A polynomial function is a function that's written under the form of $f(x) = a_{n}x^{n}+a_{n-1}x^{n-1}+\dotsb +a_{2}x^{2}+a_{1}x+a_{0}$. There can be infinite terms in a polynomial function, but we're going to look at a much simpler polynomial function: the second degree polynomial. This simply means that the highest exponent is 2, or that n = 2:
$f(x) = ax^2 + bx + c$
Here's a second degree polynomial whose parameters you can change with the sliders:
A second degree polynomial creates a parabola. It has a defining characteristic: the vertex. The vertex is simply the point at which the function passes from going down to going up or the other way around.
The vertex is the red point. It moves as the value of $b$ changes
For the square function, we can easily find the vertex as the point at which the curve "goes flat". In a more mathematical language, we could say that it's the point at which the slope of the tangent of the function is 0. That's just a fancy way of saying that if you drew a straight line as aligned as possible at the vertex, the line would be horizontal.
This is translated to:
$f'(x) = 0$
We can derive the function $f$:
$f'(x) = 2ax + b + 0$
We can then solve for $f'(x) = 0$:
$f'(x_{vertex}) = 0 \Leftrightarrow 2ax_{vertex} + b = 0 \Leftrightarrow 2ax_{vertex} = -b \Leftrightarrow$
$\boxed{x_{vertex} = \frac{-b}{2a}} $
This is simply the equation that gives us the $x$ coordinate of the vertex given the values of $a$ and $b$.
The challenge that I want to solve is the following: what's the function $g(x)$ that gives every possible position of a vertex of a function when changing the $b$ parameter. In other words, I want a function that would give a group of coordinates that correspond exactly to the group of coordinates of vertexes based on the value of $b$ while keeping $a$ and $c$ constant.
The formula gives us a couple of important pieces of data: every unique value of $b$ brings a unique value of $x$. This is a bijective function.
$f(x)$ is bijective so we can start with a value of $x$ and find the corresponding value of b to find the single function that would give a vertex with an x coordinate of $x_{vertex}$.
We can create a protocol that we could then follow in the function:
Chose a value of $x$
Find the function whose vertex (given $a$ and $c$ are constant) is at $x$ on its x position
$x=-\frac{b}{2a}\Leftrightarrow \boxed{b=-2ax}$
Rewrite the function with this value of $b$:
$g(x) = ax^2 + (-2ax)x + c$
And with a few simplifications, we have:
$g(x) = ax^2 + (-2ax)x + c$
$\Leftrightarrow g(x) = ax^2 -2ax^2 + c$
$\Leftrightarrow \boxed{ g(x) = -ax^2 + c}$
And it's that simple!
We can see that it works perfectly:
The blue function is $f(x)$ and the orange function is $g(x)$
You can play around with it here: