Parabola Property: Vertex on X-axis and Positive Value Analysis

To solve this problem, let's analyze the function:

• Step 1: Given the parabola is in vertex form $y = a(x - h)^2 + k$ where $a > 0$ for an upward opening.
• Step 2: Since the vertex is on the x-axis, $k = 0$. Thus, the equation simplifies to $y = a(x - h)^2$.
• Step 3: Analyze values:
  • At $x = h$, the vertex, $y = a(h - h)^2 = 0$.
  • For $x \neq h$, $y = a(x - h)^2 > 0$ since $a > 0$ and any square is non-negative, but only zero at $h$.
• Step 4: Therefore, the function $y$ is zero only at the vertex $(h, 0)$ and positive for all other $x$.

The statement in the problem says the function is always positive except at the vertex. As we see, the function is indeed zero only at the vertex and positive elsewhere, meaning the statement provided is incorrect in its description if one understands it as implying it should never reach zero, which technically it does only at the vertex.

Therefore, the correct answer is Incorrect.