Example  4
Convert \(y = 2x^2 + 3x + 1\) to vertex form by completing the square.

Step 1: Group the first two terms, and remove \(a\) as the common factor.

\[\begin{array}{l}
 y = \left( {2x^2 + 3x} \right) + 1 \\
 y = 2\left( {x^2 + \frac{3}{2}x} \right) + 1 \\
 \end{array}\]

Step 2: Build a perfect square trinomial. Calculate the value of the constant for the perfect square trinomial, \(c = \left( {\frac{b}{2}} \right)^2 \). Add and subtract this value within the brackets.

\[\begin{array}{l}
 c = \left( {\frac{{\frac{3}{2}}}{2}} \right)^2 =\left( {\frac{3}{4}} \right)^2  \\
 y = 2\left[ {x^2 + \frac{3}{2}x + \left( {\frac{3}{4}} \right)^2 - \left( {\frac{3}{4}} \right)^2 } \right] + 1 \\
 \end{array}\]

Step 3: Remove the subtracted value from inside the brackets by multiplying by \(a\).

\[\begin{array}{l}
 y = 2\left[ {x^2 + \frac{3}{2}x + \left( {\frac{3}{4}} \right)^2 } \right] + 1 - 2\left( {\frac{3}{4}} \right)^2  \\
 y = 2\left[ {x^2 + \frac{3}{2}x + \left( {\frac{3}{4}} \right)^2 } \right] + 1 - \frac{9}{8} \\
 \end{array}\]

Step 4: Factor the perfect square trinomial and simplify the constants.

\[\begin{array}{l}
 y = 2\left( {x + \frac{3}{4}} \right)^2 + \frac{8}{8} - \frac{9}{8} \\
 y = 2\left( {x + \frac{3}{4}} \right)^2 - \frac{1}{8} \\
 \end{array}\]

Notice that the binomial’s constant value is equal to \(\frac{b}{2}\).