1. The problem is to analyze the quadratic function $$f(x) = 2x^2 - 5x + 1$$ and understand its properties. 2. The general form of a quadratic function is $$f(x) = ax^2 + bx + c$$ where $a$, $b$, and $c$ are constants. 3. For this function, $a=2$, $b=-5$, and $c=1$. 4. To find the vertex, use the formula for the $x$-coordinate of the vertex: $$x = -\frac{b}{2a}$$. 5. Substitute the values: $$x = -\frac{-5}{2 \times 2} = \frac{5}{4}$$. 6. Find the $y$-coordinate by evaluating $f\left(\frac{5}{4}\right)$: $$f\left(\frac{5}{4}\right) = 2\left(\frac{5}{4}\right)^2 - 5\left(\frac{5}{4}\right) + 1 = 2\times \frac{25}{16} - \frac{25}{4} + 1 = \frac{50}{16} - \frac{100}{16} + \frac{16}{16} = \frac{50 - 100 + 16}{16} = \frac{-34}{16} = -\frac{17}{8}$$. 7. The vertex is at $$\left(\frac{5}{4}, -\frac{17}{8}\right)$$. 8. The parabola opens upwards because $a=2 > 0$. 9. To find the roots, use the quadratic formula: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$. 10. Calculate the discriminant: $$\Delta = b^2 - 4ac = (-5)^2 - 4 \times 2 \times 1 = 25 - 8 = 17$$. 11. The roots are: $$x = \frac{5 \pm \sqrt{17}}{4}$$. 12. These are the $x$-intercepts of the function. 13. Summary: The quadratic function $$f(x) = 2x^2 - 5x + 1$$ has vertex at $$\left(\frac{5}{4}, -\frac{17}{8}\right)$$, opens upwards, and roots at $$x = \frac{5 \pm \sqrt{17}}{4}$$.