Determine Critical Values for y = 3x² - 6x + 4: When is f(x) < 0?

Let's analyze the function $y = 3x^2 - 6x + 4$ to determine when it is negative.

First, calculate the discriminant $\Delta$:

$\Delta = (-6)^2 - 4 \cdot 3 \cdot 4 = 36 - 48 = -12$

A negative discriminant ($\Delta < 0$) indicates that there are no real roots, meaning the graph of the function does not intersect the x-axis. Since $a = 3 > 0$, the parabola opens upwards.

This means the vertex of the parabola represents the minimum point, and the entire graph is above the x-axis.

Consequently, the function $y = 3x^2 - 6x + 4$ does not attain any negative values for any real $x$.

The correct interpretation is that the function stays positive, confirming the conclusion:

The function has no negative values.