# Why does the “$bx$” term in the general quadratic $ax^2+bx+c$ shift the parabola along a parabolic arc?

I've been playing around a little in Desmos with the graphs of quadratics and trying to figure out the role each of the terms $ax^{2}$, $bx$, and $c$ in the general quadratic $ax^{2}+bx+c$ play, and I've noticed some interesting behavior with the $bx$ term. Changing $a$ and $c$ has the effect I'd expect; $a$ determines the direction/scaling of the parabola, and $c$ shifts it vertically, but changing $b$ shifts the graph in a way I found rather unexpected. Keeping $a$ and $c$ the same and sliding $b$ around seems to move the parabola along another downwards parabolic arc; I assumed that the arc might be just a flipped version of the original parabola, and sure enough, graphing $-ax^{2}+c$ showed me that the first parabola's vertex was moving perfectly along the path traced by the second. Some further experimentation in Desmos also opened up a connection for me of the tangent line of the parabola $ax^{2}+bx+c$ and the line $bx$. (Link to the Desmos graph here)

What I'm curious about is: Why is this the case? Is there an intuitive way of understanding why the $bx$ term makes the graph behave like this?

The vertex of the parabola $y=ax^2+bx+c$ is the point $\left(-\frac b{2a},a\frac{b^2}{4a^2}-b\frac{b}{2a}+c\right)=\left(-\frac{b}{2a},c-\frac{b^2}{4a}\right)$.
This describes the locus $y=c-ax^2$ as $b$, and hence $x=-\frac{b}{4a}$, range over $\Bbb R$.
We have\begin{align}ax^2+bc+c&=a\left(x^2+\frac bax+\frac ca\right)\\&=a\left(\left(x+\frac b{2a}\right)^2+\frac{4ac-b^2}{4a^2}\right).\end{align}So, the vertex of the parabola is located at $\left(-\frac b{2a},\frac{4ac-b^2}{4a}\right)$. So, what happens when we move $b$ around? The vertex follows a parabolic path, since the first coordinate is linear in $b$, whereas the second one is quadratic.
$a$, $b$, and $c$ all dictate how the polynomial works at $x=0$. $c$ says where it is vertically; $a$ describes the "openness", which remains the same for the whole parabola. $b$ says the slope at $x=0$. This has the effect of saying, approximately, where on the parabola you are, which moves the parabola along another parabola for the same reason rolling a circle around a point makes the center of the circle trace another circle.
Mathematically, we can locate the vertex of a parabola by converting to vertex form $a(x-h)^2+k$, which gives $h=\frac{-b}{2a}$ and $k=-\frac{b^2}{4a}+c$, which describes a parabola in $b$.