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:

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.
   

   

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

   

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

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