If x is real, the minimum value of $x^2-8x+17$ is :
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 1
$y=x^2-8x+17$
$y=(x-4)^2+1$
min $(x-4)^2=0$
$y_{min}=1$
If x is real, the minimum value of $x^2-8x+17$ is :
Correct answer: option 3
The correct answer is Option (3) → 1
$y=x^2-8x+17$
$y=(x-4)^2+1$
min $(x-4)^2=0$
$y_{min}=1$