Converting standard form to vertex form, parental homework help I'm trying to help my son with his homework but am having trouble feeling confident that I know what the assignment is asking for. 
I've been learning (maybe relearning) about standard vs vertex form quadratic equations.
The first part of the homework states:


Rewrite the standard form $f(x) = ax^2 + bx + c$. Once you've done that set the vertex form equal to zero and solve for x. Show your work:


All of the examples I can find on how to convert a standard form to vertex are using actual numbers for $a$ and $b$. Because the steps involve converting to a perfect trinomial, I'm not sure how to "rewrite" the equation without using actual values as an example.
It then asks:


You've now discovered the ___  ____! Use it to solve this: $x^2 + 4x - 11 = 0$


I'm not sure what the underlines are asking for - "perfect trinomial"? I can figure out how to the solve the equation but I want to be sure I know how using the expected method.

This is not entirely straightforward, but we'll walk through it.
You start with an equation in standard form: $y = ax^2 + bx + c$. To convert this to 'vertex form' we must complete the square. 
$$y = ax^2 + bx + c$$
$$y - c = ax^2 + bx$$
$$y - c = a(x^2 + \frac{b}{a}x)$$
$$y - c + a(\frac{b^2}{4a^2}) = a(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}) = a (x + \frac{b}{2a})^2$$
Hence we get,
$$y = a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a})$$
If unfamiliar with the process of completing the square as above see here. (You can also always multiply this out and check that it is, indeed, the same as $y = ax^2 + bx + c$).
Now, the problem is telling you to set this equal to 0 and solve for $x$. It is much easier to solve for $x$ from vertex form (hence the conversion) as we will see:
$$0 = a(x + \frac{b}{2a})^2 + (c - \frac{b^2}{4a})$$
$$(x + \frac{b}{2a})^2 = \frac{-c + \frac{b^2}{4a}}{a} = \frac{-4ac + b^2}{4a^2}$$
Take the square root of both sides and we get,
$$x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
which becomes the familiar quadratic equation:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
