This is a two step question, so can someone answer the first part ?

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?  ;)

Where is the diagram