Printable · GCSE Higher · ages 14-16
Algebra worksheet — GCSE Higher
Fifteen questions across the algebra statements at Higher tier. Choose the non-calculator filter to rehearse Paper 1, which counts for a third of the marks.
Algebra worksheet — GCSE Higher
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- 1.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.
- 2.A sequence is defined by aₙ = aₙ₋₁ + aₙ₋₂ for n ≥ 3. Given that a₃ = 11 and a₅ = 29, work out a₁.
- 3.Work out the value of m² + 3m when m = −2
- 4.For the graph of y = 1/x, where x cannot be zero, which of these statements is correct?
- 5.The formula connecting distance (d), speed (s) and time (t) is d = st. Make s the subject of the formula.
- 6.Point P has coordinates (5, 2). Point P is rotated 180° about the origin to point Q. Write down the coordinates of Q.
- 7.The graph of y = f(x) has a maximum turning point at (−1, 6). Write down the coordinates of the maximum turning point of the graph of y = f(x − 3).
- 8.Solve x² − 8x + 3 = 0 by completing the square. Give your answers in surd form.
- 9.Describe the single transformation that maps the graph of y = x² onto the graph of y = x² + 3.y = x²y = x² + 3
- 10.Three consecutive integers add up to 72. Using n for the smallest integer, form an equation and solve it to find the smallest of the three integers.
- 11.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
- 12.The 2nd term of a geometric sequence is 6 and the 4th term is 24. Given that all the terms of the sequence are positive, work out the common ratio.
- 13.A circle has centre (0, 0) and equation x² + y² = 3721. The point (11, 60) lies on the circle. One of these is the gradient of the tangent to the circle at (11, 60). Work out which one.
- 14.The simple interest, £I, earned on a savings account is given by the formula I = PRT/100, where P is the amount invested in pounds, R is the interest rate per year and T is the time in years. Rearrange the formula to make P the subject.
- 15.A curve has equation y = x³ − 4x. Work out the coordinates of the point where the curve crosses the y-axis.y = x
Answer key
- (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.
- (a) 4 — a₅ = a₄ + a₃, so a₄ = a₅ − a₃ = 29 − 11 = 18. a₄ = a₃ + a₂, so a₂ = a₄ − a₃ = 18 − 11 = 7. a₃ = a₂ + a₁, so a₁ = a₃ − a₂ = 11 − 7 = 4. Checking forwards: 4, 7, 11, 18, 29. Answering 7 stops one step early, reporting a₂ = 7 instead of continuing back one more step to a₁ — wrong, because the question asks for a₁, not a₂. Answering 18 reports a₄ = 18, an intermediate value found along the way, instead of a₁ — wrong, because a₄ is a term used to reach the answer, not the term the question asks for. Answering 3 takes one backward step too many, working out a further term a₀ = a₂ − a₁ = 7 − 4 = 3 — wrong, because the sequence starts at a₁, so a₁ = 4 is as far back as the question goes.
- (d) −2 — Method: substitute the value into both terms, working out the index and the multiplication before the addition. Working: m² = (−2) × (−2) = 4 and 3m = 3 × (−2) = −6, so the expression becomes 4 + (−6), which is −2. Answer: −2. The distractors: −10 comes from squaring −2 as −4, giving −4 + (−6); 10 comes from working out 3m as +6 and losing the minus sign, giving 4 + 6; −14 comes from working from left to right instead of multiplying first, giving (4 + 3) × (−2).
- (d) The graph never crosses either axis — Since x ≠ 0, there is no point on the graph where x = 0, so it cannot cross the y-axis; likewise 1/x is never equal to 0 for any x, so it cannot cross the x-axis either — the graph never touches either axis. A candidate who forgets the restriction x ≠ 0 might think the graph behaves like other graphs and passes through the origin, (0, 0). A candidate who correctly rules out the x-axis but forgets that x = 0 is also excluded might say the graph crosses the y-axis but never the x-axis. A candidate who only pictures the branch where x and y are both positive might say the graph has only one branch, in quadrant 1, forgetting the second branch where x and y are both negative.
- (b) s = d / t — Method: undo the multiplication by t by dividing both sides by t. Working: d = st, so dividing both sides by t gives s = d / t. The value s = dt comes from multiplying by t instead of dividing. The value s = t / d comes from inverting the fraction, dividing t by d instead of d by t. The value s = d − t comes from subtracting t instead of dividing by it.
- (b) (−5, −2) — A rotation of 180° about the origin reverses the sign of both coordinates, so Q = (−5, −2). A candidate who reverses the sign of only the y-coordinate, as if reflecting in the x-axis, gets (5, −2). A candidate who reverses the sign of only the x-coordinate, as if reflecting in the y-axis, gets (−5, 2). A candidate who swaps the coordinates instead of reversing their signs gets (2, 5).
- (b) (2, 6) — y = f(x − 3) translates y = f(x) horizontally by 3 units to the RIGHT — inside the brackets, subtracting moves the graph in the positive x-direction. Turning point (−1, 6) → (−1 + 3, 6) = (2, 6). The common slip is to move LEFT instead, since the sign inside the bracket is negative — that gives (−4, 6). Changing the y-coordinate instead of the x-coordinate, as in (−1, 3) or (−1, 9), treats this as a vertical shift, which y = f(x − 3) is not.
- (a) x = 4 ± √13 — Method: halve the coefficient of x to form the bracket, subtract the square of that number to keep the expression equal to the original, then rearrange and take the square root of both sides. Working: half of −8 is −4, so x² − 8x + 3 = (x − 4)² − 16 + 3 = (x − 4)² − 13; setting this equal to zero gives (x − 4)² = 13, so x − 4 = ±√13 and x = 4 ± √13. Answer: x = 4 ± √13. The distractors: x = 8 ± √13 comes from putting the whole coefficient 8 inside the bracket instead of half of it; x = 4 ± √19 comes from adding 16 and 3 rather than subtracting 3 from 16; x = −4 ± √13 comes from writing the bracket as (x + 4)², which reverses the sign of the number that comes out of it.
- (a) A translation by vector (0, 3) — y = x² + 3 adds a constant outside the squaring, so it is a vertical translation of y = x² — every point moves the same distance parallel to the y-axis, with no change in x. The vector is (0, 3), matching the +3. A vector of (3, 0) confuses this with a horizontal shift; (0, −3) has the right axis but the wrong sign, since the graph moves up, not down; a stretch changes the shape of the curve, which adding a constant term never does.
- (d) 23 — Method: let the smallest integer be n, so the three consecutive integers are n, n + 1 and n + 2. Form the equation n + (n + 1) + (n + 2) = 72. Working: simplify the left side: 3n + 3 = 72, so 3n = 69, giving n = 23. Answer: 23. 24 comes from dividing 72 by 3 directly, 72 ÷ 3 = 24, which finds the middle integer rather than realising the three numbers differ. 25 comes from correctly finding n = 23 but reading off the largest integer, n + 2, instead of the smallest as asked. 21 comes from dividing first and subtracting after, (72 ÷ 3) − 3, instead of subtracting 3 before dividing by 3.
- (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.
- (c) 2 — For a geometric sequence, a₄ = a₂ × r². So r² = a₄ ÷ a₂ = 24 ÷ 6 = 4, giving r = 2 or r = −2. Since all the terms are positive, r = 2. 4 comes from computing a₄ ÷ a₂ = 4 and stopping there, without taking the square root to find r. 18 comes from subtracting instead of dividing: a₄ − a₂ = 24 − 6 = 18, taken directly as the common ratio. −2 is the other square root of 4, but it is rejected because all the sequence's terms are given as positive.
- (d) −11/60 — The radius from the origin to (11, 60) has gradient 60/11. The tangent is perpendicular to this radius, so its gradient is the negative reciprocal: −1 ÷ (60/11) = −11/60. Choosing 60/11 uses the radius's gradient unchanged, without applying perpendicularity. Choosing −60/11 negates the radius's gradient but forgets to take its reciprocal. Choosing 11/60 takes the reciprocal correctly but keeps the gradient positive instead of negative.
- (a) P = 100I / (RT) — Method: undo the operations done to P in reverse order — multiply by 100, then divide by R and by T. Working: I = PRT/100, so 100I = PRT, so P = 100I / (RT). Answer: P = 100I / (RT). P = IRT/100 comes from leaving R and T in the numerator instead of moving them to the denominator. P = 100RT/I comes from swapping P and I when rearranging. P = 100I/R comes from dividing by R only and forgetting to also divide by T.
- (c) (0, 0) — A curve crosses the y-axis where x = 0. Substituting x = 0 into y = x³ − 4x gives y = 0³ − 4(0) = 0 − 0 = 0, so the curve crosses the y-axis at (0, 0). A candidate who reads off the coefficient of x as the y-intercept, instead of working out the constant term, might write (0, −4). A candidate who swaps the coordinates might write (4, 0). A candidate who takes the coefficient of x but drops its sign might write (0, 4).
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