If y = -4x^2 - 7x + 2, and we want it in vertex form, we'll have to use "completing the square." Do you know that approach?
y = -4x^2 - 7x + 2 can be re-written as y-2 = -4x^2 - 7x , which looks a bit like the "vertex form" y-k = a(x-h)^2
We must complete the square of -4(x^2 + (7/4)x).
Here's what I'd do: -4(x^2 + (7/4)x + (7/8)^2 - (7/8)^2 )
This becomes -4( (x+7/4)^2 - (7/8)^2 ) (remember that this is equal to y-2)
Then y-2 = -4(x+7/4)^2 -(49/64) ) = -4(x+7/4)^2 + 4(49/64)
or y-2 = -4(x+7/4)^2 + 49/16
Subtract 49/16 from both sides, obtaining y - (2+49/16) = -4(x+7/4)^2
Note that 2+49/16 = (32+49)/16 = 81/16. Thus, we have
y - 81/16 = -4(x+7/4)^2. Compare this to y-k = a(x-h)^2, and see that h=-7/4 and k = 81/16. The vertex is at (-7/4, 81/16).
Having fun yet? ;)
