# Assignment 13, Q3(iii)

I'm having some trouble locating the stationary point for $y=x^4-6x^2+ax$.

I presume I need to solve $y'=0$, ie $4x^3-12x+a=0$. And so the location of the TP is going to be a function of $a$, but I can't figure out its exact value. Wolfram Alpha gives me something horrible.

### Turning points

You don't really want to find the turning points.

Start by considering the case when $a=8$. You can show that this has two stationary points (at $x=1$ and $x=-2$ and two points where the second derivative is zero (at $x=1$ and $x=-1$). The points where the second derivative is zero cannot change (as these are not affected by changes in $a$) but the stationary points can.

Consider the graph of the gradient when $a=8$, i.e. the graph of $4x^3-12x+8$. You can show that this crosses the x-axis at -2 and 1, and sketch it (very useful). For $a>8$ the graph of the gradient will be $4x^3=12x +a$, so is a translation of $a-8$ up from the previous one. This means that the graph has shifted up, and therefore there is only one point where it crosses the x-axis. This means that there can only be one stationary point of $y=x^4-6x^2+ax$ and you can use your graph to roughly place it. The important point is that the point where $x=1$ which was a stationary point of inflection when $a=8$ is a non-stationary point of inflection for $a>8$.

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