Question

If $x$ is real, then the minimum value of $x^2-8x+17$ is

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

Answer: (A)

$1$

$x^2-8x+17=(x-4{)}^2+1$
$\therefore$ The minimum value of $x^2-8x+17=(x-4{)}^2+1$ is $1$.