The function \( f(x) = 2x^2 - 8x + 5 \) has a minimum value. What is this minimum value?

["Title: Finding the Minimum Value of the Quadratic Function ( f(x) = 2x^2 - 8x + 5 )", "---", "Understanding the Minimum Value of Quadratic Functions", "When analyzing quadratic functions of the form ( f(x) = ax^2 + bx + c ), one key feature is their parabolic shape — either opening upward (if ( a > 0 )) or downward (if ( a < 0 )). In this case, the function\n[\nf(x) = 2x^2 - 8x + 5\n]\nforms a parabola that opens upward since the coefficient of ( x^2 ) is positive (( a = 2 > 0 )). This guarantees the function has a minimum value at its vertex.", "---", "Finding the Vertex and Minimum Value", "The vertex of a quadratic function ( f(x) = ax^2 + bx + c ) occurs at\n[\nx = -\frac{b}{2a}\n]\nSubstituting ( a = 2 ) and ( b = -8 ):\n[\nx = -\frac{-8}{2 \cdot 2} = \frac{8}{4} = 2\n]", "Now, substitute ( x = 2 ) back into the original function to find the minimum value:\n[\nf(2) = 2(2)^2 - 8(2) + 5 = 2 \cdot 4 - 16 + 5 = 8 - 16 + 5 = -3\n]", "---", "Conclusion", "The quadratic function ( f(x) = 2x^2 - 8x + 5 ) reaches its minimum value at ( x = 2 ), and this minimum value is:", "[\n\boxed{-3}\n]", "This insight is essential in optimization problems, physics, economics, and many real-world applications where minimizing cost, maximizing efficiency, or finding lowest points is critical.", "---", "Keywords:\nquadratic function minimum, minimum of ( f(x) = 2x^2 - 8x + 5 ), vertex of a parabola, vertex formula, parabola opening upward, calculus-based minimum, algebra quadratic minimum", "Read more about how to find the vertex and analyze quadratic functions for maximum/minimum values.", "---", "Optimize your understanding of parabolas — their shape, vertex, and real-world implications—today."]