If the range of the function \(f(x)=x^2-8x+20\), defined for real x, is \([m,\infty)\), what is the value of \(m\)?

Correct answer: \(4\)

Completing the square gives \(f(x)=x^2-8x+20=(x-4)^2+4\). Since \((x-4)^2\geq 0\), the minimum value of the function is \(4\), attained at \(x=4\). Hence the range is \([4,\infty)\), so \(m=4\). In an exam, rewrite a quadratic as \((x-h)^2+k\) to identify its minimum value quickly; \(8\) is only related to the coefficient of x, not the minimum.