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)

Let $y=x^2-8x+17$
$=(x-4{)}^2-16+17$
$=(x-4{)}^2+1$
$\Rightarrow y\ge 1$ for all real values of x as
$(x-4{)}^2\ge 0$
Hence , minimum value of y is 1