Parabola Analysis: Finding Positive Values Where f(x) > 0 at Vertex Point

The graph of the function below intersects the X-axis at one point A (the vertex of the parabola).

Find all values of $x$

where $f\left(x\right) > 0$.

To identify the conditions where $f(x) > 0$, we need to analyze the nature of the quadratic function as represented on the provided graph.

Based on the problem, the graph intersects the x-axis exactly at one point, recognized as point A, the vertex. In a quadratic function $ax^2 + bx + c$, if the vertex intersects at the x-axis and nowhere else, it means the graph is tangent to the x-axis at that vertex.

To determine if the function is positive, examine the orientation: - If $a > 0$, the parabola opens upwards, making it have a minimum at the vertex. - If $a < 0$, the parabola opens downwards, making it have a maximum at the vertex. Given that the problem states the parabola intersects the x-axis only at the vertex, the parabola opens downward. This is inferred from the phrased graph where no areas reach above the x-axis.

Therefore, the function never reaches a value greater than zero, as the parabola is concave down, and the vertex sits on the x-axis.

Conclusively, the range where $f(x) > 0$ is nonexistent given the parameters of the problem.

Therefore, the solution is that there are no such values.