[tex]\begin{gathered} \text{If a parabola "opens" up, the minimum is the vertex, and it has no maximum} \\ \text{If a parabola "opens" down, the maximum is the vertex, and it has no minimum} \\ \\ \text{ This parabola opens up, so it has no maximum} \\ \frac{4}{3}x^2-4x=\frac{4}{3}(x-\frac{3}{2})^2-3 \\ \\ \text{Thus the minimum point is } \\ (\frac{3}{2},-3) \\ \\ Now,\text{ the minimum value is }-3 \end{gathered}[/tex]
