We can put this form of a “completing square” learned in algebra.
First, put a space in-between the 2nd and 3rd terms,
\(y = {x^2} + 6x\;\,\;\; + 4\)
Second, take\(\frac{1}{2}\)of middle coefficient, square it, and add and subtract from the right side,
\(y = {x^2} + 6x + {3^2} + \;\;\;\;4 - {3^2}\)
Third, the first three terms become the square of a binomial\({\left( {x + 3} \right)^2}\),
hence\(y = {\left( {x + 3} \right)^2} - 5\).
The transformed graph is given in purple below.
We can now see that the basic graph is given by\(f\left( x \right) = {x^2}\)as seen in red below.
Next the \( - 5\) shifts the blue graph by 5 units down. The final transformed graph has the colour purple.