1. The problem is to find the vertex of the quadratic function $$y = -3x^2 + 12x + 29$$. 2. The vertex of a quadratic function in the form $$y = ax^2 + bx + c$$ can be found using the formula for the x-coordinate of the vertex: $$x = -\frac{b}{2a}$$ 3. After finding the x-coordinate, substitute it back into the original equation to find the y-coordinate of the vertex. 4. Important rules: - The coefficient $$a$$ determines the direction of the parabola (if $$a < 0$$, it opens downward; if $$a > 0$$, it opens upward). - The vertex represents the maximum point if the parabola opens downward, and the minimum point if it opens upward. 5. Steps to find the vertex: - Identify $$a = -3$$, $$b = 12$$, and $$c = 29$$ from the equation. - Calculate $$x = -\frac{b}{2a} = -\frac{12}{2 \times -3}$$. - Simplify the fraction, showing cancellation: $$x = -\frac{\cancel{12}}{2 \times \cancel{-3}} = -\frac{12}{-6}$$ - Evaluate $$x$$. - Substitute this $$x$$ value into the original equation to find $$y$$. 6. The vertex coordinates $$ (x, y) $$ give the maximum or minimum point of the parabola. Final answer: The steps to find the vertex of $$y = -3x^2 + 12x + 29$$ are to use $$x = -\frac{b}{2a}$$, then substitute $$x$$ back into the equation to find $$y$$.