What is the minimum value of y = frac x ^ 2 2-3x ?

The correct answer is Option (2) → $-9/2$

$f(x)=\frac{x^2}{2}-3x$

for extreme values, $f'(c)=0$

$⇒2x-3=0$

$⇒x=3/2$

$∴f(\frac{3}{2})=\frac{9}{8}-\frac{9}{2}$

$=-\frac{9}{2}$