[tex]\begin{gathered} f(x)=2x^2-12x+19 \\ a) \\ The\text{ given function has the form} \\ f(x)=ax^2+bx+c \\ when\text{ a>0, the function opens upward. In this case a=2,} \\ \text{hence, the function has a minimum} \\ b) \\ The\text{ minimum of this fuction will be its vertex} \\ V=(\frac{-b}{2a},f(\frac{-b}{2a})) \\ a=2 \\ b=-12 \\ \frac{-b}{2a}=\frac{-(-12)}{2(2)}=\frac{12}{4}=3 \\ f(\frac{-b}{2a})=f(3)=2(3)^2-12(3)+19=2(9)-36+19=18-36+19 \\ f(3)=1 \\ Hence \\ The\text{ function^^b4s minimum value is 1} \\ \\ c) \\ The\text{ minimun value occurs at x=3} \end{gathered}[/tex]
