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I have been playing around with desmos and found this strange idea

(I encourage you to view here to visualise what I am talking about: https://www.desmos.com/calculator/dwqp8jmp7v)

In high school I learnt that if given a parabola $Ax^2 + Bx + C$, that:

  • Varying $C$ varies the height (this is intuitive, each point is shifted vertically)

  • Varying $A$ stretches the Parabola (also intuitive, each point scales appropriately)

I have never for the life of me understood what the $B$ does however.

Yes, you can factorise it and find where the roots are and form the parabola from there. Yes I understand that. But it still leaves me stumped because I feel like I am manipulating a black box. (that is, I can get the result and form a parabola but I never really understand what $B$ does)

Desmos has shown me that varying $B$ makes the parabola's vertex trace out another parabola underneath it. My question is, why is this, and is there a simple explanation? (In the sense that explaining $A$ and $C$ are simple)

(I should note, I have found another answer here to a similar question, and an answer has been provided. But it does not satisfy me because it feels like attacking a nail with a jackhammer. There has got to be a simpler explanation for something so simple as quadratics.. I hope??)

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    $\begingroup$ Parameterisation is actually a very useful tool in mathematics. However, since you are only talking about varying $b$ and nothing else, then there is a simple answer based on transformations of functions. $\endgroup$
    – Toby Mak
    Oct 8, 2019 at 9:46
  • $\begingroup$ The placement of the vertex in terms of left/right $\endgroup$
    – wilsonw
    Oct 8, 2019 at 14:42

3 Answers 3

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Completing the square, we have:

$$x^2+bx= (x+0.5b)^2 - 0.25b^2$$

which is equivalent to first translating the parabola $y =x^2$ left by $0.5b$ units, and then down by $0.25b^2$ units.


In fact, we can find the equation of the parabola that the vertex traces. Given that the vertex of $y=x^2$ is $(0,0)$, the vertex of $y=x^2+bx$ is $(-0.5b, -0.25b^2)$.

We can immediately notice that $-x^2 = y$, where $x = -0.5b$ and $y = -0.25b^2$. Therefore the vertex traces out the parabola $y=-x^2$.

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The two roots of $x^2+bx$ are $0$ and $b$, so the $x$-coordinate of the vertex is their average $b/2$. When you plug this in to get the $y$-coordinate of the vertex, you get $y = (b/2)^2 +b(b/2) = 3b^2/4.$

So as $b$ varies, the $x$-coordinate of the vertex varies linearly with $b$, while the $y$-coordinate varies "quadraticly" with $b$.

If you think of $b$ as a parameter, you can write parametric equations for the coordinates of the vertex:

$$x= \frac{1}{2}b$$

$$y = \frac{3}{4}b^2$$

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    $\begingroup$ Hello, please check again about the roots. They are $0$ and $-b$. Thank you. $\endgroup$
    – autodavid
    Oct 15, 2019 at 19:05
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Maybe this is not an answer about what $B$ "is". There are already good answers demonstrating how does $B$ affect the position of vertex. I would just like to provide some more geometric viewpoints here.


Derivative

This question is tagged precalculus so I won't assume that you have calculus knowledge. But if you do, then we can just say $$B=f'(0)$$ where $f(x)=x^2+Bx$. Therefore $B$ indicates the slope at which the parabola meets the $y$-axis.

slope of parabola

In general, every coefficient in a polynomial $g(x)$ has something to do with the $n$-th derivative $g^{(n)}$ of $g$, according to the Taylor expansion formula.


Why does vertex trace out a parabola

Here is an easy reason, which however does not provide any insight about $B$.

As you indicates, "$A$" decides the scale of a parabola. So varying $B$ in $x^2+Bx$ only moves the parabola and does not change shape. Also, $x^2+Bx$ always pass through $0$. So we are moving a parabola on the plane and decide which point on the parabola should pass through the origin (the green dot in the left picture below).

moving parabola

The picture in the right marks that green dot on a non-moving parabola. Observe that the relative coordinates of the green and red (vertex) dot obeys the quadratic (of course). Therefore in the left picture, the red dot should trace the same (but reversed) quadratic when the green dot is fixed.

Or we can say the green-red vector in the left is just the negative of the red-green vector in the right.

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