### Cubic Spin

Prove that the graph of f(x) = x^3 - 6x^2 +9x +1 has rotational symmetry. Do graphs of all cubics have rotational symmetry?

### Sine Problem

In this 'mesh' of sine graphs, one of the graphs is the graph of the sine function. Find the equations of the other graphs to reproduce the pattern.

### More Parabolic Patterns

The illustration shows the graphs of twelve functions. Three of them have equations y=x^2, x=y^2 and x=-y^2+2. Find the equations of all the other graphs.

# Parabolic Patterns

##### Stage: 4 and 5 Challenge Level:

We have had solutions from Fiona, a Year 10 student from Stamford High School (Lincolnshire), and from Bei Guo, Kristin and Ryan from Riccarton High School in Christchurch (New Zealand). Well done to you all

They all noticed that the parabolas could be treated as three separate sets of five parabolas.

Starting with $y = x^2$, the parabola going through the origin, they noticed that it could be raised or lowered to produce the rest of the parabolas in the middle set by either adding or subtracting a $2$ or a $4$.

Therefore the solutions for the middle set are:

$y = x^2$
$y = x^2 + 2$
$y = x^2 + 4$
$y = x^2 - 2$
$y = x^2 - 4$

Then they used the other equation that had been given, $y = - (x - 4)^2$ , and found that it gave one of the parabolas in the right hand set.

$y = - (x^2 )$ is a reflection of $y = x^2$ in the horizontal axis, so that is why the new parabola was an inverted version of the original one.

$y = (x - 4)^2$ is a translation of $y = x^2$ by $4$ units to the right, so that is why the new parabola was the inverted parabola shifted $4$ units to the right.

As before, they noticed that $y = - (x - 4)^2$ could be raised or lowered to produce the rest of the parabolas in the right hand set by either adding or subtracting a $2$ or a $4$.

Therefore the solutions for the right hand set are:

$y = - (x - 4)^2$
$y = - (x - 4)^2 + 2$
$y = - (x - 4)^2 + 4$
$y = - (x - 4)^2 - 2$
$y = - (x - 4) ^2 - 4$

Finally, they noticed that the left hand set of parabolas were a reflection of the right hand set in the vertical axis. Therefore, they reasoned that the parabola that went through $(- 4, 0)$ would be $y = - (x + 4)^2$ : the negative sign in front of the brackets produces the inverted parabola, and the $+ 4$ inside the bracket translates it $4$ units to the left.

As before, they noticed that $y = - (x + 4)^2$ could be raised or lowered to produce the rest of the parabolas in the left hand set by either adding or subtracting a $2$ or a $4$.

Therefore the solutions for the left hand set are:

$y = - (x + 4)^2$
$y = - (x + 4)^2 + 2$
$y = - (x + 4)^2 + 4$
$y = - (x + 4)^2 - 2$
$y = - (x + 4)^2 - 4$