[ Investigation of a quadratic trinomial ] [ Methods for solving problems with parameters ]

For the quadratic trinomial $f(x)=a x^{2}-a x+1$, it is known that $|f(x)| \leq 1$ for $0 \leq x \leq 1$. Find the greatest possible value of $a$.
Since $f(0)=f(1)=1$, the graph of the quadratic polynomial is a parabola symmetric with respect to the line $x=0.5$. From the condition $|f(x)| \leq 1$ for

$0 \leq x \leq 1$, it follows that the branches of the parabola are directed upwards. The minimum value of $f(x)$ is $f(0.5)=1-a / 4$.

The maximum possible value of $a$ is achieved when this number equals -1, that is, when $a=8$.