f(x) should be in canonical form. So it must have the form

Where a is the main coefficient and is the vertex
Explanation:
A quadratic function has a unique extreme value in its vertex. That value might be a maximum or a minimum depending on the sign of the main coefficient of the quadratic function. In order to quickly obtain the vertex, the quadratic must be written in canonical form. That means that f(x) must have the form

Where a is the main coefficient (which should be negative so that a minimum exists in the first place) and is the vertex. If f(x) is written in that form, then it will be easier to find the minimum of f(x), which is the vertex
Hence for the quadratic function below

Hence there is only one x- intercept and answer is (2, -9)