A-Level Curve sketching Practice Questions

Free A-Level Curve sketching practice questions with full step-by-step worked solutions. Covers cubic, sketch, intercepts, y-intercept. Practise exam-style problems and check your method.

cubicsketchinterceptsy-interceptbasic curvereciprocal
A-Level70 questionsStep-by-step solutions
Question 1
3 markseasy
Sketch the graph of y=(x+1)(x−2)(x−3)y = (x+1)(x-2)(x-3), showing clearly the coordinates of the points where the curve meets the coordinate axes.
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Worked solution

  1. Identify the shape

    y=x3−4x2+x+6y = x^3 - 4x^2 + x + 6

    This is a cubic (highest power is x3x^3) with a positive leading coefficient, so it has the classic cubic shape. Recognising the degree first tells you how many bends and roots to expect.

  2. Find where the curve crosses the x-axis

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

    The curve is already in factor form, so set y=0y=0. A product is zero when any bracket is zero, which gives the x-intercepts.

  3. State the roots

    x=−1, x=2, x=3x=-1,\ x=2,\ x=3

    These are the x-values where the curve meets the x-axis. Plotting them first gives the skeleton of the sketch.

  4. Find the y-intercept

    x=0:y=x3−4x2+x+6∣x=0=6x=0:\quad y = x^3 - 4x^2 + x + 6 \Big|_{x=0} = 6

    Substitute x=0x=0 to find where the curve crosses the y-axis. Every graph crosses the y-axis at the constant term here.

  5. Describe the end behaviour

    asx→∞, y→∞andasx→−∞, y→−∞as x\to\infty,\ y\to\infty and as x\to-\infty,\ y\to-\infty

    Because the leading coefficient is positive, the tails of the cubic point in opposite directions. Knowing this stops you drawing the ends upside down.

  6. Sketch the curve

    y=x3−4x2+x+6y = x^3 - 4x^2 + x + 6

    Join the features smoothly: pass through each root, the y-intercept, and (for the harder version) the turning points, following the end-behaviour arrows. The finished sketch is shown.

Answer
Cubic crossing the x-axis at (-1,\ 0), (2,\ 0), (3,\ 0) and the y-axis at (0, 6).
Question 2
3 markseasy
Sketch y=(x+3)(x+1)(x−2)y=(x+3)(x+1)(x-2), labelling the axis intercepts.
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Worked solution

  1. Roots

    x=−3, −1, 2x=-3,\ -1,\ 2

    Set each factor to zero for the x-intercepts.

  2. y-intercept

    y=(3)(1)(−2)=−6y=(3)(1)(-2)=-6

    Substitute x=0x=0.

  3. Shape

    positive cubic\text{positive cubic}

    Leading term x3x^3 is positive, so tails go down-left and up-right.

  4. Sketch

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

    Positive cubic through the three roots.

Answer
Positive cubic through (-3,0), (-1,0), (2,0), y-intercept (0, -6).
Question 3
3 marksintermediate
How many stationary points does the curve y=x3+xy=x^3+x have? Justify using calculus.
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Worked solution

  1. Differentiate

    dydx=3x2+1\frac{dy}{dx}=3x^2+1

    Stationary points need the gradient to be zero, so differentiate first.

  2. Set to zero

    3x2+1=03x^2+1=0

    Solve for where the gradient vanishes.

  3. Analyse

    3x2=−1 has no real solution3x^2=-1 \text{ has no real solution}

    A square can never be negative, so there is no real x.

  4. 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.

  5. 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.

  6. Conclude

    00

    The curve has no stationary points; it is always increasing.

Answer
00
Question 4
5 markshard
Find the coordinates of the local maximum of y=x3−12xy=x^3-12x.
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Worked solution

  1. Differentiate

    dydx=3x2−12\frac{dy}{dx}=3x^2-12

    Differentiate to find the gradient function.

  2. Set to zero

    3x2−12=03x^2-12=0

    Stationary points occur where the gradient is zero.

  3. Solve

    x2=4⇒x=±2x^2=4\Rightarrow x=\pm2

    Two stationary points.

  4. Classify

    d2ydx2=6x; 6(−2)=−12<0\frac{d^2y}{dx^2}=6x;\ 6(-2)=-12<0

    Negative second derivative at x=−2x=-2 means a local maximum.

  5. Find y

    f(−2)=−8+24=16f(-2)=-8+24=16

    Substitute into the original equation.

  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 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.

  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. State

    (−2, 16)(-2,\ 16)

    The local maximum point.

Answer
(−2,16)(-2, 16)
Question 5
6 markschallenging
Describe fully the graph of y=2x+1−3y=\dfrac{2}{x+1}-3, including its asymptotes and the axes it crosses.
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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).

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