To find the correct equation for a parabola that opens upward, has a minimum value of 3, and has an axis of symmetry at [tex]\(x = 3\)[/tex], we need to use the standard form of the quadratic equation representing a parabola:
[tex]\[ f(x) = a(x - h)^2 + k \][/tex]
Here:
- [tex]\((h, k)\)[/tex] is the vertex of the parabola.
- [tex]\(a\)[/tex] determines the direction the parabola opens (if [tex]\(a > 0\)[/tex], it opens upward).
Given the conditions:
- The parabola opens upward, so [tex]\(a > 0\)[/tex].
- The minimum value (vertex's y-coordinate) is 3, so [tex]\(k = 3\)[/tex].
- The axis of symmetry is [tex]\(x = 3\)[/tex], so [tex]\(h = 3\)[/tex].
Therefore, we substitute [tex]\(h = 3\)[/tex] and [tex]\(k = 3\)[/tex] into the vertex form equation:
[tex]\[ f(x) = a(x - 3)^2 + 3 \][/tex]
Since the parabola opens upward, [tex]\(a\)[/tex] must be positive. For simplicity, we can take [tex]\(a = 1\)[/tex] (as there is no indication otherwise). Thus:
[tex]\[ f(x) = (x - 3)^2 + 3 \][/tex]
Now, we match this equation with the given options:
- Option A: [tex]\(f(x) = (x+3)^2 + 3\)[/tex]
- Option B: [tex]\(f(x) = (x+3)^2 - 6\)[/tex]
- Option C: [tex]\(f(x) = (x-3)^2 - 6\)[/tex]
- Option D: [tex]\(f(x) = (x-3)^2 + 3\)[/tex]
Clearly, the equation [tex]\((x - 3)^2 + 3\)[/tex] matches Option D.
So, the correct answer is:
[tex]\[ \boxed{D} \][/tex]
