Hard A-Level Curve sketching Questions

Challenging, exam-style A-Level Curve sketching questions with worked solutions. Stretch yourself on the hardest cubic, turning points, sketch, differentiation problems.

cubicturning pointssketchdifferentiationstationary pointreciprocal
A-Level34 questionsStep-by-step solutions
Question 1
6 markschallenging
Describe fully the graph of y=2x+1−3y=\dfrac{2}{x+1}-3, including its asymptotes and the axes it crosses.
Show worked solution

Worked solution

  1. Base curve

    y=2xy=\dfrac{2}{x}

    Start from a reciprocal curve scaled by 2.

  2. Horizontal shift

    x→x+1⇒left 1x\to x+1 \Rightarrow \text{left 1}

    Replacing xx with x+1x+1 shifts the graph one unit left: vertical asymptote x=−1x=-1.

  3. Vertical shift

    −3⇒down 3-3 \Rightarrow \text{down 3}

    Subtracting 3 lowers the curve: horizontal asymptote y=−3y=-3.

  4. y-intercept

    x=0: y=2−3=−1x=0:\ y=2-3=-1

    Substitute x=0x=0.

  5. x-intercept

    0=2x+1−3⇒x=−130=\dfrac{2}{x+1}-3\Rightarrow x=-\tfrac{1}{3}

    Set y=0y=0 and solve.

  6. Understand what is being asked

    identify the required feature\text{identify the required feature}

    Read the question slowly and underline exactly what it wants — an intercept, a turning point, an asymptote or a full sketch. Knowing the target keeps the working focused.

  7. Plan the method

    choose the technique before calculating\text{choose the technique before calculating}

    Decide which tool fits: factorising for intercepts, differentiation for turning points, or limits for asymptotes. Planning first prevents wasted work.

  8. Recall how reciprocals behave

    1small→large, 1large→0\tfrac{1}{\text{small}}\to\text{large},\ \tfrac{1}{\text{large}}\to 0

    Remember that dividing by a number close to zero gives a huge value (a vertical asymptote) and dividing by a huge number gives almost zero (a horizontal asymptote). This shapes every reciprocal graph.

  9. Link the algebra to the picture

    each result corresponds to a point on the curve\text{each result corresponds to a point on the curve}

    Every number we find matches something you could see on the graph. Tying the algebra to the sketch helps the whole method make sense.

  10. Check the sign / direction of the curve

    test the sign of y in each region\text{test the sign of } y \text{ in each region}

    Pick a simple test value in each region and check whether yy is positive or negative. This confirms the curve is on the correct side of the axis.

  11. Verify the result

    substitute back to confirm\text{substitute back to confirm}

    Put the answer back into the original equation (or read it off the sketch) to make sure it fits. Checking is a quick way to catch a slip.

  12. Watch a common mistake

    repeated root⇒touch, not cross\text{repeated root} \Rightarrow \text{touch, not cross}

    A frequent error is treating a squared factor like an ordinary root. Remember a repeated factor makes the curve touch the axis and turn back, not pass through.

  13. Consider the end behaviour

    as x→±∞ the highest power dominates\text{as } x\to\pm\infty \text{ the highest power dominates}

    For very large positive or negative xx, the term with the biggest power controls the graph. This fixes how the two tails point.

  14. Consider any special values

    check where the function is undefined\text{check where the function is undefined}

    Look out for values that make a denominator zero or that need excluding. These create asymptotes or gaps rather than points on the curve.

  15. Summary

    asymptotes x=−1, y=−3\text{asymptotes } x=-1,\ y=-3

    State the full description with both intercepts.

Answer
A reciprocal curve with vertical asymptote x=-1 and horizontal asymptote y=-3, crossing the y-axis at (0,-1) and the x-axis at (-1/3, 0).
Question 2
8 markschallenging
Fully analyse and sketch y=(x+5)(x+2)(x−1)y=(x+5)(x+2)(x-1), giving intercepts, turning points, point of inflection and end behaviour.
Show worked solution

Worked solution

  1. State the degree and leading behaviour

    y=x3+6x2+3x−10y = x^3 + 6x^2 + 3x - 10

    This is a cubic with a positive leading coefficient. Identifying the degree tells us to expect up to three roots and two turning points.

  2. Set y = 0 for the x-intercepts

    y=0y=0

    The curve meets the x-axis where the output is zero, so we solve the factorised equation.

  3. Root from factor (x +5)

    x=−5x = -5

    Setting this factor to zero gives the intercept (−5,0)(-5,0). Listing roots one at a time keeps the working clear.

  4. Root from factor (x +2)

    x=−2x = -2

    Setting this factor to zero gives the intercept (−2,0)(-2,0). Listing roots one at a time keeps the working clear.

  5. Root from factor (x -1)

    x=1x = 1

    Setting this factor to zero gives the intercept (1,0)(1,0). Listing roots one at a time keeps the working clear.

  6. Find the y-intercept

    x=0: y=−10x=0:\ y=-10

    Substitute x=0x=0 to find where the curve meets the y-axis.

  7. Differentiate

    dydx=3x2+12x+3\frac{dy}{dx}=3x^2 + 12x + 3

    Differentiate to find the gradient function, needed to locate the turning points.

  8. Solve dy/dx = 0

    3x2+12x+3=0⇒x=−3.73, −0.273x^2 + 12x + 3=0 \Rightarrow x=-3.73,\ -0.27

    Setting the gradient to zero and solving the quadratic gives the x-coordinates of the turning points.

  9. Turning-point coordinates

    (−3.73, 10.39), (−0.27, −10.39)(-3.73,\ 10.39),\ (-0.27,\ -10.39)

    Substitute each x back into the original equation to find the matching heights.

  10. Second derivative

    d2ydx2=6x+12\frac{d^2y}{dx^2}=6x + 12

    The second derivative tells us the nature of each turning point and locates the inflection.

  11. Classify the turning points

    x=−3.73: <0⇒maximum;x=−0.27: >0⇒minimumx=-3.73:\ <0\Rightarrow \text{maximum}; x=-0.27:\ >0\Rightarrow \text{minimum}

    A positive second derivative gives a local minimum; a negative one gives a local maximum.

  12. Point of inflection

    d2ydx2=0⇒x=−2, y=0\frac{d^2y}{dx^2}=0 \Rightarrow x=-2,\ y=0

    Where the second derivative is zero the concavity switches, giving the point of inflection.

  13. Describe end behaviour

    x→∞: y→∞; x→−∞: y→−∞x\to\infty:\ y\to\infty;\ x\to-\infty:\ y\to-\infty

    Because the leading coefficient is positive, the two tails head off in opposite vertical directions.

  14. Check a point between roots

    sign of y between consecutive roots alternates\text{sign of } y \text{ between consecutive roots alternates}

    Testing the sign between roots confirms whether the curve is above or below the axis in each interval.

  15. Produce the sketch

    y=x3+6x2+3x−10y=x^3 + 6x^2 + 3x - 10

    Combine every feature — roots, y-intercept, turning points, inflection and end behaviour — into one smooth curve, shown here.

Answer
Cubic through (-5,\ 0), (-2,\ 0), (1,\ 0), y-intercept (0,-10), turning points (-3.73, 10.39), (-0.27, -10.39).
Question 3
7 markschallenging
Find the number of real solutions of 6x=x2\dfrac{6}{x}=x^2.
Show worked solution

Worked solution

  1. Interpret graphically

    intersections of y=6x and y=x2\text{intersections of } y=\dfrac{6}{x} \text{ and } y=x^2

    Solutions are where the reciprocal curve meets the parabola.

  2. Clear the fraction

    6=x36=x^3

    Multiply both sides by xx (valid for x≠0x\neq0).

  3. Solve the cubic

    x3=6⇒x=63x^3=6\Rightarrow x=\sqrt[3]{6}

    A cube root gives exactly one real value.

  4. Consider other branches

    x3=6 has only one real rootx^3=6 \text{ has only one real root}

    The other two roots of x3=6x^3=6 are complex, so there is a single real x.

  5. Check validity

    x=63≠0x=\sqrt[3]{6}\neq0

    This value is in the domain, so it is a genuine intersection.

  6. Understand what is being asked

    identify the required feature\text{identify the required feature}

    Read the question slowly and underline exactly what it wants — an intercept, a turning point, an asymptote or a full sketch. Knowing the target keeps the working focused.

  7. Plan the method

    choose the technique before calculating\text{choose the technique before calculating}

    Decide which tool fits: factorising for intercepts, differentiation for turning points, or limits for asymptotes. Planning first prevents wasted work.

  8. Recall how reciprocals behave

    1small→large, 1large→0\tfrac{1}{\text{small}}\to\text{large},\ \tfrac{1}{\text{large}}\to 0

    Remember that dividing by a number close to zero gives a huge value (a vertical asymptote) and dividing by a huge number gives almost zero (a horizontal asymptote). This shapes every reciprocal graph.

  9. Link the algebra to the picture

    each result corresponds to a point on the curve\text{each result corresponds to a point on the curve}

    Every number we find matches something you could see on the graph. Tying the algebra to the sketch helps the whole method make sense.

  10. Check the sign / direction of the curve

    test the sign of y in each region\text{test the sign of } y \text{ in each region}

    Pick a simple test value in each region and check whether yy is positive or negative. This confirms the curve is on the correct side of the axis.

  11. Verify the result

    substitute back to confirm\text{substitute back to confirm}

    Put the answer back into the original equation (or read it off the sketch) to make sure it fits. Checking is a quick way to catch a slip.

  12. Watch a common mistake

    repeated root⇒touch, not cross\text{repeated root} \Rightarrow \text{touch, not cross}

    A frequent error is treating a squared factor like an ordinary root. Remember a repeated factor makes the curve touch the axis and turn back, not pass through.

  13. Consider the end behaviour

    as x→±∞ the highest power dominates\text{as } x\to\pm\infty \text{ the highest power dominates}

    For very large positive or negative xx, the term with the biggest power controls the graph. This fixes how the two tails point.

  14. Consider any special values

    check where the function is undefined\text{check where the function is undefined}

    Look out for values that make a denominator zero or that need excluding. These create asymptotes or gaps rather than points on the curve.

  15. Conclude

    11

    There is exactly one real solution.

Answer
11
Question 4
8 markschallenging
Fully analyse and sketch y=−(x+1)(x−2)(x−4)y=-(x+1)(x-2)(x-4), showing intercepts, turning points, inflection and end behaviour.
Show worked solution

Worked solution

  1. State the degree and leading behaviour

    y=−x3+5x2−2x−8y = -x^3 + 5x^2 - 2x - 8

    This is a cubic with a negative leading coefficient. Identifying the degree tells us to expect up to three roots and two turning points.

  2. Set y = 0 for the x-intercepts

    y=0y=0

    The curve meets the x-axis where the output is zero, so we solve the factorised equation.

  3. Root from factor (x +1)

    x=−1x = -1

    Setting this factor to zero gives the intercept (−1,0)(-1,0). Listing roots one at a time keeps the working clear.

  4. Root from factor (x -2)

    x=2x = 2

    Setting this factor to zero gives the intercept (2,0)(2,0). Listing roots one at a time keeps the working clear.

  5. Root from factor (x -4)

    x=4x = 4

    Setting this factor to zero gives the intercept (4,0)(4,0). Listing roots one at a time keeps the working clear.

  6. Find the y-intercept

    x=0: y=−8x=0:\ y=-8

    Substitute x=0x=0 to find where the curve meets the y-axis.

  7. Differentiate

    dydx=−3x2+10x−2\frac{dy}{dx}=-3x^2 + 10x - 2

    Differentiate to find the gradient function, needed to locate the turning points.

  8. Solve dy/dx = 0

    −3x2+10x−2=0⇒x=0.21, 3.12-3x^2 + 10x - 2=0 \Rightarrow x=0.21,\ 3.12

    Setting the gradient to zero and solving the quadratic gives the x-coordinates of the turning points.

  9. Turning-point coordinates

    (0.21, −8.21), (3.12, 4.06)(0.21,\ -8.21),\ (3.12,\ 4.06)

    Substitute each x back into the original equation to find the matching heights.

  10. Second derivative

    d2ydx2=−6x+10\frac{d^2y}{dx^2}=-6x + 10

    The second derivative tells us the nature of each turning point and locates the inflection.

  11. Classify the turning points

    x=0.21: >0⇒minimum;x=3.12: <0⇒maximumx=0.21:\ >0\Rightarrow \text{minimum}; x=3.12:\ <0\Rightarrow \text{maximum}

    A positive second derivative gives a local minimum; a negative one gives a local maximum.

  12. Point of inflection

    d2ydx2=0⇒x=1.67, y=−2.07\frac{d^2y}{dx^2}=0 \Rightarrow x=1.67,\ y=-2.07

    Where the second derivative is zero the concavity switches, giving the point of inflection.

  13. Describe end behaviour

    x→∞: y→−∞; x→−∞: y→∞x\to\infty:\ y\to-\infty;\ x\to-\infty:\ y\to\infty

    Because the leading coefficient is negative, the two tails head off in opposite vertical directions.

  14. Check a point between roots

    sign of y between consecutive roots alternates\text{sign of } y \text{ between consecutive roots alternates}

    Testing the sign between roots confirms whether the curve is above or below the axis in each interval.

  15. Produce the sketch

    y=−x3+5x2−2x−8y=-x^3 + 5x^2 - 2x - 8

    Combine every feature — roots, y-intercept, turning points, inflection and end behaviour — into one smooth curve, shown here.

Answer
Cubic through (-1,\ 0), (2,\ 0), (4,\ 0), y-intercept (0,-8), turning points (0.21, -8.21), (3.12, 4.06).
Question 5
7 markschallenging
The curve y=x3+ax2+bxy=x^3+ax^2+bx has a stationary point at (1,−2)(1,-2). Find the value of aa.
Show worked solution

Worked solution

  1. Use the point on the curve

    1+a+b=−21+a+b=-2

    Substitute (1,−2)(1,-2) into y=x3+ax2+bxy=x^3+ax^2+bx to get one equation.

  2. Simplify

    a+b=−3a+b=-3

    Rearrange the equation from step 1.

  3. Differentiate

    dydx=3x2+2ax+b\frac{dy}{dx}=3x^2+2ax+b

    We need the gradient condition at the stationary point.

  4. Gradient is zero at x=1

    3+2a+b=03+2a+b=0

    Substitute x=1x=1 into the derivative and set it to zero.

  5. Simplify

    2a+b=−32a+b=-3

    Rearrange the gradient equation.

  6. Subtract the equations

    (2a+b)−(a+b)=−3−(−3)(2a+b)-(a+b)=-3-(-3)

    Eliminate bb by subtracting.

  7. Understand what is being asked

    identify the required feature\text{identify the required feature}

    Read the question slowly and underline exactly what it wants — an intercept, a turning point, an asymptote or a full sketch. Knowing the target keeps the working focused.

  8. Plan the method

    choose the technique before calculating\text{choose the technique before calculating}

    Decide which tool fits: factorising for intercepts, differentiation for turning points, or limits for asymptotes. Planning first prevents wasted work.

  9. Recall the differentiation link

    ddx(xn)=nxn−1\frac{d}{dx}(x^n)=nx^{n-1}

    Remember from the differentiation topic that the gradient function is found by multiplying by the power and lowering it by one. Turning points are exactly where this gradient is zero.

  10. Link the algebra to the picture

    each result corresponds to a point on the curve\text{each result corresponds to a point on the curve}

    Every number we find matches something you could see on the graph. Tying the algebra to the sketch helps the whole method make sense.

  11. Check the sign / direction of the curve

    test the sign of y in each region\text{test the sign of } y \text{ in each region}

    Pick a simple test value in each region and check whether yy is positive or negative. This confirms the curve is on the correct side of the axis.

  12. Verify the result

    substitute back to confirm\text{substitute back to confirm}

    Put the answer back into the original equation (or read it off the sketch) to make sure it fits. Checking is a quick way to catch a slip.

  13. Watch a common mistake

    repeated root⇒touch, not cross\text{repeated root} \Rightarrow \text{touch, not cross}

    A frequent error is treating a squared factor like an ordinary root. Remember a repeated factor makes the curve touch the axis and turn back, not pass through.

  14. Consider the end behaviour

    as x→±∞ the highest power dominates\text{as } x\to\pm\infty \text{ the highest power dominates}

    For very large positive or negative xx, the term with the biggest power controls the graph. This fixes how the two tails point.

  15. Solve

    a=0a=0

    So a=0a=0.

Answer
00

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