What is the minimum value of y = x ^ 2 -4x ?

The correct answer is Option (1) → -4

$y=x^2-4x$

for critical points, $f'(c)=0$

$⇒2c-4=0$

$⇒c=\frac{4}{2}=2$

and,

$f''(c)=2>0$

∴ At $x=2$, $f(x)=y=x^2-4x$ attains it's minimum value.