Vertex Intersection with X-Axis: Find X in the Parabolic Function

To solve this problem, we will look at the behavior of the quadratic function and determine when it is greater than zero:

• Step 1: The intersection point A is the vertex, which means $f(x) = a(x - A)^2 + k$ for some constants $a$ and $k=0$. This implies $f(x)$ changes sign at its vertex.
• Step 2: Determine if the parabola opens upwards or downwards. Since the graph of the function intersects the $x$-axis at the vertex, there are no additional real roots, which indicates either $f(x) \geq 0$ or $f(x) \leq 0$ throughout. As $f(x) > 0$ requires parts of the parabola above the $x$-axis, the parabola must open upwards.
• Step 3: For $f(x) > 0$, the graph being a parabola indicates positive $x$ intervals are outside of the vertex, i.e., $x < A$ and $x > A$.
• Step 4: The answers fitting this description are (b) $x < A$ and (c) $x > A$, which combined correspond to option (d) "Answers (b) + (c) are correct".

Therefore, the correct intervals for $f(x) > 0$ are both $x < A$ and $x > A$, leading to:

Answers (b) + (c) are correct.