[tex]To \ convert \ the \ standard \ form \ y = ax^2 + bx + c \ of \ a \ function \ into \ vertex \ form \\ \\ y = a(x - h)^2 + k ,\\ \\ we \ have \ to \ write \ the \ equation \ in \ the \ complete \ square \ form \ and \ vertex(h, k) \ is \ given \ by: [/tex]
[tex]h = \frac{-b}{2a} , \ \ k = c -\frac{b^2}{4a} \\ \\y = a(x - h)^2+k[/tex]
opens up for a > 0, and down for a < 0
[tex] y=x^2 +64x + 12\\ \\a=1,\ b=64 \ c = 12 \\ \\h = \frac{-64}{2}=-32 \\ \\ k = 12 -\frac{64^2}{4 }=12-\frac{4096}{4}=12-1024= -1012\\ \\y = (x+32)^2-1012 \\ \\ This \ means \ the \ vertex \ of \ the \ parabola \ is \ at \ the \ point \ (-32, -1012)\\ \\ and \ the \ parabola \ is \ concave \ up. [/tex]
