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