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Roots, intercepts and turning points of quadratics worksheet — GCSE Foundation

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.

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Roots, intercepts and turning points of quadratics worksheet — GCSE Foundation

Maths · GCSE Foundation · A11 · Roots, intercepts and turning points of quadratics · Non-calculator · 15 questions

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1.The graph of y = x² − 2x − 8 crosses the x-axis at x = −2 and x = 4. Work out the x-coordinate of the turning point of the curve, using the symmetry of the graph.
y = x² − 2x − 8
(a)x = −1
(b)x = 3
(c)x = 2
(d)x = 1
2.A charity's fundraising total, T pounds, over d days follows T = (d − 3)(30 − d) for 3 ≤ d ≤ 30, where T = 0 marks the start and end of the campaign. Work out how many days the campaign runs for, from start to end.
(a)24 days
(b)30 days
(c)27 days
(d)33 days
3.A quadratic curve has a root at x = −2 and its turning point has x-coordinate 3. Work out the curve's other root, using the symmetry of the graph.
(a)x = −8
(b)x = 1
(c)x = 8
(d)x = 5
4.A curve has equation y = x² − 9. Which statement about its graph is correct?
y = x² − 9
(a)It crosses the x-axis only once, at x = 9.
(b)It does not cross the x-axis, since −9 is negative.
(c)It crosses the x-axis at x = 9 and x = −9.
(d)It crosses the x-axis at x = 3 and x = −3.
5.The graph of y = x² − 7x + 2 crosses the x-axis at two points. One root, read from the graph, is approximately x = 0.30. Using the fact that the sum of the two roots of x² − 7x + 2 = 0 is 7, estimate the other root, correct to 2 decimal places.
y = x² − 7x + 2
(a)x ≈ 6.70
(b)x ≈ 6.30
(c)x ≈ 7.30
(d)x ≈ 0.70
6.A table shows values of y = x² − x − 6 for x from −3 to 4: at x = −3, y = 6; x = −2, y = 0; x = −1, y = −4; x = 0, y = −6; x = 1, y = −6; x = 2, y = −4; x = 3, y = 0; x = 4, y = 6. Using the table, write down the two roots of x² − x − 6 = 0.
y = x²
(a)x = −1 and x = 2
(b)x = −3 and x = 4
(c)x = 0 and x = 1
(d)x = −2 and x = 3
7.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.
(a)t = 1 second and t = 4 seconds
(b)t = 1 second and t = −5 seconds
(c)t = 1 second and t = 5 seconds
(d)t = −1 second and t = 5 seconds
8.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
(a)x = −1 and x = −6
(b)x = 2 and x = −3
(c)x = 2 and x = 3
(d)x = −2 and x = −3
9.The graph of y = (x − 4)(x + 1) crosses the x-axis. Which pair gives the correct roots and reasoning?
(a)x = −4 and x = −1, because the roots are always the negative of the numbers shown in the brackets.
(b)x = 4 and x = 1, because both brackets must be positive at the roots.
(c)x = 4 and x = −1, because the graph crosses the x-axis where x − 4 = 0 or x + 1 = 0.
(d)x = −4 and x = 1, because the signs inside the brackets must be reversed to solve them.
10.A quadratic graph has roots at x = −3 and x = 5, and it crosses the y-axis at (0, −15). Work out the equation of the curve in the form y = (x − a)(x − b).
(a)y = (x − 3)(x − 5)
(b)y = (x − 3)(x + 5)
(c)y = (x + 3)(x + 5)
(d)y = (x + 3)(x − 5)
11.A quadratic graph has equation y = (x − 4)². A student says this graph crosses the x-axis at two different points. Explain why the student is wrong.
(a)It touches the x-axis once, only at x = 4.
(b)It never touches the x-axis at all.
(c)It crosses twice, at x = 4 and x = −4.
(d)It crosses twice, at x = 2 and x = −2.
12.The graph of y = (x − 2)(x + 5) crosses the x-axis at two points. Work out the x-coordinates of these two points.
(a)x = −2 and x = −5
(b)x = 2 and x = −5
(c)x = −2 and x = 5
(d)x = 2 and x = 5
13.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
(a)x = 1
(b)x = 5
(c)x = 6
(d)x = 3
14.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
(a)The student is correct, because the turning point's x-coordinate always equals the coefficient of x.
(b)The turning point has x-coordinate −6, because the sign of the coefficient must be reversed.
(c)The turning point has x-coordinate 5, because that is the larger of the equation's two roots.
(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.
15.Which of these quadratic graphs does NOT cross the x-axis at all?
(a)y = (x − 2)² + 3
(b)y = (x + 4)(x − 1)
(c)y = (x − 2)(x + 3)
(d)y = x² − 9

Answer key

(d) x = 1 — A quadratic graph is symmetrical about its turning point, so the turning point's x-coordinate is the midpoint of the two roots: (−2 + 4) ÷ 2 = 2 ÷ 2 = 1. The option x = 2 comes from adding the two roots but forgetting to divide by 2. The option x = −1 makes a sign error when adding the roots, treating −2 + 4 as −2. The option x = 3 comes from subtracting the roots instead of adding them: (4 − (−2)) ÷ 2 = 3.
(c) 27 days — The campaign starts at d = 3 and ends at d = 30, so it runs for 30 − 3 = 27 days. Getting 33 days comes from adding the two values, 3 + 30 = 33, instead of subtracting them. Getting 30 days uses only the end day and ignores that the campaign did not start at day 0. Getting 24 days comes from subtracting the start day twice, 30 − 3 − 3 = 24, instead of once.
(c) x = 8 — The turning point lies exactly halfway between the two roots. If the other root is r, the midpoint of −2 and r must be 3, so (−2 + r) ÷ 2 = 3, giving r = 8. Choosing x = 5 comes from adding 2 and 3 rather than using the midpoint relationship correctly. Choosing x = 1 comes from subtracting 2 from 3 instead of reflecting −2 across the turning point. Choosing x = −8 finds the right distance but then reflects in the y-axis instead of in the line of symmetry x = 3, so the sign of the answer is flipped.
(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.
(a) x ≈ 6.70 — Since the two roots sum to 7, the other root is 7 − 0.30 = 6.70. The option 7.30 comes from adding the given root to 7 instead of subtracting it. The option 6.30 comes from subtracting 0.70 (one minus the given root) rather than the given root itself. The option 0.70 confuses the required root with the amount by which the given root falls short of 1.
(d) x = −2 and x = 3 — The roots are the x-values where y = 0. Reading the table, y = 0 at x = −2 and at x = 3, so these are the two roots. Choosing x = −3 and x = 4 picks the endpoints of the table, where y = 6, not where y = 0. Choosing x = −1 and x = 2 picks values near the curve's lowest points, where y = −4, not where the curve crosses the axis. Choosing x = 0 and x = 1 picks the two x-values in the middle of the table without checking their y-values, which are both −6, not 0.
(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.
(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.
(c) x = 4 and x = −1, because the graph crosses the x-axis where x − 4 = 0 or x + 1 = 0. — The graph crosses the x-axis where y = 0, which happens when either bracket equals zero. Solving x − 4 = 0 gives x = 4, and solving x + 1 = 0 gives x = −1. The option x = −4 and x = 1 incorrectly reverses both signs. The option x = 4 and x = 1 misreads the second bracket, ignoring that x + 1 = 0 requires x to be negative. The option x = −4 and x = −1 wrongly assumes both roots must be the negative of the constants shown, which only matches the second bracket, not the first.
(d) y = (x + 3)(x − 5) — A root at x = −3 means the matching bracket must be zero when x = −3, so the bracket is (x − (−3)) = (x + 3). A root at x = 5 means the other bracket is (x − 5). So the equation is y = (x + 3)(x − 5); checking the y-intercept, (0 + 3)(0 − 5) = 3 × (−5) = −15, which matches the given value. The option (x − 3)(x + 5) swaps the signs of both roots. The option (x + 3)(x + 5) keeps the correct sign for the first root but gets the second wrong. The option (x − 3)(x − 5) gets the first root's sign wrong.
(a) It touches the x-axis once, only at x = 4. — (x − 4)² is a square, so it equals zero only when x − 4 = 0, that is at x = 4 — the curve just touches the x-axis there rather than crossing it, since a square cannot be negative on either side to cross through. Saying it crosses at x = 4 and x = −4 wrongly introduces a plus-or-minus, as if taking a square root of x, rather than recognising the bracket is already squared and only zero once. Saying it never touches the x-axis forgets that a squared term CAN equal zero, even though it can never be negative. Saying it crosses at x = 2 and x = −2 confuses (x − 4)² with the different expression x² − 4.
(b) x = 2 and x = −5 — Set each factor equal to zero: x − 2 = 0 gives x = 2, and x + 5 = 0 gives x = −5, so the graph crosses the x-axis at x = 2 and x = −5. Writing x = −2 and x = 5 flips the sign of both roots. Writing x = 2 and x = 5 keeps the first root correct but forgets to flip the sign for the second factor, using +5 instead of solving x + 5 = 0. Writing x = −2 and x = −5 flips the sign of the first root only, from solving x − 2 = 0 as x = −2.
(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) 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.
(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.

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