Find the minimum value of $x^2-6x+10$ for real $x$.
A0
B1
C2
D4
Answer & Solution
Correct answer: B. 1
Complete the square: $x^2-6x+10=(x-3)^2+1$. Since $(x-3)^2\ge 0$ for all real $x$, the minimum value occurs when $(x-3)^2=0$, that is at $x=3$. Therefore the minimum value is $1$.
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