Answer :
[tex]y=4x^2+40x+96\\ \\ The \ vertex \ form \ of \ the \ function \ is: \\ \\y = a(x - h)^2 + k \\ \\vertex =(h, k) \ is \ given \ by: \\ \\h = \frac{-b}{2a} , \ \ \ k = c -\frac{b^2}{4a} \\ \\a=4 , \ \ b=40 , \ \ c=96[/tex]
[tex]h = \frac{-40}{2 \cdot 4}=\frac{-40}{8}=-5 \\ \\ k = 96 -\frac{40^2}{4 \cdot 4} =96-\frac{1600}{16} =96-100=-4 \\ \\ y = 4(x - (-5) )^2 + (-5)=4(x +5 )^2 -5[/tex]
[tex]h = \frac{-40}{2 \cdot 4}=\frac{-40}{8}=-5 \\ \\ k = 96 -\frac{40^2}{4 \cdot 4} =96-\frac{1600}{16} =96-100=-4 \\ \\ y = 4(x - (-5) )^2 + (-5)=4(x +5 )^2 -5[/tex]
