = 4; 3 ≥ 1, so f(3) uses the second: 3(3) − 2 = 7; the sum is 11. A) uses the wrong branch for both inputs (−8 + 9). B) applies 3x − 2 to −2, getting −8 + 7 = −1. D) squares the 3, using the wrong branch for the second input (4 + 9). E) reports f(3) alone and forgets the first term. Fix: for each input, check the condition FIRST, then substitute into that branch only.
Q10. A. Vertex (2, −3) with upward opening and unit slopes is y = |x − 2| − 3; check (0, −1): |−2| − 3 = −1 ✓ and (5, 0): |3| − 3 = 0 ✓. B) shifts left instead of right — inside the bars, (x − 2) moves the corner to +2. C) shifts up 3 instead of down, putting the vertex at (2, 3). D) opens downward. E) swaps the two shift amounts, moving the vertex to (3, −2). Fix: read absolute-value graphs like vertex form — corner at (h, k) from a|x − h| + k, with the inside sign reversed from the shift direction.
Q11. D. Required features: vertical asymptote x = 1 (denominator zero at 1), horizontal asymptote y = 2 (equal degrees, leading ratio 2/1), and passing through (0, 0). y = 2x/(x − 1) delivers all three. A) puts the vertical asymptote at x = 2 and levels off at y = 1. B) has the asymptote at x = −1, a sign error. C) has the right vertical asymptote but the 2 in the denominator makes the horizontal asymptote y = 1/2. E) swaps numerator and denominator roles: asymptote at x = 0, zero at x = 1. Fix: match graph to formula feature-by-feature — VA from the denominator's zero, HA from the leading ratio, then confirm one plotted point.
Q12. E. Since x² − 1 = (x − 1)(x + 1), the factor cancels: r agrees with y = x + 1 everywhere EXCEPT x = 1, where the original denominator is zero and r is undefined — a hole at (1, 2). The calculator can't display a one-point gap, which is precisely why the conclusion fails. A) forgets that cancelling a factor doesn't repair the domain. B) invents a sign condition; the hole is at x = 1 regardless. C) misclassifies the cancelled factor — matching multiplicities give a hole, not an asymptote. D) botches the simplification; (x² − 1)/(x − 1) = x + 1, not x − 1. Fix: record excluded x-values BEFORE cancelling — the cancelled factor leaves a hole the calculator will never show.