1. **Problem Statement:** Given the quadratic function in vertex form $f(x) = 3(x - 4)^2 + 2$, find the vertex, determine if it has a maximum or minimum value, find the axis of symmetry, and state the range. 2. **Formula and Rules:** The vertex form of a quadratic function is: $$f(x) = a(x - h)^2 + k$$ where $(h, k)$ is the vertex. - The axis of symmetry is the vertical line $x = h$. - If $a > 0$, the parabola opens upwards and the vertex is a minimum point. - If $a < 0$, the parabola opens downwards and the vertex is a maximum point. - The range depends on the vertex and the direction the parabola opens. 3. **Identify the vertex:** From $f(x) = 3(x - 4)^2 + 2$, we see: - $a = 3$ (positive) - $h = 4$ - $k = 2$ So, the vertex is at $(4, 2)$. 4. **Determine maximum or minimum:** Since $a = 3 > 0$, the parabola opens upwards, so the vertex is a minimum point. 5. **Axis of symmetry:** The axis of symmetry is the vertical line: $$x = 4$$ 6. **Range:** Because the parabola opens upwards and the minimum value is $k = 2$, the range is: $$[2, \infty)$$ 7. **Summary:** - Vertex: $(4, 2)$ - Minimum value: $2$ - Axis of symmetry: $x = 4$ - Range: $[2, \infty)$ This completes the solution for the first problem.