Printable · GCSE Higher · ages 14-16
Roots, intercepts and turning points of quadratics worksheet — GCSE Higher
Fifteen questions on "roots, intercepts and turning points of quadratics" — DfE statement A11. Print it, or print three versions so neighbours cannot copy by letter; the key gives the letter for each version.
part Higher
Roots, intercepts and turning points of quadratics worksheet — GCSE Higher
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- 1.By completing the square, find the turning point of the curve y = x² − 10x + 30.y = x² − 10x + 30
- 2.Which of these quadratic graphs does NOT cross the x-axis at all?
- 3.By completing the square, find the turning point of the curve y = x² + 8x − 3.y = x² + 8x − 3
- 4.By completing the square, find the turning point of the curve y = x² + 6x + 2.y = x² + 6x + 2
- 5.The graph of y = x² − 5x + 6 crosses the x-axis at two points. By factorising, work out the x-coordinates of these two points.y = x² − 5x + 6
- 6.The curve y = 2x² − 12x + 7 has a minimum point. By completing the square, find the value of x at which the minimum occurs.y = 2x² − 12x + 7
- 7.A table shows y = x² − 6x + 5 at these points (x, y): (0, 5), (1, 0), (2, −3), (3, −4), (4, −3), (5, 0), (6, 5). Using the symmetry shown, write down the x-coordinate of the turning point.y = x² − 6x + 5
- 8.A quadratic graph y = ax² + bx + c has its turning point on the y-axis. Which statement about its roots must be true?
- 9.By completing the square, find the turning point of the curve y = 2x² − 8x + 3.y = 2x² − 8x + 3
- 10.A curve has equation y = x² − 9. Which statement about its graph is correct?y = x² − 9
- 11.Write y = x² + 12x + 40 in the form (x + a)² + b, and hence write down the minimum value of y.y = x² + 12x + 40
- 12.A graph has equation y = x² − 6x + 5. A student says its turning point has x-coordinate 6, because that's the coefficient of x. Which statement corrects the student's mistake?y = x² − 6x + 5
- 13.By completing the square, find the turning point of the curve y = 3x² + 12x + 7.y = 3x² + 12x + 7
- 14.A ball's height, h metres, t seconds after being thrown follows h = (t − 1)(9 − t). Given that the ball is at ground level at t = 1 and t = 9, work out at what time t the ball reaches its maximum height, using symmetry.
- 15.The height, h metres, of a ball t seconds after a stopwatch is started follows h = (t − 1)(5 − t) for 1 ≤ t ≤ 5, where h = 0 means the ball is at ground level. Work out the two times at which the ball is at ground level.
Answer key
- (c) x = 5, y = 5 — x² − 10x + 30 = (x − 5)² − 5² + 30 = (x − 5)² + 5. Substituting x = 5: 5² = 25, 10 × 5 = 50, so 25 − 50 + 30 = 5, confirming the minimum value 5 at x = 5: turning point (5, 5). Using −10 instead of half of it inside the bracket gives (x − 10)² − 70, turning point (10, −70) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (−5, 5) — wrong, because (x − 5)² is zero at x = 5, not x = −5. Computing 25 − 30 = −5 instead of 30 − 25 = 5 flips the sign of the constant, giving (5, −5) — wrong, since the completed square's constant must be evaluated as 30 minus 25, not 25 minus 30. Always check a turning point by substituting its x-value back into the original equation.
- (a) y = (x − 2)² + 3 — Since (x − 2)² is never negative, (x − 2)² + 3 is always at least 3, so y can never equal 0 and the graph never crosses the x-axis. The other three graphs are all given in a factorised or difference-of-squares form that shows two real roots: y = (x − 2)(x + 3) crosses at x = 2 and x = −3; y = x² − 9 = (x − 3)(x + 3) crosses at x = 3 and x = −3; y = (x + 4)(x − 1) crosses at x = −4 and x = 1.
- (d) x = −4, y = −19 — x² + 8x − 3 = (x + 4)² − 4² − 3 = (x + 4)² − 19. Substituting x = −4: (−4)² = 16, 8 × (−4) = −32, so 16 − 32 − 3 = −19, confirming the minimum value −19 at x = −4: turning point (−4, −19). Using 8 instead of half of it inside the bracket gives (x + 8)² − 67, turning point (−8, −67) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (4, −19) — wrong, because (x + 4)² is zero at x = −4, not x = 4. Computing 16 + 3 = 19 instead of −3 − 16 = −19 flips the sign of the constant, giving (−4, 19) — wrong, since the completed square's constant must be evaluated as −3 minus 16, not 16 plus 3. Check by substitution whenever the sign of a constant feels uncertain.
- (b) x = −3, y = −7 — x² + 6x + 2 = (x + 3)² − 3² + 2 = (x + 3)² − 7. Substituting x = −3: (−3)² = 9, 6 × (−3) = −18, so 9 − 18 + 2 = −7, confirming the minimum value −7 at x = −3: turning point (−3, −7). Using 6 instead of half of 6 inside the bracket gives (x + 6)² − 34, turning point (−6, −34) — wrong, because only half the coefficient of x belongs inside the bracket. Reading the bracket's sign directly as the turning point's x-coordinate gives (3, −7) — wrong, because (x + 3)² is zero at x = −3, not x = 3. Computing 9 − 2 = 7 instead of 2 − 9 = −7 flips the sign of the constant, giving (−3, 7) — wrong, since the completed square's constant must be evaluated as 2 minus 9, not 9 minus 2. Always check a turning point by substituting its x-value back into the original equation.
- (c) x = 2 and x = 3 — x² − 5x + 6 factorises as (x − 2)(x − 3), since −2 and −3 multiply to give 6 and add to give −5. The graph crosses the x-axis where each bracket equals zero: x − 2 = 0 gives x = 2, and x − 3 = 0 gives x = 3. The option x = −2 and x = −3 comes from reading the signs inside the brackets directly instead of solving x − 2 = 0 and x − 3 = 0. The option x = 2 and x = −3 mixes up the sign of only one root. The option x = −1 and x = −6 comes from picking the wrong pair of factors of 6 (1 and 6 instead of 2 and 3) and then reading their signs directly from the brackets.
- (d) 3 — 2x² − 12x + 7 rewrites as 2(x² − 6x) + 7, then as 2[(x − 3)² − 9] + 7, which simplifies to 2(x − 3)² − 11, since −2 × 9 + 7 = −11. The bracket (x − 3)² is zero when x = 3, so the minimum occurs at x = 3. Treating the shift as b/a instead of b/(2a) — using 6 instead of 3 — gives x = 6, which is wrong. Reading the bracket's sign directly without negating it gives x = −3, wrong, because (x − 3)² is zero at x = 3, not x = −3. Reading off the coefficient of x itself, −12, and calling that the answer skips the completing-the-square process entirely and gives x = −12, which is wrong because b is not the turning point's x-coordinate under any circumstance. Always check: substituting your value of x should make the bracketed term equal to zero, and nothing else.
- (d) x = 3 — The table is symmetrical about the turning point: y = 0 at both x = 1 and x = 5, and the lowest value, y = −4, occurs exactly halfway between them, at x = 3. Choosing x = 5 picks one of the roots rather than the midpoint between them. Choosing x = 1 picks the other root for the same reason. Choosing x = 6 picks the x-value where y returns to its starting value of 5, which is not the turning point.
- (d) If the graph has two real roots, they are equal and opposite in value, so they sum to zero. — A turning point on the y-axis means the graph's axis of symmetry is the line x = 0, so any two roots must be symmetrical about x = 0 — equal in size but opposite in sign, summing to zero. The graph could still have no real roots, but that is not guaranteed just from the turning point's position, so the option claiming it must have none is too strong. The roots do not have to both be positive — if real, one is positive and one negative (or both are zero). The graph does not have to touch the x-axis at exactly one point either; it could cross at two symmetrical points, touch at one point, or miss the x-axis entirely.
- (b) x = 2, y = −5 — 2x² − 8x + 3 rewrites as 2(x² − 4x) + 3, then as 2[(x − 2)² − 4] + 3, which simplifies to 2(x − 2)² − 5, since −2 × 4 + 3 = −5. Substituting x = 2: 2 × 2² = 8, 8 × 2 = 16, so 8 − 16 + 3 = −5, confirming the minimum value −5 at x = 2: turning point x = 2, y = −5. Halving b instead of halving b/a — using 4 as the shift instead of 2 — lands on turning point x = 4, y = −29, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = −2, y = −5 — wrong, because (x − 2)² is zero at x = 2, not x = −2. Computing 8 − 3 = 5 instead of 3 − 8 = −5 flips the sign of the constant, giving x = 2, y = 5 — wrong, since the completed square's constant must be evaluated as 3 minus 8, not 8 minus 3. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (d) It crosses the x-axis at x = 3 and x = −3. — y = x² − 9 factorises as (x − 3)(x + 3), since 9 = 3², so the curve crosses the x-axis at x = 3 and x = −3. Saying it crosses once at x = 9 mistakes the constant term for a root directly, without taking its square root. Saying it crosses at x = 9 and x = −9 makes the same mistake but adds a sign either way. Saying it does not cross the x-axis confuses the y-intercept, which is negative at (0, −9), with the number of times the curve meets the x-axis — a negative y-intercept combined with an upward-opening curve guarantees it crosses the x-axis twice.
- (c) 4 — x² + 12x + 40 = (x + 6)² − 6² + 40 = (x + 6)² + 4, so a = 6 and b = 4. Since (x + 6)² can never be negative, y = (x + 6)² + 4 is smallest when the bracket is zero, so the minimum value of y is 4. Giving 40 instead reads off the ORIGINAL constant term and ignores the completing-the-square step entirely — wrong, because 40 is the value of y when x = 0, not the minimum value. Giving −6 instead answers with the x-coordinate of the turning point (where the bracket is zero) rather than the minimum y-value itself — wrong, because the question asks for the minimum value of y, not the value of x that produces it. Giving 36 instead stops after squaring half the coefficient, 6² = 36, without combining it with the 40 already in the expression — wrong, because the minimum value is 40 minus 36, not 36 on its own.
- (d) The roots of x² − 6x + 5 = 0 are x = 1 and x = 5 (since it factorises to (x − 1)(x − 5)), so by symmetry the turning point has x-coordinate 3, not 6. — Factorising, x² − 6x + 5 = (x − 1)(x − 5), so the roots are x = 1 and x = 5. The turning point lies midway between the roots by symmetry: (1 + 5) ÷ 2 = 3. The coefficient of x has no direct role in locating the turning point this way. The option giving −6 makes an arbitrary sign change with no mathematical basis. The option giving 5 wrongly takes just one of the two roots instead of their midpoint.
- (d) x = −2, y = −5 — 3x² + 12x + 7 rewrites as 3(x² + 4x) + 7, then as 3[(x + 2)² − 4] + 7, which simplifies to 3(x + 2)² − 5, since −3 × 4 + 7 = −5. Substituting x = −2: 3 × (−2)² = 12, 12 × (−2) = −24, so 12 − 24 + 7 = −5, confirming the minimum value −5 at x = −2: turning point x = −2, y = −5. Halving b instead of halving b/a — using −4 as the shift instead of −2 — lands on turning point x = −4, y = −41, which is wrong because when a ≠ 1 the shift inside the bracket is b/(2a), not b/a alone. Reading the bracket's sign directly as the turning point's x-coordinate gives x = 2, y = −5 — wrong, because (x + 2)² is zero at x = −2, not x = 2. Computing 12 − 7 = 5 instead of 7 − 12 = −5 flips the sign of the constant, giving x = −2, y = 5 — wrong, since the completed square's constant must be evaluated as 7 minus 12, not 12 minus 7. Whenever a ≠ 1, factor a out of the x² and x terms first, and always check a turning point by substituting back into the original equation.
- (d) t = 5 — The maximum height occurs halfway between the two times when the ball is at ground level: the midpoint of t = 1 and t = 9 is (1 + 9) ÷ 2 = 5, so the ball reaches its maximum height at t = 5 seconds. Getting t = 4 comes from halving the DIFFERENCE between the times, 9 − 1 = 8, then 8 ÷ 2 = 4, instead of finding their midpoint. Getting t = 8 uses that difference, 9 − 1 = 8, as if the gap between the two times were itself the time of the maximum. Getting t = 10 adds the two times, 1 + 9 = 10, but forgets to divide by 2.
- (c) t = 1 second and t = 5 seconds — The ball is at ground level when h = 0, which happens when either bracket is zero. t − 1 = 0 gives t = 1, and 5 − t = 0 gives t = 5. The option with t = −1 makes a sign error solving the first bracket. The option with t = −5 makes a sign error solving the second bracket, treating 5 − t = 0 as if it gave a negative solution. The option with t = 4 seconds is an arithmetic slip in solving 5 − t = 0.
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