Skip to main content

Curve sketching

You can sketch an accurate graph of a function if you know some of its key characteristics. Use this resource to learn how to bring together everything to sketch a curve.

To sketch a curve, it is helpful to know the key points of the function, like:

  • maxima, minima, or turning points
  • \(x\)- and \(y\)-intercepts
  • regions where the gradient is positive or negative.

Stationary points

A stationary point is a point on a graph of a function \(y=f(x)\) where the tangent to the curve is horizontal. From Maxima and minima, you should recognise that this is where \(y'=f'(x)=0\).

These are often also called turning points.

Maximum stationary point

A maximum stationary point occurs at \(x=a\) if \(f'(a)=0\) and \(f'(x)>0\) for \(x<a\) and \(f'(x)<0\) for \(x>a\).

Maximum stationary point.

Minimum stationary point

A minimum stationary point occurs at \(x=a\) if \(f'(a)=0\) and \(f'(x)<0\) for \(x<a\) and \(f'(x)>0\) for \(x>a\).

Minimum stationary point.

Sketching a curve

To sketch a curve, we can follow these steps:

  1. Find the \(x\)- and \(y\)-intercepts.
  2. Find the stationary point/s.
  3. Determine whether the stationary points are maximum or minimum values.
  4. Plot the intercepts and stationary points and join them with a smooth curve.

Example – sketching a curve

Sketch the graph of \(y=x^{3}-x\).

  1. Find the \(x\)-intercepts by letting \(y=0\) and solving for \(x\).
    \[\begin{align*} 0 & = x^{3}-x\\
    0 & = x(x^{2}-1)\\
    x=0 & \textrm{ or }0=x^{2}-1\\
    x=0 & \textrm{ or }x=\pm1
    \end{align*}\]

    This means the \(x\)-intercepts are at \((-1,0)\), \((0,0)\) and \((1,0)\). For this function, the \(y\)-intercept occurs at the origin.

  2. Find the stationary point/s by letting \(y'=0\) and solving for \(x\).
    \[\begin{align*} y' & = 3x^{2}-1\\
    & = 0\\
    1 & = 3x^{2}\\
    \frac{1}{3} & = x^{2}\\
    x & = \pm \sqrt{\frac{1}{3}}\\
    & = \pm \frac{1}{\sqrt{3}}\\
    & \approx \pm0.58
    \end{align*}\]

    We need to substitute \(x=-0.58\) and \(x=0.58\) back into \(y\) to find the coordinates for the stationary points.
    \[\begin{align*} y & = x^{3}-x\\
    & = (-0.58)^{3}-(-0.58)\\
    & = 0.38
    \end{align*}\] \[\begin{align*} y & = x^{3}-x\\
    & = (0.58)^{3}-(0.58)\\
    & = -0.39
    \end{align*}\]

    So, the stationary points occur at around \((-0.58, 0.38)\) and \((0.58,-0.39)\).

  3. Determine whether the stationary points are maximum or minimum values by taking the second derivative and evaluating whether it is \(<0\), \(>0\) or \(=0\).
    \[y''=6x\]

    For \(x=-0.58\):
    \[\begin{align*} y'' & = 6(-0.58)\\
    & = -3.46\\
    & <0\quad\textrm{local maximum}
    \end{align*}\]

    For \(x=0.58\):
    \[\begin{align*} y'' & = 6(0.58)\\
    & = 3.46\\
    & >0\quad\textrm{local minimum}
    \end{align*}\]

    We can also look at values on either side of the stationary point to decide whether it is a maximum or minimum. For \(x=-0.58\):

    \(x\) \(-0.6\) \(-0.58\) \(-0.5\)
    \(f'(x)\) Positive \(0\) Negative
    Gradient / \

    For \(x=0.58\):

    \(x\) \(0.5\) \(0.58\) \(0.6\)
    \(f'(x)\) Negative \(0\) Positive
    Gradient \ /
  4. Put it all together by first plotting the \(x\)-intercepts and stationary points.
    x intercepts and stationary points

    We know that point \(E\) is a maximum and point \(D\) is a minimum, so we can graph the curve.

    Graph of x cubed minus x.

Exercise – sketching a curve

Sketch the graphs of the following functions, showing all intercepts and turning points.

  1. \(y=x^{2}-4x\)
  2. \(y=x^{3}-2x^{2}+x\)
  3. \(y=6-x-x^{2}\)
  4. \(y=(x+1)^{4}\)

  1. \(y=x^{2}-4x\)
    Graph of x squared minus four x.
  2. \(y=x^{3}-2x^{2}+x\)
    Graph of x cubed minus two x squared plus x.
  3. \(y=6-x-x^{2}\)
    Graph of six minus x minus x minus x squared.
  4. \(y=(x+1)^{4}\)
    Graph of x plus one to the power of four.

Curve sketching in context

In urban planning, visualising the interplay between infrastructure and natural landscapes is vital for a balanced development. The path of a proposed city park is modelled using the function \(f(x)=-x^{3}+3x^{2}-2x\), where \(x\) represents the horizontal positions along the park's main pathway.

Find the coordinates of the intercepts to inform the key positions where the pathway intersects with ground level.

To determine the \(y\)-intercept, let \(x=0\).
\[\begin{align*} f(0) & = -(0)^{3}+3(0)^{2}-2(0)\\
& = 0
\end{align*}\]

The \(y\)-intercept is as \((0,0)\).

To determine the \(x\)-intercept, let \(y=0\).
\[\begin{align*} -x^{3}+3x^{2}-2x & = 0\\
-x(x^{2}-3x+2) & = 0\\
-x(x-1)(x-2) & = 0\\
x = 0\quad x & = 0\quad x = 2
\end{align*}\]

The \(x\)-intercepts are \((0,0)\), \((1,0)\) and \((2,0)\).

Images on this page by RMIT, licensed under CC BY-NC 4.0