1. **State the problem:** We are given the quadratic function $f(x) = -x^2 + 6x + 3$ and need to find its range, opening direction, vertex, and domain. 2. **Recall the general form and properties:** A quadratic function is generally written as $f(x) = ax^2 + bx + c$. - The parabola opens upwards if $a > 0$ and downwards if $a < 0$. - The vertex $(h, k)$ can be found using the formula $h = -\frac{b}{2a}$ and $k = f(h)$. - The domain of any quadratic function is all real numbers, $(-\infty, \infty)$. - The range depends on the vertex and the direction of opening. 3. **Identify coefficients:** Here, $a = -1$, $b = 6$, and $c = 3$. 4. **Determine the opening:** Since $a = -1 < 0$, the parabola opens downward. 5. **Find the vertex:** $$ h = -\frac{b}{2a} = -\frac{6}{2 \times (-1)} = -\frac{6}{-2} = 3 $$ Calculate $k = f(3)$: $$ f(3) = -(3)^2 + 6 \times 3 + 3 = -9 + 18 + 3 = 12 $$ So the vertex is at $(3, 12)$. 6. **Determine the range:** Since the parabola opens downward and the vertex is the maximum point, $$ \text{Range} = (-\infty, 12] $$ 7. **Domain:** The domain of any quadratic function is all real numbers: $$ (-\infty, \infty) $$