Can we sketch $y=ax/(x^2+x+1)$ when $a$ is positive?

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If \[y=\frac{ax}{x^2+x+1} \qquad (a>0)\] show that for real values of \(x\) \[-a\leq y \leq \frac{1}{3}a.\]

Sketch the graph of \(y\) in the case \(a=6\).

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.