<s>[INST] Is it really necessary to learn how to write a quadratic in standard to vertex form? How crucial is this skill or form of writing and polynomial function? Can't we just always use the $-b/2a$ trick for the $x$ intercept and just plug it back into the function to find the $y$?
 [/INST] To answer your question, it isn't crucial in any serious way. However, it is as handy as, say, "point-slope form" or "slope-intercept form" for lines, it gives you a way to immediately recognize certain bits of information. Additionally, as Daniel M. mentions, it covers completing the square which is more often than not, the desired way to solve for the roots etc. of a quadratic (it's also just a handy algebraic manipulation.
Perhaps to demonstrate some utility of completing the square, we'll derive the quadratic formula starting from standard form:
$$ax^2+bx+c=0 \implies x^2+\frac{b}{a}x=-\frac{c}{a}$$
but now, we really need to complete the square:
$$x^2+\frac{b}{a}x+\frac{b^2}{4a^2}=\frac{b^2}{4a^2}-\frac{c}{a} \implies(x+\frac{b}{2a})^2=\frac{b^2-4ac}{4a^2} \implies x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Notice how one could deduce the $\frac{-b}{2a}$ trick from the middle implication! You're right to notice that the two are intimately related, I think it's best to know both.
Let's use a similar technique to examine what's going on.
vertex form: $y=a_1(x-h)^2+k$. But, given a quadratic in standard form, $y=ax^2+bx+c$, we can deduce the following by adding and subtracting the same thing, while factoring out an $a$, and then completing the square:
\begin{align*}y&=ax^2+bx+c\\
&=a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)+\left(\frac{b^2}{4a}-\frac{b^2}{4a}\right)\\
&=a\left(x^2+\frac{b}{a}x+\frac{b^2}{4a^2}+\frac{c}{a}-\frac{b^2}{4a^2}\right)\\
&=a\left((x+\frac{b}{2a})^2+\frac{c}{a}-\frac{b^2}{4a^2} \right)\\
&=a(x+\frac{b}{2a})^2+c-\frac{b^2}{4a}.
\end{align*}
Notice that this is a way to always get to "vertex form" and tells you exactly what $a,h,k$ are supposed to be in vertex form.
Namely:
$a_1=a$ (Hoorah!)
$h=-\frac{b}{2a}$ (As you have noticed)
$k=c-\frac{b^2}{4a}$
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