## General Cubic Function $y=ax^{3}+bx^{2}+cx+d$

In this form the y intercept is easy to find since if we set x=0, y =d.

To calculate the x intercepts is much more difficult. Since we know that a cubic will always end up heading in the direction that it starts in, either through a point of inflection or two turning points, we know there will be at least one x intercept. The easiest way to find the intercepts is through substitution.

The turning points can only be found through use of differentiation or using a calculator.

### Example $f(x)=-x^{3}+19x-30$

Step 1: Find y-intercept, let x=0 $f(0)=-(0)^{3}+19(0)-30=-30$

Step 2: Find x intercept, let y=0 $f(x)=-x^{3}+19x-30=0$

The easiest way is to factorise the equation is to start substituting in values of x and seeing if the equation equals zero.

Let’s try x=1 $f(1)=-1^{3}+19\times 1-30=-12$

Since the function does not equal zero then we know that one is not a factor so we try another number. $f(2)=-2^{3}+19\times 2-30=0$

Therefore two is a factor of the function. If we take out x minus two from the function then we get $f(x)=(x-2)(-x^{2}-2x+15)$

We can then use either the quadratic formula or continue to guess and check other factors.

Let’s try 3 given it is a factor of 15 $f(3)=-3^{3}+19\times 3-30\newline = -27+57-30=0$

Then we can take out (x-3) from the equation $f(x)=(x-2)(-x+3)(x+5)$

Therefore the x-intercepts are 2,3 and -5

#### Sign Diagram

Step 3: Determine the Shape

A sign diagram gives a general idea of the shape of a cubic by telling us where the function is positive or negative.

We start by substituting in points between the x intercepts $f(\frac{5}{2})=-(\frac{5}{2})^{3}+19(\frac{5}{2})-30 \newline \newline = -\frac{125}{8}+\frac{95}{2}-30=\frac{15}{8}$

This tells us the function will be positive for 2>x>3.

We already know a point between -5 and 2, the y intercept, which is negative 30. Therefore we know that the function will be negative between these two points.

Now we just need to work out whether the function is positive or negative before and after the x intercepts. $f(-6)=-(-6)^{3}+19(-6)-30 \newline = 216-114-30=72$ $f(4)=-(4)^{3}+19(4)-30 \newline = -64+76-30=-18$

Therefore it is positive when x<-5 and negative for x>4

Now we can can draw the sign diagram to get an idea of the shape Step 4: Find the local minimum and maximums

They can be found through the use of the min and max functions on your calculators or through power differentiation shown below $f'(x)=-3\times x^{3-1}+19\newline =-3x^{2}+19$

Find the turning points by setting f'(x)=0 and solve for x $-3x^{2}+19=0\newline x=\pm\sqrt{\frac{19}{3}}\newline=\pm2.52$

Find the y co ordinates by substituting in the x values above into the original equation $f(2.52)=-2^{2.52}+19\times 2.52-30=1.88$ $f(-2.52)=-2^{-2.52}+19\times -2.52-30=-61.88$

Therefore the turning points are (2.52,1.88) and (-2.52,-61.88)

So the graph can be sketched below: 