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.
 A: 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:
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
A: Now that you've seen the standard way to derive the formula, it might be interesting to see how you can do it "backwards."
That is, convert the vertex form into the standard form, then figure out how to reverse that procedure so that you start with the standard form and end up with the vertex form.
You can get a sense of how  this works from the second half of this answer. We just need to avoid plugging in actual numbers so that we get the quadratic formula.
Let's write the vertex form this way:
$$ f(x) = a(x - h)^2 + k. $$
Expanding the square and collecting terms we find that this is the same function as
$$ f(x) = ax^2 − 2ah x + ah^2 + k. $$
What we wanted was something in the form $ax^2 + bx + c.$ Fortunately we already have $ax^2$ in both formulas, so we just need to match the $bx$ and $c,$ which we can do by letting
\begin{align}
 b &= -2ah, \\ c &= ah^2 + k.
\end{align}
If we stipulate those conditions for $b$ and $c$ then indeed
$a(x - h)^2 + k = ax^2 + bx + c$ and we have found how to turn the vertex form into the standard form.
Now to reverse the process, we already know the coefficient $a$ in the standard form is the same as the coefficient $a$ in the vertex form.
Next we can use the equation $b = -2ah$ to solve for $h$ in terms of the known coefficients $a$ and $b$:
$$ h = -\frac{b}{2a}.$$
Finally we want the value of $k,$ which we can get by solving for $k$
in $c = ah^2 + k$:
$$ k = c - ah^2 = c - a\left(-\frac{b}{2a}\right)^2
 = c - \frac{b^2}{4a}.
$$
And now we have found how to get the vertex form from the standard form:
$$
f(x) = ax^2 + bx + c 
 = a\left(x - \left(-\frac{b}{2a}\right)\right)^2 + \left(c - \frac{b^2}{4a}\right),
$$
which, apart from trivial differences in the formulas such as whether to
write $x - \left(-\frac{b}{2a}\right)$ or $x + \frac{b}{2a}$,
is the same result we get from "completing the square."
T. Ford's answer gets us the rest of the way to the quadratic formula, so I see no need to repeat those steps.
I just think it's neat how different approaches to the same problem can work together in mathematics.
