Tell whether f(x) = (1/5)x^2 - 5x + 27 has a minimum value or a maximum value, and find the value.

Steps to Solve

1. Determine if the function has a minimum or maximum value

   The quadratic function is given by $f(x) = \frac{1}{5}x^2 - 5x + 27$. Since the coefficient of the $x^2$ term, which is $\frac{1}{5}$, is positive, the parabola opens upwards. Therefore, the function has a minimum value.

2. Find the x-coordinate of the vertex

   The x-coordinate of the vertex of a parabola given by $f(x) = ax^2 + bx + c$ is given by the formula $x = -\frac{b}{2a}$. In this case, $a = \frac{1}{5}$ and $b = -5$, so the x-coordinate of the vertex is: $$x = -\frac{-5}{2(\frac{1}{5})} = \frac{5}{\frac{2}{5}} = 5 \cdot \frac{5}{2} = \frac{25}{2}$$

3. Find the y-coordinate of the vertex (minimum value)

   To find the minimum value, substitute the x-coordinate of the vertex into the function: $$f(\frac{25}{2}) = \frac{1}{5}(\frac{25}{2})^2 - 5(\frac{25}{2}) + 27$$ $$f(\frac{25}{2}) = \frac{1}{5}(\frac{625}{4}) - \frac{125}{2} + 27$$ $$f(\frac{25}{2}) = \frac{125}{4} - \frac{250}{4} + \frac{108}{4}$$ $$f(\frac{25}{2}) = \frac{125 - 250 + 108}{4} = \frac{-125 + 108}{4} = \frac{-17}{4}$$