Graph-identification items ("Which of the following could be an equation of the function graphed in the xy-plane above?") appear throughout both sections of the CLEP exam. Everything in this lesson runs by hand — these are staple no-calculator skills for Section 2, though Section 1's online graphing calculator can confirm a vertex or zero when one is available.
The slope of the line through (x₁, y₁) and (x₂, y₂) is
m = (y₂ − y₁)/(x₂ − x₁)
Three interchangeable forms, chosen by what the problem hands you:
| Form | Statement | Use when given |
|---|---|---|
| Slope-intercept | y = mx + b | slope and y-intercept |
| Point-slope | y − y₁ = m(x − x₁) | slope and any point |
| Standard | Ax + By = C | intercepts are wanted |
Given two points, compute m first, then use point-slope with either point. Two classic slips to guard against: reversing the subtraction order in only one of the two coordinates (which flips the sign of m), and back-substituting for b while dropping the slope's sign. A linear function has the same rate of change m over every interval — that constancy is its defining feature.
Given a table with equally spaced inputs (step h):
| x | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| f(x) | 0 | 1 | 4 | 9 | 16 |
| Δ | 1 | 3 | 5 | 7 | |
| Δ² | 2 | 2 | 2 |
Increasing first differences alone prove nothing — exponential data also grow by increasing amounts. The quadratic signature is specifically that the differences of the differences are constant.
For f(x) = ax² + bx + c:
Building a quadratic from its zeros: f(x) = a(x − z₁)(x − z₂), then pin a with one more point (often the y-intercept or the stated max/min value).
[GRAPH: Parabola f(x) = 2x² − 16x + 30 in the xy-plane, opening upward. Vertex (4, −2) marked as the minimum; dashed axis of symmetry x = 4; x-intercepts (3, 0) and (5, 0) symmetric about the axis; y-intercept (0, 30) off the top of the frame, arrow indicating it. Annotation: "axis x = −b/(2a) = 4; minimum value f(4) = −2".]
For a polynomial p, these say the same thing: a is a zero of p ⟺ (x − a) is a factor ⟺ the graph has an x-intercept at (a, 0).
If (x − a) appears to the power m (the multiplicity):
Anchor the rule to two pictures: y = (x − 2)² is a shifted parabola (touches); y = (x − 2)³ is a shifted cubic (flattens and crosses). The multiplicities across all factors add up to the degree.
For large |x|, a polynomial behaves like its leading term aₙxⁿ — every lower-degree term becomes a rounding error, no matter how large its coefficient. (0.001x³ eventually outgrows 1,000,000x².) Four cases:
| Degree | Leading coefficient | Far left | Far right |
|---|---|---|---|
| even | positive | rises | rises |
| even | negative | falls | falls |
| odd | positive | falls | rises |
| odd | negative | rises | falls |
Don't memorize the table — memorize the shapes of x², −x², x³, −x³ and match. For a factored polynomial, the degree is the sum of the factors' exponents and the leading coefficient is the product of the constants out front and each factor's leading 1.
Given a graph and five candidate equations, don't expand anything. Check in order:
One candidate survives; the whole routine takes under a minute.
Q1. E. m = (2 − 10)/(7 − 3) = −2; then b = 10 − (−2)(3) = 16, so y = −2x + 16 (both points check: −2·3 + 16 = 10 and −2·7 + 16 = 2). A) drops the sign of the slope; y = 2x + 4 happens to pass through (3, 10), which makes the error feel confirmed, but it misses (7, 2). B) has the correct slope with b computed as 10 − 2·3 = 4 — the slope's negative sign was dropped during back-substitution. C) inverts the slope, computing Δx/Δy = 4/(−8) = −1/2. D) drops the slope's sign and fits only the second point. Fix: after writing any line, plug in BOTH given points — a sign error always fails one of them.
Q2. B. For 2, 5, 12, 23, 38: first differences 3, 7, 11, 15; second differences 4, 4, 4 — constant, the quadratic signature. A) has constant ratios (each output doubles): exponential, not quadratic — its second differences 3, 6, 12 grow. C) has constant first differences (3): linear. D) has second differences 1, 1, 2 — close but not constant, so no family is confirmed. E) is constant (a degenerate linear function, slope 0). Fix: run the two-step test — first differences constant means linear; only constant SECOND differences mean quadratic.
Q3. A. For a quadratic sampled at step h, the constant second difference equals 2ah², so 8 = 2a(2²) = 8a and a = 1. B) uses h instead of h², computing 8/(2·2). C) divides by 2 only, ignoring the step size entirely. D) reports the second difference itself as the coefficient. E) multiplies by the step instead of dividing. Fix: a = Δ²/(2h²) — the step size enters squared, and it matters whenever h ≠ 1.
Q4. E. The vertex is at x = −b/(2a) = 16/4 = 4, and h(4) = 2(16) − 64 + 30 = −2; since a = 2 > 0 the parabola opens upward, so −2 is the minimum value. A) reports the x-coordinate of the vertex, not the minimum value — the most common error on this item type. B) is h(0), the y-intercept. C) drops the sign of the final arithmetic. D) is −b/(2a) computed with a sign slip. Fix: "minimum value" asks for the OUTPUT at the vertex — find the vertex x, then evaluate the function there.
Q5. D. Zeros −1 and 7 force factors (x + 1)(x − 7); the vertex sits at their midpoint x = 3, and f(3) = a(4)(−4) = −16a = 32 gives a = −2, so f(x) = −2(x + 1)(x − 7). A negative leading coefficient is also required for a maximum, and −2 delivers it. A) has a > 0, which opens upward and produces a minimum, not a maximum. B) flips the signs inside the factors, moving the zeros to 1 and −7. C) has the right zeros and opens downward, but its maximum is −(4)(−4) = 16, not 32. E) combines both errors. Fix: build from zeros as a(x − z₁)(x − z₂), then use the stated max/min at the midpoint of the zeros to solve for a.
Q6. D. The factor (x + 2)² has even multiplicity (2), so the graph touches the x-axis at x = −2 and turns back without crossing. A) x = 1 has multiplicity 1 (odd) — the graph crosses there. B) x = 5 has multiplicity 3 (odd) — the graph flattens but still crosses. C) x = 3 is not a zero at all; 3 is the leading constant. E) x = 2 flips the sign of the factor (x + 2). Fix: even multiplicity touches, odd multiplicity crosses — and the zero of (x + a) is x = −a, not x = a.
Q7. A. The leading term is −4x⁶: even degree makes both ends point the same way, and the negative coefficient points them down — the graph falls on both the far left and the far right. B) reverses the sign of the leading coefficient. C) and D) treat the visible 50x⁵ as the leading term; a degree-5 term never outruns a degree-6 term, no matter how large its coefficient. E) confuses the constant term −3 with a horizontal asymptote — polynomials have none. Fix: end behavior is decided by the single highest-degree term; degree beats coefficient, always, eventually.
Q8. C. A polynomial of degree n has at most n − 1 turning points: 5 − 1 = 4. A) is an undercount with no rule behind it. B) applies n − 2, the bound for inflection-type behavior, not turning points. D) uses the degree itself. E) uses n + 1. Fix: turning points ≤ degree − 1 — and remember it's "at most," not "exactly."
Q9. D. The graph crosses at −3 (factor (x + 3), odd multiplicity), touches at 2 (factor (x − 2)², even multiplicity), giving degree 3; falling-left/rising-right matches an odd degree with positive leading coefficient, and the y-intercept checks: (3)(−2)² = 12. A) crosses at both intercepts — it can't produce the touch at x = 2. B) swaps the multiplicities, touching at −3 and crossing at 2. C) flips the signs of the zeros to 3 and −2. E) has the right factors but a negative leading coefficient, which rises left and falls right and gives y-intercept −12. Fix: check end behavior first, then cross-vs-touch at each intercept, then the y-intercept — the three-check routine.
Q10. E. Four turning points require degree ≥ 5; both ends falling requires an even degree (with negative leading coefficient). The smallest degree that is both even and ≥ 5 is 6. A) degree 3 allows at most 2 turning points. B) degree 4 allows at most 3. C) and D) are odd degrees, whose ends must point in opposite directions. Fix: turning points set a floor (degree ≥ turning points + 1); matching ends force even degree — both constraints must hold at once.
Q11. B. Quadratic data over equal steps must have constant second differences; here the first differences are 4, 8, 16, so the second differences are 4 and 8 — not constant. (The constant ratio of differences actually suggests exponential-type growth.) A) "increasing outputs" describes many families and confirms none. C) states the linear test, not the quadratic test — constant first differences would rule OUT a quadratic. D) positive second differences indicate upward bending, not quadratic form; they must be constant, not merely positive. E) is false — an upward-opening quadratic increases forever on the right. Fix: to claim "quadratic" from a table, compute second differences and check that they are constant, not just growing or positive.
Q12. C. x = 4 is indeed a zero, but the factor (x − 4) appears squared — even multiplicity — so the graph touches the axis at (4, 0) and turns back up; (x − 4)² ≥ 0 on both sides, so there is no crossing. A) overgeneralizes: crossing happens only at odd-multiplicity zeros. B) wrongly denies the zero — the student's premise was fine; the inference failed. D) the direction of opening explains which way it turns, not whether it crosses. E) confuses the zero of (x − 4) with x = −4. Fix: identifying a zero is step one; its multiplicity's parity — not the zero itself — decides cross versus touch.
Q1. E. m = (2 − 10)/(7 − 3) = −2; then b = 10 − (−2)(3) = 16, so y = −2x + 16 (both points check: −2·3 + 16 = 10 and −2·7 + 16 = 2). A) drops the sign of the slope; y = 2x + 4 happens to pass through (3, 10), which makes the error feel confirmed, but it misses (7, 2). B) has the correct slope with b computed as 10 − 2·3 = 4 — the slope's negative sign was dropped during back-substitution. C) inverts the slope, computing Δx/Δy = 4/(−8) = −1/2. D) drops the slope's sign and fits only the second point. Fix: after writing any line, plug in BOTH given points — a sign error always fails one of them.
Q2. B. For 2, 5, 12, 23, 38: first differences 3, 7, 11, 15; second differences 4, 4, 4 — constant, the quadratic signature. A) has constant ratios (each output doubles): exponential, not quadratic — its second differences 3, 6, 12 grow. C) has constant first differences (3): linear. D) has second differences 1, 1, 2 — close but not constant, so no family is confirmed. E) is constant (a degenerate linear function, slope 0). Fix: run the two-step test — first differences constant means linear; only constant SECOND differences mean quadratic.
Q3. A. For a quadratic sampled at step h, the constant second difference equals 2ah², so 8 = 2a(2²) = 8a and a = 1. B) uses h instead of h², computing 8/(2·2). C) divides by 2 only, ignoring the step size entirely. D) reports the second difference itself as the coefficient. E) multiplies by the step instead of dividing. Fix: a = Δ²/(2h²) — the step size enters squared, and it matters whenever h ≠ 1.
Q4. E. The vertex is at x = −b/(2a) = 16/4 = 4, and h(4) = 2(16) − 64 + 30 = −2; since a = 2 > 0 the parabola opens upward, so −2 is the minimum value. A) reports the x-coordinate of the vertex, not the minimum value — the most common error on this item type. B) is h(0), the y-intercept. C) drops the sign of the final arithmetic. D) is −b/(2a) computed with a sign slip. Fix: "minimum value" asks for the OUTPUT at the vertex — find the vertex x, then evaluate the function there.
Q5. D. Zeros −1 and 7 force factors (x + 1)(x − 7); the vertex sits at their midpoint x = 3, and f(3) = a(4)(−4) = −16a = 32 gives a = −2, so f(x) = −2(x + 1)(x − 7). A negative leading coefficient is also required for a maximum, and −2 delivers it. A) has a > 0, which opens upward and produces a minimum, not a maximum. B) flips the signs inside the factors, moving the zeros to 1 and −7. C) has the right zeros and opens downward, but its maximum is −(4)(−4) = 16, not 32. E) combines both errors. Fix: build from zeros as a(x − z₁)(x − z₂), then use the stated max/min at the midpoint of the zeros to solve for a.
Q6. D. The factor (x + 2)² has even multiplicity (2), so the graph touches the x-axis at x = −2 and turns back without crossing. A) x = 1 has multiplicity 1 (odd) — the graph crosses there. B) x = 5 has multiplicity 3 (odd) — the graph flattens but still crosses. C) x = 3 is not a zero at all; 3 is the leading constant. E) x = 2 flips the sign of the factor (x + 2). Fix: even multiplicity touches, odd multiplicity crosses — and the zero of (x + a) is x = −a, not x = a.
Q7. A. The leading term is −4x⁶: even degree makes both ends point the same way, and the negative coefficient points them down — the graph falls on both the far left and the far right. B) reverses the sign of the leading coefficient. C) and D) treat the visible 50x⁵ as the leading term; a degree-5 term never outruns a degree-6 term, no matter how large its coefficient. E) confuses the constant term −3 with a horizontal asymptote — polynomials have none. Fix: end behavior is decided by the single highest-degree term; degree beats coefficient, always, eventually.
Q8. C. A polynomial of degree n has at most n − 1 turning points: 5 − 1 = 4. A) is an undercount with no rule behind it. B) applies n − 2, the bound for inflection-type behavior, not turning points. D) uses the degree itself. E) uses n + 1. Fix: turning points ≤ degree − 1 — and remember it's "at most," not "exactly."
Q9. D. The graph crosses at −3 (factor (x + 3), odd multiplicity), touches at 2 (factor (x − 2)², even multiplicity), giving degree 3; falling-left/rising-right matches an odd degree with positive leading coefficient, and the y-intercept checks: (3)(−2)² = 12. A) crosses at both intercepts — it can't produce the touch at x = 2. B) swaps the multiplicities, touching at −3 and crossing at 2. C) flips the signs of the zeros to 3 and −2. E) has the right factors but a negative leading coefficient, which rises left and falls right and gives y-intercept −12. Fix: check end behavior first, then cross-vs-touch at each intercept, then the y-intercept — the three-check routine.
Q10. E. Four turning points require degree ≥ 5; both ends falling requires an even degree (with negative leading coefficient). The smallest degree that is both even and ≥ 5 is 6. A) degree 3 allows at most 2 turning points. B) degree 4 allows at most 3. C) and D) are odd degrees, whose ends must point in opposite directions. Fix: turning points set a floor (degree ≥ turning points + 1); matching ends force even degree — both constraints must hold at once.
Q11. B. Quadratic data over equal steps must have constant second differences; here the first differences are 4, 8, 16, so the second differences are 4 and 8 — not constant. (The constant ratio of differences actually suggests exponential-type growth.) A) "increasing outputs" describes many families and confirms none. C) states the linear test, not the quadratic test — constant first differences would rule OUT a quadratic. D) positive second differences indicate upward bending, not quadratic form; they must be constant, not merely positive. E) is false — an upward-opening quadratic increases forever on the right. Fix: to claim "quadratic" from a table, compute second differences and check that they are constant, not just growing or positive.
Q12. C. x = 4 is indeed a zero, but the factor (x − 4) appears squared — even multiplicity — so the graph touches the axis at (4, 0) and turns back up; (x − 4)² ≥ 0 on both sides, so there is no crossing. A) overgeneralizes: crossing happens only at odd-multiplicity zeros. B) wrongly denies the zero — the student's premise was fine; the inference failed. D) the direction of opening explains which way it turns, not whether it crosses. E) confuses the zero of (x − 4) with x = −4. Fix: identifying a zero is step one; its multiplicity's parity — not the zero itself — decides cross versus touch.