[tex]y=2x^2-6x-7\\\\You \ have \ correctly \ found \ the \ x-coordinate \ of \ the \ vertex: \\\\x=\frac{-b}{2a}=\frac{6}{4}=\frac{3}{2}\\\\y-coordinate \ by \ substituting \ x =\frac{3}{2} \ into \ the \ given \ quadratic \ equation \ to \ solve \ for \ y \\\\y=2\cdot (\frac{3}{2})^2-6 \cdot \frac{3}{2}-7=2\cdot \frac{9}{4} -3*3-7= \frac{9}{2} -9-7= 4.5-16=-11.5[/tex]
[tex]The \ range \ consists \ of \ all \ of \ the \ valid \ y-values .\\ Since \ the \ y=coordinate \ of \ the \ vertex \ (a \ minimum) \\ is \ y =-11.5 \ it \ is \ clear \ that\ the\ range \ is \ -11.5 \ and \ above.\\\\Answer: \ \ So \ the \ range \ can \ be \ written \ as:\\\\ y\geq -11.5[/tex]
