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D9 Curve sketching


If you find some key points of a function such as: maxima, minima, or turning points; x and y axis intercepts; and regions where the gradient is positive or negative, you can put together a sketch of a curve. Read this section to find examples of this being done.

To sketch a curve it is helpful to find the

  1. \(x\) and \(y\) intercepts

  2. Maximum and minimum points.

This module describes how to do this.

Definition of Maximum and Minimum

A stationary point is a point on a graph of a function \(y=f\left(x\right)\) where the tangent to the curve is horizontal. At a stationary point the derivative function \(y=f'\left(x\right)=0.\)

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

Maximum stationary point.

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

Minimum stationary point.

Example 1

Find the turning point of the parabola defined by \(y=x^{2}+4x+5\) and determine if it is a maximum or minimum.



\[\begin{align*} f\left(x\right)=x^{2}+4x+5 \end{align*}\] then \[\begin{align*} f'\left(x\right) & =2x+4. \end{align*}\] At a stationary point, \(f'\left(x\right)=0,\) so \[\begin{align*} 2x+4 & =0\\ 2x & =-4\\ x & =-2 \end{align*}\] is the \(x-\)coordinate of a stationary point. When \(x=-2,\) \[\begin{align*} f\left(-2\right) & =(-2)^{2}+4(-2)+5\\ & =1. \end{align*}\] Hence the coordinates of the stationary point are \(\left(-2,1\right).\) Now we ascertain if it is a maximum or minimum.

A sign test can be used to determine whether the stationary point is minimum or a maximum by checking the slope of the tangent on each side of the stationary point. Consider the table below.

Table shows change in gradient at a minimum.

As we move from the left to the right of the stationary point at \(x=-2\), the gradient changes from negative to positive. This indicates there is a minimum at \(\left(-2,1\right)\).

Hence the turning point of the parabola is a minimum and occurs at \(\left(-2,1\right)\).

Example 2

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

Solution strategy:

  1. Find intercepts on \(x-\)axis.

  2. Find stationary points.

  3. Establish if stationary points are maximum or minimum values.

  4. Plot intercepts and stationary points and sketch the graph.


For \(x-\)axis intercepts, we set \(y=0,\) so \[\begin{align*} 0 & =x^{3}-x\\ & =x\left(x^{2}-1\right). \end{align*}\] Consequently, \(x=0\) or \[\begin{align*} x^{2}-1 & =0\\ x & =1\text{ or $-1.$ } \end{align*}\] Hence the \(x-\)axis intercepts are \(\left(-1,0\right),\ \left(0,0\right)\ \text{and $\left(1,0\right)$ .}\)

For stationary points, \[\begin{align*} \frac{dy}{dx} & =0 \end{align*}\] that is \[\begin{align*} 3x^{2}-1 & =0\\ x^{2} & =\frac{1}{3}\\ x & =\pm\frac{1}{\sqrt{3}}\\ & \approx\pm0.58. \end{align*}\]

Substituting these \(x\) values back in \(y=x-x^{3}\) gives for \(x=1/\sqrt{3}\), \[\begin{align*} y & =\left(\frac{1}{\sqrt{3}}\right)^{3}-\frac{1}{\sqrt{3}}\\ & \approx-0.39.\\ \end{align*}\] For \(x=-1/\sqrt{3}\) \[\begin{align*} y & =\left(-\frac{1}{\sqrt{3}}\right)^{3}-\left(-\frac{1}{\sqrt{3}}\right)\\ & \approx0.38.\\ \end{align*}\] Hence the stationary points occur at approximately, \(\left(0.58,-0.39\right)\) and \(\left(-0.58,0.38\right)\). Now we decide if these points are maxima or minima. Consider the tables below.

For \(x\approx0.58\) we have

Table shows change in gradient at x equals 0.58.

and so the point \(x=1/\sqrt{3}\approx0.58\) is a minimum.

For \(x\approx-0.58\) we have

Table shows change in gradient at a maximum.

and so the point \(x=-1/\sqrt{3}\approx-0.58\) is a maximum.

We can now graph \(y=x^{3}-x\). First plot the \(x-\)intercepts and the stationary points:

x intercepts and stationary points

We know \(E\) is a maximum and \(D\) is a minimum and so can graph \(y=x^{3}-x\) as shown below.

Graph of x cubed minus x.


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. \(y=\left(x+1\right)^{4}\)

Download this page, D9 Curve Sketching (PDF 779KB)

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