Hard GCSE Cubic and reciprocal graphs Questions

Challenging, exam-style GCSE Cubic and reciprocal graphs questions with worked solutions. Stretch yourself on the hardest roots of a cubic, factor theorem, factorising, intersection of curve and line problems.

roots of a cubicfactor theoremfactorisingintersection of curve and linereciprocal graphsolving equations
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Solve x3−19x+30=0x^3 - 19x + 30 = 0. Give the smallest solution.
Show worked solution

Worked solution

  1. Write down the equation

    x3−19x+30=0x^3 - 19x + 30 = 0

    The solutions are the x-values where the curve crosses the x-axis.

  2. Try x=1x = 1

    (1)3−19(1)+30=1−19+30=12(1)^3 - 19(1) + 30 = 1 - 19 + 30 = 12

    This is not 0, so x=1x = 1 is not a solution.

  3. Try x=2x = 2

    (2)3−19(2)+30=8−38+30=0(2)^3 - 19(2) + 30 = 8 - 38 + 30 = 0

    The value is 0, so x=2x = 2 is a solution and (x - 2) is a factor.

  4. Divide by (x - 2)

    x3−19x+30=(x−2)(x2+2x−15)x^3 - 19x + 30 = (x - 2)(x^2 + 2x - 15)

    Dividing the cubic by the known factor leaves a quadratic.

  5. Check the division by expanding

    (x−2)(x2+2x−15)=x3+2x2−15x−2x2−4x+30=x3−19x+30(x - 2)(x^2 + 2x - 15) = x^3 + 2x^2 - 15x - 2x^2 - 4x + 30 = x^3 - 19x + 30

    Expanding gives back the original cubic.

  6. Factorise the quadratic

    x2+2x−15=(x+5)(x−3)x^2 + 2x - 15 = (x + 5)(x - 3)

    5 × (-3) = -15 and 5 + (-3) = 2.

  7. Write the cubic fully factorised

    (x−2)(x+5)(x−3)=0(x - 2)(x + 5)(x - 3) = 0

    The cubic is a product of three linear factors.

  8. Set each factor equal to zero

    x−2=0,x+5=0,x−3=0x - 2 = 0, \quad x + 5 = 0, \quad x - 3 = 0

    A product is zero only when one of the factors is zero.

  9. Solve each equation

    x=2,x=−5,x=3x = 2, \quad x = -5, \quad x = 3

    These are the three solutions.

  10. Put them in order

    −5<2<3-5 < 2 < 3

    Ordering the solutions makes it easy to pick the smallest.

  11. Choose the smallest solution

    x=−5x = -5

    The smallest of the three solutions is -5.

  12. Check x=−5x = -5

    (−5)3−19(−5)+30=−125+95+30=0(-5)^3 - 19(-5) + 30 = -125 + 95 + 30 = 0

    Substituting x=−5x = -5 makes the equation true.

  13. Check x=3x = 3

    (3)3−19(3)+30=27−57+30=0(3)^3 - 19(3) + 30 = 27 - 57 + 30 = 0

    This root works as well, so the factorisation is right.

  14. Interpret on the graph

    (−5, 0), (2, 0), (3, 0)(-5,\ 0), \ (2,\ 0), \ (3,\ 0)

    The cubic crosses the x-axis at three points, the leftmost being (-5, 0).

  15. State the answer

    x=−5x = -5

    The smallest solution is -5.

Answer
x=−5x = -5
Question 2
5 markschallenging
yy is inversely proportional to xx, so y=kxy = \frac{k}{x}. When x=4x = 4, y=9y = 9. Work out the value of xx when y=12y = 12.
Show worked solution

Worked solution

  1. Write down the general equation

    y=kxy = \frac{k}{x}

    y is inversely proportional to x, so y is a constant divided by x.

  2. Substitute x=4x = 4 and y=9y = 9

    9=k49 = \frac{k}{4}

    Use the pair of values given in the question.

  3. Multiply both sides by 4

    k=9×4k = 9 \times 4

    Multiplying by the denominator clears the fraction.

  4. Work out k

    k=36k = 36

    9×4=369 \times 4 = 36.

  5. Write the equation

    y=36xy = \frac{36}{x}

    This is the reciprocal graph through (4, 9).

  6. Check the given pair

    364=9\frac{36}{4} = 9

    Substituting x=4x = 4 gives y=9y = 9, which is correct.

  7. Substitute y=12y = 12

    12=36x12 = \frac{36}{x}

    Now find the x-value that gives y=12y = 12.

  8. Multiply both sides by x

    12x=3612x = 36

    Multiplying by x clears the fraction; x cannot be 0.

  9. Divide both sides by 12

    x=3612x = \frac{36}{12}

    Divide by the coefficient of x.

  10. Work out the division

    x=3x = 3

    36÷12=336 \div 12 = 3.

  11. Check the answer

    363=12\frac{36}{3} = 12

    Substituting x=3x = 3 gives y=12y = 12, as required.

  12. Check with the product rule

    xy=3×12=36xy = 3 \times 12 = 36

    For inverse proportion the product xyxy is constant, and 4×9=364 \times 9 = 36 as well.

  13. Check the direction of the change

    9→12 so x must fall9 \rightarrow 12 \text{ so } x \text{ must fall}

    y has increased, so x must decrease. 3 is smaller than 4, which is sensible.

  14. Interpret on the graph

    (3, 12)(3,\ 12)

    The point (3, 12) lies on the same branch of the reciprocal curve as (4, 9).

  15. State the answer

    x=3x = 3

    When y=12y = 12, x=3x = 3.

Answer
x=3x = 3
Question 3
5 markschallenging
Work out the coordinates of the point where the graph of y=x3−8y = x^3 - 8 crosses the xx-axis.
Show worked solution

Worked solution

  1. Use the fact that y=0y = 0 on the x-axis

    x3−8=0x^3 - 8 = 0

    The graph crosses the x-axis where y=0y = 0.

  2. Add 8 to both sides

    x3=8x^3 = 8

    Get the cube on its own.

  3. Take the cube root of both sides

    x=83x = \sqrt[3]{8}

    The cube root undoes the cubing.

  4. Work out the cube root

    x=2x = 2

    2×2×2=82 \times 2 \times 2 = 8.

  5. Write the coordinates

    (2, 0)(2,\ 0)

    On the x-axis the y-coordinate is 0.

  6. Check the point

    (2)3−8=8−8=0(2)^3 - 8 = 8 - 8 = 0

    Substituting x=2x = 2 gives y=0y = 0, so the curve does pass through (2, 0).

  7. Find the y-intercept as well

    x=0⇒y=0−8=−8x = 0 \Rightarrow y = 0 - 8 = -8

    The curve crosses the y-axis at (0, -8).

  8. Describe the transformation

    y=x3−8y = x^3 - 8

    This is the graph of y=x3y = x^3 moved down by 8 units, which is why the root has moved from x=0x = 0 to x=2x = 2.

  9. Factorise to check for other roots

    x3−8=(x−2)(x2+2x+4)x^3 - 8 = (x - 2)(x^2 + 2x + 4)

    x=2x = 2 is a root, so (x - 2) is a factor.

  10. Check the factorisation by expanding

    (x−2)(x2+2x+4)=x3+2x2+4x−2x2−4x−8=x3−8(x - 2)(x^2 + 2x + 4) = x^3 + 2x^2 + 4x - 2x^2 - 4x - 8 = x^3 - 8

    Expanding gives back the original expression.

  11. Complete the square on the quadratic

    x2+2x+4=(x+1)2+3x^2 + 2x + 4 = (x + 1)^2 + 3

    (x+1)2=x2+2x+1(x + 1)^2 = x^2 + 2x + 1, so three more units are needed to make 4.

  12. Show the quadratic has no roots

    (x+1)2+3≥3(x + 1)^2 + 3 \geq 3

    A square is never negative, so this can never be 0.

  13. Conclude there is only one crossing point

    x=2x = 2

    The curve crosses the x-axis exactly once.

  14. Sense-check either side

    x=1⇒y=−7,x=3⇒y=19x = 1 \Rightarrow y = -7, \quad x = 3 \Rightarrow y = 19

    The curve changes from below the axis to above it between x=1x = 1 and x=3x = 3, which fits a single crossing at x=2x = 2.

  15. State the coordinates

    (2, 0)(2,\ 0)

    The graph of y=x3−8y = x^3 - 8 crosses the x-axis at (2, 0).

Answer
(2, 0)(2,\ 0)
Question 4
6 markschallenging
Solve x3+2x2−5x−6=0x^3 + 2x^2 - 5x - 6 = 0. Give the largest solution.
Show worked solution

Worked solution

  1. Write down the equation

    x3+2x2−5x−6=0x^3 + 2x^2 - 5x - 6 = 0

    The solutions are the x-values where the curve crosses the x-axis.

  2. Try x=1x = 1

    (1)3+2(1)2−5(1)−6=1+2−5−6=−8(1)^3 + 2(1)^2 - 5(1) - 6 = 1 + 2 - 5 - 6 = -8

    This is not 0, so x=1x = 1 is not a solution.

  3. Try x=2x = 2

    (2)3+2(2)2−5(2)−6=8+8−10−6=0(2)^3 + 2(2)^2 - 5(2) - 6 = 8 + 8 - 10 - 6 = 0

    The value is 0, so x=2x = 2 is a solution and (x - 2) is a factor.

  4. Divide by (x - 2)

    x3+2x2−5x−6=(x−2)(x2+4x+3)x^3 + 2x^2 - 5x - 6 = (x - 2)(x^2 + 4x + 3)

    Dividing the cubic by the known factor leaves a quadratic.

  5. Check the division by expanding

    (x−2)(x2+4x+3)=x3+4x2+3x−2x2−8x−6=x3+2x2−5x−6(x - 2)(x^2 + 4x + 3) = x^3 + 4x^2 + 3x - 2x^2 - 8x - 6 = x^3 + 2x^2 - 5x - 6

    Expanding gives back the original cubic.

  6. Factorise the quadratic

    x2+4x+3=(x+1)(x+3)x^2 + 4x + 3 = (x + 1)(x + 3)

    1×3=31 \times 3 = 3 and 1+3=41 + 3 = 4.

  7. Write the cubic fully factorised

    (x−2)(x+1)(x+3)=0(x - 2)(x + 1)(x + 3) = 0

    The cubic is a product of three linear factors.

  8. Set each factor equal to zero

    x−2=0,x+1=0,x+3=0x - 2 = 0, \quad x + 1 = 0, \quad x + 3 = 0

    A product is zero only when one of the factors is zero.

  9. Solve each equation

    x=2,x=−1,x=−3x = 2, \quad x = -1, \quad x = -3

    These are the three solutions.

  10. Put them in order

    −3<−1<2-3 < -1 < 2

    Ordering the solutions makes it easy to pick the largest.

  11. Choose the largest solution

    x=2x = 2

    The largest of the three solutions is 2.

  12. Check x=2x = 2

    8+8−10−6=08 + 8 - 10 - 6 = 0

    Substituting x=2x = 2 makes the equation true.

  13. Check x=−1x = -1

    (−1)3+2(−1)2−5(−1)−6=−1+2+5−6=0(-1)^3 + 2(-1)^2 - 5(-1) - 6 = -1 + 2 + 5 - 6 = 0

    This root works too, so the factorisation is right.

  14. Check x=−3x = -3

    (−3)3+2(−3)2−5(−3)−6=−27+18+15−6=0(-3)^3 + 2(-3)^2 - 5(-3) - 6 = -27 + 18 + 15 - 6 = 0

    All three roots check out.

  15. State the answer

    x=2x = 2

    The curve crosses the x-axis at -3, -1 and 2, and the largest solution is 2.

Answer
x=2x = 2
Question 5
6 markschallenging
The graph of y=x3−6x2+9xy = x^3 - 6x^2 + 9x touches the xx-axis at one point. Work out the xx-coordinate of that point.
Show worked solution

Worked solution

  1. Set y=0y = 0

    x3−6x2+9x=0x^3 - 6x^2 + 9x = 0

    The graph touches or crosses the x-axis where y=0y = 0.

  2. Take out the common factor x

    x(x2−6x+9)=0x(x^2 - 6x + 9) = 0

    Every term has a factor of x.

  3. Factorise the quadratic

    x2−6x+9=(x−3)(x−3)x^2 - 6x + 9 = (x - 3)(x - 3)

    −3×−3=9-3 \times -3 = 9 and −3+(−3)=−6-3 + (-3) = -6.

  4. Write it as a square

    x2−6x+9=(x−3)2x^2 - 6x + 9 = (x - 3)^2

    The quadratic is a perfect square.

  5. Write the cubic fully factorised

    x(x−3)2=0x(x - 3)^2 = 0

    The cubic is now a product of the factor x and a squared bracket.

  6. Set each factor equal to zero

    x=0or(x−3)2=0x = 0 \quad \text{or} \quad (x - 3)^2 = 0

    A product is zero only when one of the factors is zero.

  7. Solve each factor

    x=0,x=3 (twice)x = 0, \quad x = 3 \ (\text{twice})

    x=3x = 3 is a repeated root because the bracket is squared.

  8. Interpret the single root

    x=0x = 0

    At a single root the sign of y changes, so the curve cuts straight through the x-axis at the origin.

  9. Interpret the repeated root

    (x−3)2≥0(x - 3)^2 \geq 0

    A squared bracket is never negative, so y does not change sign at x=3x = 3: the curve touches the x-axis there.

  10. Test a value just below 3

    x=2: 2(2−3)2=2×1=2x = 2: \ 2(2 - 3)^2 = 2 \times 1 = 2

    The curve is above the x-axis just before x=3x = 3.

  11. Test a value just above 3

    x=4: 4(4−3)2=4×1=4x = 4: \ 4(4 - 3)^2 = 4 \times 1 = 4

    The curve is above the x-axis just after x=3x = 3 as well, which confirms that it only touches at x=3x = 3.

  12. Check the value at x=3x = 3

    (3)3−6(3)2+9(3)=27−54+27=0(3)^3 - 6(3)^2 + 9(3) = 27 - 54 + 27 = 0

    The curve does pass through (3, 0).

  13. Find the y-intercept

    x=0⇒y=0x = 0 \Rightarrow y = 0

    There is no constant term, so the curve passes through the origin.

  14. Describe the sketch

    cuts at (0, 0), touches at (3, 0)\text{cuts at } (0,\ 0), \text{ touches at } (3,\ 0)

    The curve rises from the bottom left, cuts the axis at the origin, turns over, comes back down to touch the axis at (3, 0) and then rises again.

  15. State the x-coordinate of the touching point

    x=3x = 3

    The graph touches the x-axis at (3, 0).

Answer
x=3x = 3

Unlock 29 more Cubic and reciprocal graphs questions

Create a free account to work through every GCSE Cubic and reciprocal graphs question with instant step-by-step worked solutions, progress tracking and interactive lessons.

  • Full worked solutions for every question
  • Interactive lessons and instant feedback
  • Track your mastery across every topic
Create a Free Account

No card required · Free forever

More Cubic and reciprocal graphs practice

Practice QuestionsWorked Solutions

Related Algebra topics

Algebraic fractionsEquation of a circleCoordinatesExponential and trig graphsExpanding binomialsFactorising ax2 + bx + c