Final answer:
The range of the quadratic function f(x) = (2/5)x^2 + 11 is [11, ∞), as this parabola opens upwards and the minimum value is at the vertex (0,11).
Step-by-step explanation:
The range of the function f(x) = (2/5)x^2 + 11 can be determined by looking at the properties of the quadratic function. Since the coefficient of x^2 is positive, the parabola opens upwards and the vertex of the parabola represents the minimum point of the function on the graph.
Therefore, for all real values of x, the minimum value of f(x) is at the vertex. In this case, since there is no x-term that displaces the vertex horizontally, the vertex is at (0, 11). Hence, the range of f(x) is [11, ∞), meaning it includes 11 and all real numbers greater than 11.