## Sketching Cubic Graphs

Contents

**General method for sketching cubic graphs:**

- Consider the sign of \(a\) and determine the general shape of the graph.
Determine the \(y\)-intercept by letting \(x=0\).

Determine the \(x\)-intercepts by factorising \(a{x}^{3}+b{x}^{2}+cx+d=0\) and solving for \(x\).

Find the \(x\)-coordinates of the turning points of the function by letting \(f'(x) = 0\) and solving for \(x\).

Determine the \(y\)-coordinates of the turning points by substituting the \(x\)-values into \(f(x)\).

Plot the points and draw a smooth curve.

## Example

### Question

Sketch the graph of \(g(x)={x}^{3}-3{x}^{2}-4x\).

### Determine the shape of the graph

The coefficient of the \(x^{3}\) term is greater than zero, therefore the graph will have the following shape:

### Determine the intercepts

The \(y\)-intercept is obtained by letting \(x = 0\):

\begin{align*} g(0)&=(0)^{3}-3(0)^{2}-4(0)\\ & = 0 \end{align*}

This gives the point \((0;0)\).

The \(x\)-intercept is obtained by letting \(g(x) = 0\) and solving for \(x\):

\begin{align*} 0 & = {x}^{3}-3{x}^{2}-4x \\ & = x(x^{2} – 3x – 4) \\ & = x(x – 4)(x+1)\\ \therefore x=-1, \enspace & x = 0 \text{ or } x = 4 \end{align*}

This gives the points \((-1;0)\), \((0;0)\) and \((4;0)\).

### Calculate the stationary points

Find the \(x\)-coordinates of the stationary points by setting \({g}'(x)=0\):

\begin{align*} {g}'(x) & = 3x^{2} – 6x – 4 \\ 0 & = 3x^{2} – 6x – 4 \\ \text{Using the quadratic formula} \enspace x & = \cfrac{-(-6) \pm \sqrt{(-6)^{2} – 4(3)(-4)}}{2(3)} \\ & = \cfrac{6 \pm \sqrt{36 + 48}}{6} \\ \therefore x = \text{2.53} & \text{ or } x = -\text{0.53} \end{align*}

Substitute these \(x\)-coordinates into \(g(x)\) to determine the corresponding \(y\)-coordinates:

\begin{align*} g(x)&=(\text{2.53})^{3}-3(\text{2.53})^{2}-4(\text{2.53})\\ & = -\text{13.13} \end{align*}\begin{align*} g(x)&=(-\text{0.53})^{3}-3(-\text{0.53})^{2}-4(-\text{0.53}) \\ & = \text{1.13} \end{align*}

Therefore, the stationary points are \((\text{2.53}; -\text{13.13})\) and \((-\text{0.53}; \text{1.13})\).

### Draw a neat sketch

**Concavity**

Concavity indicates whether the gradient of a curve is increasing, decreasing or stationary.

- Concave up: the gradient of the curve increases as \(x\) increases.
- Concave down: the gradient of the curve decreases as \(x\) increases.
- Zero concavity: the gradient of the curve is constant.

The diagram below shows the graph of the cubic function \(k(x) = x^{3}\). The first derivative of \(k(x)\) is a quadratic function, \(k'(x) = 3x^{2}\) and the second derivative is a linear function, \(k”(x) = 6x\).

Notice the following:

- \(k”(x) > 0\), the graph is concave up.
- \(k”(x) < 0\), the graph is concave down.
- \(k”(x) = 0\), change in concavity (point of inflection).

**Points of inflection**

This is the point where the concavity of a curve changes, as shown in the diagram below. If \(a < 0\), then the concavity changes from concave up (purple) to concave down (grey) and if \(a > 0\), concavity changes from concave down (blue) to concave up (green). Unlike a turning point, the gradient of the curve on the left-hand side of an inflection point (\(P\) and \(Q\)) has the same sign as the gradient of the curve on the right-hand side.

A graph has a horizontal point of inflection where the derivative is zero but the sign of the gradient of the curve does not change. That means the graph (shown below) will continue to increase or decrease after the stationary point.

In the example above, the equation \(k'(x) = 3x^{2}\) indicates that the gradient of this curve will always be positive (except where \(x = 0\)). Therefore, the stationary point is a point of inflection.

\(f\): cubic function | \(f’\): quadratic function | \(f”\): linear function |

(blue graph) | (green graph) | (red graph) |

turning points \(arrow\) | \(x\)-intercepts | |

point of inflection | \(arrow\) turning point \(arrow\) | \(x\)-intercept |