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

Answer 2

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