Complete the square to write \(y^2 + 4y + 11\) in vertex form.
\(y^2 + 4y + 11\) is in standard form, therefore
- a = \(\answer {1}\)
- b = \(\answer {4}\)
- c = \(\answer {11}\)
This makes \(\frac {b}{2} = \answer {2}\), and its square is \(\left ( \frac {b}{2} \right )^2 = \answer {4}\), which is added and subtracted to the expression. That way we have only add \(0\) to the expression and not changed any values.
\[ y^2 + 4y + \answer {4} - \answer {4} + 11 \]
The “added” number is grouped together with the quadratic and linear terms to form a perfect square.
\[ \left ( y + \answer {2} \right )^2 - 4 + 11 \]
And, we can combine the constant terms.
\[ ( y + 2)^2 + 7 \]