Express \(3x^{2} – 6x + 10\) in the form \(a(x – b)^{2} + c\), where a, b and c are integers. Hence state the minimum value of \(3x^{2} – 6x + 10\) and the value of x for which it occurs.
Explanation
\(3x^{2} - 6x + 10\)
\(3(x^{2} - 2x) + 10\)
\(3[x^{2} - 2x + (\frac{1}{2}(-2))^{2}] + 10 - 3\)
= \(3(x - 1)^{2} + 7\)
The minimum value is 7 and this occurs at x - 1 = 0 ==> x = 1.