# Thread: Sydney Grammar 2009 help

1. ## Sydney Grammar 2009 help

Screen Shot 2017-09-17 at 8.34.25 am.png
Screen Shot 2017-09-17 at 8.34.03 am.png
I've attached the question I don't understand. specifically, I don't understand the 3rd line of the solution, which states there is a double root at the y-coordinates of the stationary points? is that a rule where y-f(x) = 0 has a double root at y=f'(x)??

2. ## Re: Sydney Grammar 2009 help

Originally Posted by mathpie
Screen Shot 2017-09-17 at 8.34.25 am.png
Screen Shot 2017-09-17 at 8.34.03 am.png
I've attached the question I don't understand. specifically, I don't understand the 3rd line of the solution, which states there is a double root at the y-coordinates of the stationary points? is that a rule where y-f(x) = 0 has a double root at y=f'(x)??
$\noindent If f(x) is that cubic and (a,b) is the coordinates of a stationary point of it, then f(a) = b\Rightarrow f(a)-b = 0 (since the point lies on the graph) and f'(a) = 0 (slope 0 at a stationary point). Hence defining the cubic C(x):= f(x) -b = x^{3}-3x-b , we have C(a) = 0 and C'(a) = 0. Therefore, a is a double root of C(x) from what you know about multiplicity of roots of polynomials. (Of course it can't be a triple root here since C(x) is not like something of the form (x-a)^{3}.)$

3. ## Re: Sydney Grammar 2009 help

Nice question
Just use the formula in a) with p=3 to yield q=plus or minus 2
This means that y=x^3-3x+2 and y=x^3-3x-2 both have double roots....ie max or min situated at the x axis
Sliding y=x^3-3x+2 down 2 units gives a stationary point for y=x^3-3x with a y-value of -2. Similarly for the other

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