As we know vertex -
[tex]y=a(x+3)^2-4[/tex]
Passes through (1,4). Substitute x = 1 and y = 4
[tex]4=a(1+3)^2-4\\4=a(4)^2-4\\4=16a-4\\4+4=16a\\16a=8\\a=\frac{8}{16}\\a=\frac{1}{2}[/tex]
Therefore the vertex equation is [tex]y=\frac{1}{2}(x+3)^2-4[/tex]
Since you want the "standard" form then..
[tex]y=\frac{1}{2}(x^2+6x+1)\\y=\frac{1}{2}x^2+3x+\frac{1}{2}[/tex]
So the standard form is 1/2x^2 + 3x + 1/2
