Parabola Property Analysis: Vertex Position and Positive Values

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

• Given a quadratic function $y = ax^2 + bx + c$, if it opens upwards, a > 0.

• The vertex form of the parabola is $y = a(x-h)^2 + k$, where $(h, k)$ is the vertex.

• If the vertex $(h, k)$ is above the x-axis, then k > 0.

First, let's restate what it means for a quadratic to be always positive:
This means the function $y = ax^2 + bx + c$ has no real roots and y > 0 for all $x$.

To ensure this, consider:

• The discriminant $\Delta = b^2 - 4ac$ helps determine if the parabola intersects the x-axis.

• If \Delta < 0, there are no real roots, meaning the parabola doesn't cross the x-axis and stays entirely above it for all $x$.

• Given a > 0 and the vertex is above the x-axis (k > 0), it ensures the function $y = a(x-h)^2 + k$ stays y > 0.

Therefore, with k > 0 and a > 0, $\Delta$ implies no x-intercepts, confirming the parabola is always positive.

The correct answer is: Correct