Analyzing y=x²: Finding Where Parabola Values Stay Positive

To solve this problem, we will use the vertex form of a quadratic function, which is $y = (x-h)^2$, where $h$ is the x-coordinate of the vertex. The problem specifies that the parabola should be zero at $x = 2$ and positive elsewhere, indicating that the vertex of the parabola is at $x = 2$.

We'll take the following steps to find the correct function:

• Step 1: Identify the required vertex: The vertex needs to be at $x = 2$.
• Step 2: Apply this x-coordinate to the vertex form: Using $h = 2$, the function becomes $y = (x-2)^2$.
• Step 3: Verify that the function meets the condition: At $x = 2$, $y = (2-2)^2 = 0$, confirming that the function is zero at this point and positive elsewhere since a square of a real number is non-negative.

Therefore, the solution to the problem is $y = (x-2)^2$.