Complete the square to write \(7 - T^2 - 6T\) in vertex form.
\(7 - T^2 - 6T\) is not quite in standard form. Let’s rearrange terms. \(-T^2 - 6T + 7\) is in standard form, therefore
- a = \(\answer {-1}\)
- b = \(\answer {-6}\)
- c = \(\answer {7}\)
\(a \ne 1\), so let’s factor it out of the quadratic and linear terms.
Now, we are just focusing on the quadratic inside the parentheses. It doesn’t have a constant term. The \(7\) will just float around outside the parentheses for a moment.
Our inside quadratic is \(T^2 + 6T\) or \(T^2 + 6T + 0\).
- a = \(\answer {1}\)
- b = \(\answer {6}\)
- c = \(\answer {0}\)
This makes \(\frac {b}{2} = \answer {3}\), and its square is \(\left ( \frac {b}{2} \right )^2 = \answer {9}\), which is added and subtracted to the expression. That way we have only add \(0\) to the expression and not changed any values.
The “added” number is grouped together with the quadratic and linear terms to form a perfect square.
That was happening inside the parentheses, so let’s replace the old inside with the new inside.
Distribute and combine the constant terms.