1. **State the problem:** Complete the square for the quadratic function $$f(x) = 4x^2 - x + 2$$ and find the vertex coordinates, maximum or minimum value, and its nature. 2. **Recall the formula:** To complete the square for $$ax^2 + bx + c$$, rewrite as $$a\left(x^2 + \frac{b}{a}x\right) + c$$, then add and subtract $$\left(\frac{b}{2a}\right)^2$$ inside the parentheses. 3. **Apply to our function:** $$f(x) = 4x^2 - x + 2 = 4\left(x^2 - \frac{1}{4}x\right) + 2$$ 4. **Complete the square inside the parentheses:** Calculate $$\left(\frac{b}{2a}\right)^2 = \left(\frac{-\frac{1}{4}}{2}\right)^2 = \left(-\frac{1}{8}\right)^2 = \frac{1}{64}$$ 5. **Add and subtract $$\frac{1}{64}$$ inside the parentheses:** $$4\left(x^2 - \frac{1}{4}x + \frac{1}{64} - \frac{1}{64}\right) + 2 = 4\left(\left(x - \frac{1}{8}\right)^2 - \frac{1}{64}\right) + 2$$ 6. **Distribute 4 and simplify:** $$4\left(x - \frac{1}{8}\right)^2 - 4 \times \frac{1}{64} + 2 = 4\left(x - \frac{1}{8}\right)^2 - \frac{1}{16} + 2$$ 7. **Combine constants:** $$-\frac{1}{16} + 2 = -\frac{1}{16} + \frac{32}{16} = \frac{31}{16}$$ 8. **Final vertex form:** $$f(x) = 4\left(x - \frac{1}{8}\right)^2 + \frac{31}{16}$$ 9. **Vertex coordinates:** The vertex is at $$\left(\frac{1}{8}, \frac{31}{16}\right)$$. 10. **Maximum or minimum value:** Since the coefficient of $$x^2$$ is positive ($$4 > 0$$), the parabola opens upwards, so the vertex is a minimum point. 11. **Minimum value:** The minimum value of $$f(x)$$ is $$\frac{31}{16}$$ at $$x = \frac{1}{8}$$. **Final answers:** - Vertex: $$\left(\frac{1}{8}, \frac{31}{16}\right)$$ - Minimum value: $$\frac{31}{16}$$ - Nature: Minimum