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 \(x>a\).

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 \(x>a\).

Sketching a curve

To sketch a curve, we can follow these steps:

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

Example – sketching a curve

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

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.
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 \((0.58,-0.39)\).
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 \ – /
Put it all together by first plotting the \(x\)-intercepts and stationary points.

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