If \[y=\frac{ax}{x^2+x+1} \qquad (a>0)\] show that for real values of \(x\) \[-a\leq y \leq \frac{1}{3}a.\]
Could we turn this into a quadratic in \(x\)?
What condition now do we need to ensure the roots for \(x\) are real?
Sketch the graph of \(y\) in the case \(a=6\).
Are there any asymptotes? How does the curve behave as \(\vert x\vert \rightarrow \infty\)? What about roots, intercepts and stationary points?
How could we use the first part in our sketch?
By drawing on the same diagram a certain straight line show that the equation \(x^3-1=6x\) has three real roots, two negative and one positive.
Can we factorise \(x^3-1\)?
