I'm learning how to convert quadratic equations from

To get the standard form from the general form, first factor out $a$, then complete the square, and finally adjust the constant term:
$\begin{array}{rl}a{x}^2+bx+c& =a(x^2+\frac{b}{a}x+\frac{c}{a})\end{array}$ $\left(1\right)$
where I don’t yet know what u is. To find out, I multiply out $\left(1\right)$ to get
$a(x^2+\frac{b}{a}x+\frac{b^2}{4a^2})+au,$
which is
$a{x}^2+bx+\frac{b^2}{4a}+au,$
If this is to be the same as the original quadratic, we must have
$\frac{b^2}{4a}+au=c,$
meaning that
$u=\frac{c-\frac{b^2}{4a}}{a}=\frac{c}{a}-\frac{b^2}{4a^2},$
and our quadratic can be written as
${\displaystyle a{(x+\frac{b}{2a})}^2+(c-\frac{b^2}{4a}),.}$
Thus, the h and k of the standard form must be
${\displaystyle h=-\frac{b}{2a}}$
and $k=c-\frac{b^2}{4a}.$
Take the quadratic ${\displaystyle 2x^2+3x+4}$ as an example. Factoring out the leading coefficient of $2$ results in
${\displaystyle 2(x^2+\frac{32}{x}+2),.}$
Then $x^2+\frac{3}{2}x+2=(x+\frac{3}{4}{)}^2+u$
for some u, and since $(x+\frac{3}{4}{)}^2=x^2+\frac{3}{2}x+\frac{9}{16}$, u must be ${\displaystyle 2-\frac{9}{16}=\frac{23}{16}}$. Thus,
$\begin{array}{rl}2x^2+3x+4& =2\left(\right(x+\frac{3}{4}{)}^2+\frac{23}{16})\\ & =2(x-(-\frac{3}{4}){)}^2+\frac{23}{4},\end{array}$
with $h=-\frac{3}{4} \text{and} k=\frac{23}{8}$

Remember that there are four basic "geometric transformations" of the graph of an equation ${\displaystyle y=f\left(x\right)}$. Specifically, for a constant $u$, we have:
- The graph of ${\displaystyle y=f(x-u)}$ is the graph of ${\displaystyle y=f\left(x\right)}$ shifted to the right by $u$.
- The graph of ${\displaystyle y=f\left(ux\right)}$ is the graph of ${\displaystyle y=f\left(x\right)}$ "compressed horizontally" by a factor of $u$ (and possibly reflected about the y-axis, depending on the sign of u).
- The graph of ${\displaystyle y=f\left(x\right)+u}$ is the graph of ${\displaystyle y=f\left(x\right)}$ shifted up by $u$.
- The graph of ${\displaystyle y=uf\left(x\right)}$ is the graph of ${\displaystyle y=f\left(x\right)}$ "stretched vertically" by a factor of $u$ (and possibly reflected about the $x$-axis, depending on the sign of $u$).
Now - granting that a parabola is a transformation of the graph of ${\displaystyle y=x^2}$ via stretching, shifting, and flipping - this tells us that every parabola has the form
${\displaystyle y=u_0{(u_{1}x-u_2)}^2+u_3}$
for some ${\displaystyle u_0,u_1,u_2,u_3}$. Moreover, at least for parabolas (this doesn't work in general) we can combine two of these variables into a single one: the expression above is equivalent to
$y=\left(u_0{u}_1^2\right)(x-\frac{u_2}{u_1}{)}^2+u_3.$
So we really only need three "transformation constants:"
Every parabola is the graph of an equation of the form
${\displaystyle y=w_0{(x-w_1)}^2+w_2}$
for some constants ${\displaystyle w_0,w_1,w_2}$.
And at this point there's just one last step remaining before the transformation into standard form is something guessable: realizing that the graph of ${\displaystyle y=a{x}^2+bx+c}$ is a parabola. This is something you can notice just by graphing a few examples (always graph a few examples!).