A

Step 1: Identify the function: \( y = 3t^2 - 4t \).
Step 2: To find the minima, we need to compute the first derivative of the function. \( \frac{dy}{dt} = 6t - 4 \).
Step 3: Set the first derivative to zero to find critical points:
\( 6t - 4 = 0 \)
\( 6t = 4 \)
\( t = \frac{4}{6} = \frac{2}{3} \).
Step 4: To confirm that this point is a minimum, we calculate the second derivative: \( \frac{d^2y}{dt^2} = 6 \). Since \( \frac{d^2y}{dt^2} > 0 \), it indicates that the function has a local minimum at \( t = \frac{2}{3} \).
Step 5: Substitute \( t = \frac{2}{3} \) back into the original function to find the minimum value of y:
\( y = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) \)
\( y = 3\left(\frac{4}{9}\right) - \frac{8}{3} \)
\( y = \frac{12}{9} - \frac{24}{9} = -\frac{12}{9} = -\frac{4}{3} \).
Therefore, the minimum value of y is -\frac{4}{3} at \( t = \frac{2}{3} \). This minimum point is significant in determining the behavior of the quadratic function.