Find the relative extrema of the function, if they exist.
\[
f(x)=x^{2}-2 x+10
\]
A. Relative minimum at $(1,9)$
B. Relative maximum at $(1,9)$
C. Relative minimum at $(9,1)$
D. Relative maximum at $(9,1)$

Step 1 ：Find the derivative of the function \(f(x) = x^{2} - 2x + 10\), which is \(f'(x) = 2x - 2\).

Step 2 ：Set the derivative equal to zero and solve for \(x\) to find the critical points: \(2x - 2 = 0\) gives \(x = 1\).

Step 3 ：Substitute \(x = 1\) back into the original function to find the corresponding \(y\)-value: \(f(1) = 1^{2} - 2*1 + 10 = 9\). So, there is a relative extrema at \((1,9)\).

Step 4 ：Find the second derivative of the function, which is \(f''(x) = 2\).

Step 5 ：Evaluate the second derivative at the critical point \(x = 1\). Since \(f''(1) = 2\) is positive, the extrema at \((1,9)\) is a relative minimum.

Step 6 ：Final Answer: \(\boxed{\text{(A) Relative minimum at }(1,9)}\)