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Quadratic Function

Let $f(x)=ax^2+bx+c$

It's obvious that when we change absolute term the vertex of the parabola graphs a vertical line with equation $x=-b/2a$


When we change $b$, the vertex traces a parabola with equation $y=-ax^2+c$ as shown here:

Define $b$ as the parameter $t$ so the parametric equation traces the vertex is $$\left({-t\over2a},f\left({-t\over2a}\right)\right)$$ $$\left({-t\over2a},{-t^2+4ac\over4a}\right)$$ Now, eliminate the parameter: $$x={-t\over2a}\iff t=-2ax\\y={-t^2+4ac\over4a}={-4a^2x^2+4ac\over 4a}=-ax^2+c\space\square$$

By the same method, when we change $a$ it traces: $$y={b\over 2}x+c$$


My question is "Does other polynomials have this pattern? Does this tracing have a name? Is there a generalization?"

NB. for cubic we can trace the inflection point. Or the mid-point between the two extrema.

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  • $\begingroup$ Your statement "It's obvious that when we change absolute term the vertex of the parabola graphs a vertical line with equation $x=−b/2a$" is not at all obvious. The standard solution for a quadratic never divides $x=−b/2a$. It is unclear what you are asking. Does "change" mean set the coefficient to $0$? $\endgroup$ – Weather Vane May 5 '18 at 17:58
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    $\begingroup$ @WeatherVane I think he refers to $c$ as the absolute term, and so a vertical shift in the parabola is indeed along the line $x=-b/2a$, which refers to the vertex of the parabola. $\endgroup$ – Christopher Marley May 5 '18 at 18:16

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