To find the minimum value of the quadratic function \(f(x) = 2x^2 - 5x - 3\), we can use the vertex formula or complete the square. The vertex of a parabola given by \(ax^2 + bx + c\) represents either the maximum or minimum point, depending on whether \(a\) is positive or negative. In this case, \(a = 2>0\), indicating a parabola that opens upwards, meaning the vertex represents the minimum value.

Method 1: Completing the Square

We can rewrite the quadratic equation in vertex form, \(a(x - h)^2 + k\), where \((h, k)\) is the vertex.

1. Factor out the coefficient of \(x^2\) from the \(x^2\) and \(x\) terms: \(2(x^2 - \frac{5}{2}x) - 3\).
2. Complete the square inside the parentheses: Take half of the coefficient of \(x\) (\(-\frac{5}{4}\)), square it (\(\frac{25}{16}\)), and add and subtract it inside the parentheses: \(2(x^2 - \frac{5}{2}x + \frac{25}{16} - \frac{25}{16}) - 3\).
3. Rewrite as a perfect square: \(2((x - \frac{5}{4})^2 - \frac{25}{16}) - 3\).
4. Simplify: \(2(x - \frac{5}{4})^2 - \frac{25}{8} - 3 = 2(x - \frac{5}{4})^2 - \frac{49}{8}\).

The vertex is at \((h, k) = (\frac{5}{4}, -\frac{49}{8})\). Therefore, the minimum value is \(-\frac{49}{8} = -6.125\), and this occurs at \(x = \frac{5}{4} = 1.25\).

Method 2: Vertex Formula

The x-coordinate of the vertex of a parabola \(ax^2 + bx + c\) is given by \(x = -\frac{b}{2a}\). In this case, \(a = 2\) and \(b = -5\), so \(x = -\frac{-5}{2(2)} = \frac{5}{4} = 1.25\).

Substitute this value of \(x\) back into the original equation to find the minimum value:

\(f(1.25) = 2(1.25)^2 - 5(1.25) - 3 = 3.125 - 6.25 - 3 = -6.125\).

Therefore, the minimum value of the function is -6.125, and this occurs at x = 1.25.