Vertex Form of Parabola - Why does it work? Recently, I have been trying to plot parabolas of quadratic equations.
First, I have to convert them to vertex form and then we can easily plot them.
This makes me wonder why the vertex form of a parabola works. What's so special about it? 
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Standard Form: $ax^2 + bx + c$
Vertex Form: $a(x-h)^2+k$
 A: The validity of the formula stems from an interesting fact about parabolas. Every parabola represented by the equation $y=ax^2+bx+c$ can be obtained by vertically stretching and translating the graph of $y=x^2$. 
You may have learned how to make basic transformation on the graph of a function. For instance to shift the graph of $f(x)$ to the right by $h$ units you substitute $x-h$ for $x$,i.e., the graph of $f(x-h)$ is the graph of $f(x)$ shifted to the right by an amount $h$. 
The basic transformations are as follows:


*

*$f(x-h)$, shifts the graph of $f$ to the right by $h$ units. $\color{blue}{ \text{(If $h$ is negative the graph shifts to the left.)}}$ 

*$f(x)+k$, shifts the graph up by $k$ units. $\color{blue}{\text{(If $k$ is negative the graph shifts down.)}}$  

*$af(x)$, vertically stretches the graph by a factor of $a$. $\color{blue}{\text{(If $a<1$ if performs a compression.)}}$


We can get the vertex formula by applying these transformations to $y=x^2$. 
First we'll scale $x^2$ vertically by a factor of $a$, then we will translate it to the right by an amount $h$, finally we'll shift it up by an amount $k$.  
$$ x^2 \rightarrow \color{blue}{\text{(scale) }}ax^2 \rightarrow \color{blue}{\text{(shift right) }} a(x-h)^2 \rightarrow \color{blue}{\text{(shift up) }}a(x-h)^2+k$$

To see that any parabola can be written this way, we can simplify the vertex formula. 
$$ y=a(x-h)^2+k$$
$$ y=a(x^2-2xh+h^2)+k$$
$$ y=ax^2-2axh+ah^2+k$$
$$ y=(a)x^2+(-2ah)x+(ah^2+k)$$
If we make the following identifications we get the standard form, 
$$a=a, \qquad b=-2ah, \qquad c=ah^2+k;$$
notice that as we go from left to right we gain a new variable on the right hand side. This means that making a choice for the value of $a$ doesn't restrict the possible values available for $b$. Similarly making a choice for $b$ doesn't restrict $c$ because it is the only variable which depends on $k$. 
To explain what I mean above consider the following. Suppose we wanted to get $a=5$, $b=10$, and $c=100023$ by choosing values for $a$,$h$, and $k$.  
To get $a=5$ we simply choose $a=5$ thats simple enough. 
To get $b=10$ we look at the formula $b=-2ah$. Now $a=5$ so this is really $b=-10h$. To get $b=10$ we choose $h=-1$.
Now to get $c=100023$ we need to look at the formula $c=ah^2+k$. We know what $a$ and $h$ are so we should plug them in. $c=(5)(-1)^2+k=-5+k$. So to get $c=100023$ we need to chose $k$ to be $100028$.
If you try this out for different values, you should become convinced that you can get any three values for $a$,$b$, and $c$ that you want. We can always find a version of the vertex formula that is algebraically equivalent to the standard form.   
