1. Problem: Find the maximum value of the function $f(x) = -6x^2 + 3x - 4$. 2. This is a quadratic function in the form $f(x) = ax^2 + bx + c$ where $a = -6$, $b = 3$, and $c = -4$. 3. Since $a < 0$, the parabola opens downward, so the function has a maximum value at its vertex. 4. The $x$-coordinate of the vertex is given by the formula $$x = -\frac{b}{2a} = -\frac{3}{2 \times (-6)} = -\frac{3}{-12} = \frac{1}{4}.$$ 5. Substitute $x = \frac{1}{4}$ back into the function to find the maximum value: $$f\left(\frac{1}{4}\right) = -6\left(\frac{1}{4}\right)^2 + 3\left(\frac{1}{4}\right) - 4 = -6\times \frac{1}{16} + \frac{3}{4} - 4 = -\frac{6}{16} + \frac{3}{4} - 4.$$ 6. Simplify the terms: $$-\frac{6}{16} = -\frac{3}{8}, \quad \frac{3}{4} = \frac{6}{8},$$ so $$f\left(\frac{1}{4}\right) = -\frac{3}{8} + \frac{6}{8} - 4 = \frac{3}{8} - 4 = \frac{3}{8} - \frac{32}{8} = -\frac{29}{8}.$$ 7. Therefore, the maximum value of the function is $$-\frac{29}{8}$$ at $$x = \frac{1}{4}.$$