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Completion requirements
Convert \(f\left( x \right) = -2x^2 + 5x + 3\) to vertex form by completing the square.
\[\begin{array}{l}
f\left( x \right) = -2x^2 + 5x + 3 \\
f\left( x \right) = \left( { -2x^2 + 5x} \right) + 3 \\
f\left( x \right) = -2\left( {x^2 - \frac{5}{2}x} \right) + 3 \\
f\left( x \right) = -2\left( {x^2 - \frac{5}{2}x + \left( {\frac{5}{4}} \right)^2 - \left( {\frac{5}{4}} \right)^2 } \right) + 3 \\
f\left( x \right) = -2\left( {x^2 - \frac{5}{2}x + \left( {\frac{5}{4}} \right)^2 } \right) + 3 + 2\left( {\frac{5}{4}} \right)^2 \\
f\left( x \right) = -2\left( {x - \frac{5}{4}} \right)^2 + 3 + \frac{{25}}{8} \\
f\left( x \right) = -2\left( {x - \frac{5}{4}} \right)^2 + \frac{{24}}{8} + \frac{{25}}{8} \\
f\left( x \right) = -2\left( {x - \frac{5}{4}} \right)^2 + \frac{{49}}{8} \\
\end{array}\]
f\left( x \right) = -2x^2 + 5x + 3 \\
f\left( x \right) = \left( { -2x^2 + 5x} \right) + 3 \\
f\left( x \right) = -2\left( {x^2 - \frac{5}{2}x} \right) + 3 \\
f\left( x \right) = -2\left( {x^2 - \frac{5}{2}x + \left( {\frac{5}{4}} \right)^2 - \left( {\frac{5}{4}} \right)^2 } \right) + 3 \\
f\left( x \right) = -2\left( {x^2 - \frac{5}{2}x + \left( {\frac{5}{4}} \right)^2 } \right) + 3 + 2\left( {\frac{5}{4}} \right)^2 \\
f\left( x \right) = -2\left( {x - \frac{5}{4}} \right)^2 + 3 + \frac{{25}}{8} \\
f\left( x \right) = -2\left( {x - \frac{5}{4}} \right)^2 + \frac{{24}}{8} + \frac{{25}}{8} \\
f\left( x \right) = -2\left( {x - \frac{5}{4}} \right)^2 + \frac{{49}}{8} \\
\end{array}\]