Along the x axis the potential is V = 2.0x³ - 9.0x, in volt, with x in metres…

Step 1. The field. $E_x = -\dfrac{dV}{dx} = -(6.0x^2 - 9.0)\ \text{V/m}$.

Step 2. At $x = 0.50$ m. $E_x = -(6.0 \times 0.25 - 9.0)\ \text{V/m} = +7.5\ \text{V/m}$. The field points in the $+x$ direction.

Step 3. Where it is zero. $6.0x^2 = 9.0$, so $x^2 = 1.5$ and $x = 1.22\ \text{m}$.

Step 4. The potential there. $V = 2.0 \times (1.2247)^3 - 9.0 \times 1.2247 = 3.674 - 11.023 = -7.35\ \text{V}$.