smc

The acceleration must be zero at the $x$-intercept; this intercept must be an inflection point for the minimum $a$ value. Derive $f(x)$ so that the acceleration $f^{\prime \prime }(x)=0$. Using the power rule, $$\begin{array}{rl}f(x)& =x^3-a{x}^2+bx-a\\ f^{\prime }(x)& =3x^2-2ax+b\\ f^{\prime \prime }(x)& =6x-2a\end{array}$$ So $x=\frac{a}{3}$ for the inflection point/root. Furthermore, the slope of the function must be zero - maximum - at the intercept, thus having a triple root at $x=a/3$ (if the slope is greater than zero, there will be two complex roots and we do not want that). The function with the minimum $a$: $$f(x)={\left(x-\frac{a}{3}\right)}^3$$ $$x^3-a{x}^2+\left(\frac{a^2}{3}\right)x-\frac{a^3}{27}$$ Since this is equal to the original equation $x^3-a{x}^2+bx-a$, equating the coefficients, we get that $$\frac{a^3}{27}=a\to a^2=27\to a=3\sqrt{3}$$ $$b=\frac{a^2}{3}=\frac{27}{3}=\boxed{9}$$ The actual function: $f(x)=x^3-\left(3\sqrt{3}\right)x^2+9x-3\sqrt{3}$ $f(x)=0\to x=\sqrt{3}$ triple root. "Complete the cube."