Derivation of $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point Derive $y=a(x-h)^2+k$ from $y=ax^2+bx+c$ given a vertex and a point.
Recently I have been solving a problem to which I could not find a solution. Google search of "quadratic equation given vertex and a point," yielded what I have been looking for. However, although I have already solved the problem, I am still wondering how to derive the equation I was looking for before I googled it.
I was given a vertex $V(-3, -2)$ and a point $P(-4, 0)$ of a parabola. Using $y=ax^2+bx+c$, I have derived an equation for finding the vertex for each parabola with $V(\frac{-b}{2a}, -a(\frac{b}{2a})^2+c)$.
I knew that given the same vertex, the parabola $y=x^2+6x+7$ was close to what I've been looking for. However, this parabola didn't go through $P(-4, 0)$, because it was too wide.
At this point it seemed as if I had enough information to derive an equation, that given a vertex and a point I would obtain a parabola according to the restrictions. However, from this point on I didn't know how to proceed further.
 A: Since you know that the parabola has vertex $V(-3, -2)$, the vertex form of its equation is 
$$y = a[x - (-3)]^2 + (-2) = a(x + 3)^2 - 2$$
Since the parabola passes through the point $P(-4, 0)$, we can substitute $-4$ for $x$ and $0$ for $y$ to determine $a$.
\begin{align*}
0 & = a(-4 + 3)^2 - 2\\
2 & = a(-1)^2\\
2 & = a
\end{align*}
Hence, $y = 2(x + 3)^2 - 2$.  You can expand this expression to obtain the standard form of the equation.
Addendum:  In general, if you know the vertex $V(h, k)$ and a point $P(u, v) \neq V(h, k)$ on the parabola, you can write 
$$y = a(x - h)^2 + k$$
then substitute $u$ for $x$ and $v$ for $y$ to determine $a$.
\begin{align*}
v & = a(u - h)^2 + k\\
v - k & = a(u - h)^2\\
\frac{v - k}{(u - h)^2} & = a
\end{align*}
Then 
$$y = a(x - h)^2 + k = \frac{v - k}{(u - h)^2}(x - h)^2 + k$$
Again, expanding the expression to obtain its standard form enables you to determine the coefficients $a$, $b$, and $c$.
A: Generalized Form:
Given $y=ax^2+bx+c$, we can move the loose number $c$ to the other side and try completing the square! So from that, we get $$y-c=ax^2+bx\tag{1}$$
Factoring out $a$ gives us $y-c=a\left(x^2+\frac bax\right)$. Completing the square by dividing the coefficient of the $x$ term by $2$ and squaring it gives us $$y-c+a\left(\frac {b^2}{4a^2}\right)=a\left(x+\frac {b}{2a}\right)^2\tag{2}$$
Simplifying the left hand side and move it to the right hand side to obtain $$y=a\left(x-\left(-\frac {b}{2a}\right)^2\right)+\left(\frac {4ac-b^2}{4a}\right)\tag{3}$$
And since the Vertex follows the formula $(h,k)=\left(-\frac {b}{2a},\frac {4ac-b^2}{4a}\right)$, you get the Vertex formula by plugging it in.
A: $$y = ax^2 + bx + c$$
$$y - c = a\left(x^2 + \frac{b}{a}x\right)$$
$$y - c + a\frac{b^2}{4a^2}= a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right)$$
$$y - c + \frac{b^2}{4a}= a\left(x+\frac{b}{2a}\right)^2$$
$$y = a\left(x-\left(-\frac{b}{2a}\right)\right)^2 + \left(c- \frac{b^2}{4a}\right)$$
$$y = a\left(x-h\right)^2 + k$$
where $h = -\frac{b}{2a}$ and $k = c- \frac{b^2}{4a}$
If you're going in reverse (going from vertex $(h,k)$ and point $(x,y)$ to quadratic parameters $a,b,c$), then you can take the last few equations, isolate $a,b,c$, and translate them into terms of $h,k,x,y$:
$$a = \frac{y-k}{(h-x)^2}$$
$$b = -2ha = \frac{-2h(y-k)}{(h-x)^2}$$
$$c = k + \frac{b^2}{4a} = k + \frac{h^2(y-k)}{(h-x)^2}$$
