<s>[INST] Getting the quadratic function given the vertex and one point. 
Find out the quadratic function for the parable that contains the
  point $(1,1)$ and the vertex $(-2,3)$.

The notes I got are pretty vague:

$$b = 4\frac{-2}{9} = \frac{-8}{9}\\ c = -\frac{19}{9}\\ a =
 -\frac{2}{9}$$
So
$$f(x) = -\frac{2}{9}x^2-\frac{8}{9}+\frac{19}{9}$$

Didn't really help me. Why did they multiply $4$ by $\frac{-2}{9}$ to get $b$? And of course, how were $c$ and $a$ calculated?
 [/INST] If a quadratic function has vertex $(h,k)$, then it can be written as
$$
f(x) = a (x - h)^2 + k
$$
where $a$ is some other real number. (This is one of the things they usually expect you to just memorize.)
So since the vertex is $(-2, 3)$, you get
$$
f(x) = a ( x - (-2))^2 + (3)
= a ( x + 2)^2 + 3.
$$
Now, we use $f(1) = 1$ to solve for $a$.
Plugging in $x = 1$, we get
\begin{align*}
(1) &= a((1) + 2)^2 + 3 \\
1 &= 9a + 3 \\
-2 &= 9a \\
a &= -\frac{2}{9}
\end{align*}
Therefore, we have
$$
f(x) = - \frac{2}{9} (x + 2)^2 + 3.
$$
We can also expand this out to get the usual form $ax^2 + bx + c$:
\begin{align*}
- \frac{2}{9} (x + 2)^2 + 3
&= -\frac{2}{9} (x^2 + 4x + 4) + 3 \\
&= -\frac{2}{9} x^2 - \frac{8}{9} x - \frac{8}{9} + 3 \\
&= -\frac{2}{9} x^2 - \frac{8}{9} x - \frac{8}{9} + \frac{27}{9} \\
&= \boxed{-\frac{2}{9} x^2 - \frac{8}{9} x + \frac{19}{9}}.
\end{align*}
</s>