If \(x\) is real, the minimum value of \(x^{2}-8x+17\) is

\(-1\) \(0\) \(1\) \(2\)

\(1\)

\(\begin{aligned}f^{\prime}(x)&=2x-8\\ f^{\prime}(x)&=0\\ x&=4\\ f^{\prime \prime}&=2>0\end{aligned}\hspace{8cm}\) Minimum value \(=16-32+17=1\)