Curve sketching Worked Solutions — A-Level Maths

Fully worked, step-by-step solutions to A-Level Curve sketching questions. See exactly how to solve problems on cubic, sketch, intercepts, y-intercept.

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.

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
1 markeasy
The curve y=(x−1)(x+2)(x+4)y=(x-1)(x+2)(x+4) crosses the yy-axis at one point. Write down the coordinates of that point.

Worked solution

  1. Use the y-axis condition

    x=0x = 0

    Every point on the y-axis has an x-coordinate of 0, so we substitute x=0x=0.

  2. Substitute

    y=(0−1)(0+2)(0+4)=(−1)(2)(4)=−8y=(0-1)(0+2)(0+4) = (-1)(2)(4) = -8

    Multiply the three bracket values together. This is quicker than expanding the whole cubic.

  3. State the point

    (0, −8)(0,\ -8)

    So the curve meets the y-axis here.

Answer
(0, −8)(0,\ -8)
Question 3
2 markseasy
Sketch the graph of y=x3y = x^3, indicating its behaviour near the origin.

Worked solution

  1. Recognise the parent cubic

    y=x3y = x^3

    This is the basic cubic. It passes through the origin and has rotational symmetry about it.

  2. Check key values

    x=1⇒y=1,x=−1⇒y=−1x=1\Rightarrow y=1,\quad x=-1\Rightarrow y=-1

    Substituting a couple of points fixes the direction of the curve on each side of the origin.

  3. Describe the flat point

    dydx=3x2=0 at x=0\frac{dy}{dx}=3x^2 = 0 \text{ at } x=0

    The gradient is zero at the origin, giving a point of inflection where the curve momentarily flattens.

  4. Sketch

    y=x3y=x^3

    The curve rises from bottom-left to top-right, flattening at the origin.

Answer
Increasing cubic through the origin with a point of inflection at (0, 0).
Question 4
2 markseasy
The diagram shows a curve with the xx-axis and yy-axis as asymptotes, lying in the first and third quadrants. Identify its equation.

Worked solution

  1. Read the asymptotes

    asymptotes: x=0, y=0\text{asymptotes: } x=0,\ y=0

    The curve never touches either axis, so both axes are asymptotes. This is the signature of a reciprocal graph.

  2. Read the quadrants

    branches in quadrants I and III\text{branches in quadrants I and III}

    Positive-over-positive and negative-over-negative both give positive-then-negative branches, matching y=1/xy=1/x.

  3. Conclude

    y=1xy = \dfrac{1}{x}

    The shape and asymptotes uniquely identify the reciprocal function.

Answer
y=1/xy = 1/x
Question 5
1 markeasy
Write down the equation of the vertical asymptote of the curve y=1x−3y=\dfrac{1}{x-3}.

Worked solution

  1. Find where the denominator is zero

    x−3=0x-3 = 0

    A reciprocal blows up where its denominator is zero, because you cannot divide by zero. That value gives the vertical asymptote.

  2. Solve

    x=3x = 3

    So the curve shoots up/down either side of this line.

  3. State the asymptote

    x=3x = 3

    The vertical asymptote is the vertical line x=3x=3.

Answer
x=3x=3

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