# How can I create a quadratic equation that flips the parabola, starting at a certain arbitrary point near the vertex?

Given an arbitrary quadratic polynomial $$ax^2+bx+c$$and a corresponding graph

where gray lines are axis, black is the parabola, red is an arbitrary line near the vertex,

How would you go about creating an equation for the left-side of the parabola to flip vertically?

That is, I am seeking to create a new quadratic equation, one that will describe the green parabola, where green parabola joins the black on at the intersection point with the red line.

So what you're looking for sounds like a parametric approach. If you write your equation in the form $$a(x-b)^2+c$$, then the minimum/maximum occurs at $$x=b$$. So if you want to create your flipped equation, just do something like $$g(x)=a(x-b)^2+c$$ for $$x\geq b$$ and $$g(x)=-a(x-b)^2+c$$ for $$x<$$b.
Note that there won't be a single quadratic equation to describe the full curve (on both sides). A simple proof of this fact is that if $$f(x)=ax^2+bx+c$$, then $$f''(x)=2a$$, meaning that the second derivative is constant, and hence never changes sign. Whereas in what you desire, the second derivative would necessarily be negative in the left-half, and positive in the right-half.