Obtain the critical points as follows. \(f'(x)=0 \frac{d}{dx}(2x^2-12x)=0\ 4x-12=0 x=\frac{12}{4} x=3\) Thus, the critical point is \(x=3\) \(f''(x)=4>0\) at \(x=3\) Therefore, minimum exist at \(x=3\) Substitute \(x=3\) in \(f(x)=2x^2-12x\) and obtain that, \(f(3)=2(3)^2-12(3) =2(9)-36 =18-36 =-18\) Thus, the quadratics function has a minimum value, the value is −18.