How do you get to the general equation of $y = ax^2 + bx + c$ for a parabola?
I have not found any resources that show how to get to that equation.
This equation is shown 1 minute 30 into this video on Simpsons Rule https://www.youtube.com/watch?v=vpfy3sGw8tI.
On
One way to show that the general quation$\,y=ax^2+bx+c\,$ is a parabola is to show that any such equation (with $a\ne 0$) is transformed by a unique homothety-translation into $\,y=x^2\,$ and verify that its focus is $\,(0,d)\,$ and directrix is $\,y=-d\,$ as in the article parabola by showing that $\,x^2+(y-d)^2 = (y+d)^2\,$ where $\,d=1/4.\,$
NOTE: the general equation is only for parabolas whose directrix is parallel to the $x$-axis. Also the recent MSE question 3342798 "Why does the focus-directrix definition of the parabola work?" has an answer to a converse question.
On
The relationship is basic coming in next after the linear relation. It better to learn fundamentals as such. If anyone asks to prove that $ y=ax+b$ for linear relations, then...you see?
It is the integration step coming next after the straight line
$$ y'= 2 a x + b $$
$$ y = a x^2+ bx +c \tag1 $$
where c is a constant of integration.
If two real or complex (p,q) values for $x$ are both valid then we must have the product of zeros to be a zero
$$ x^2- (p+q)x + pq = (x-p) (x-q) =0 $$
Setting the sum $ p+q = -b/a $ and product $pq= c/a $ in the above we obtain (1) in alternate form.
In uniform motion ( going back to Newton understanding dynamic parabolic relations) if a particle attempts to move after annulling/compensating whatever length or distance $c$ gained moving to left with starting velocity to right $ b $ along while uniformly accelerating $a/2 $ towards right, then after $x$ seconds the particle will stay put always at the same location with zero displacement...i.e., never moves.. indicated by zero of the quadratic equation.
$$ c + b x + a x^2 =0 $$
Since you show little work of your own, I'll just give an outline and let you fill in the details. If you need more detail, show some more work of your own then ask.
Your question involves two parts. First, given a parabola with a horizontal directrix, show that an equation of the form $y = ax^2+bx+c$ with $a\neq 0$ determines it. Let's say that the parabola's focus is at $(r,s)$ and the directrix is the line $y=t$. Note that we must have $s\neq t$, which means the focus is not on the directrix. The geometric definition of the parabola is that the distance from any point on the parabola $(x,y)$ to the focus equals the distance from that point to the directrix. The distance formulas are easy, so equate the two and we get
$$\sqrt{(x-r)^2+(y-s)^2}=|y-t|$$
We can convert that equation to an equivalent one by squaring both sides (I'll leave it to you to show this does not add any new solutions) and solving for $y$. We end up with
$$y = \left(\frac{1}{2(s-t)}\right)x^2+\left(\frac{-r}{s-t}\right)x+\left(\frac{r^2+s^2-t^2}{2(s-t)}\right)$$
Since we know that $s-t$ is not zero and it is clear that $\frac{1}{2(s-t)}$ is also not zero, this equation has the desired form $y=ax^2+bx+c$ with $a\neq 0$.
The next part is to show that the curve defined by $ax^2+bx+c$ where $a\neq 0$ is a geometric parabola. First complete the square to get the form
$$y=a(x-h)^2+k$$
where again $a\neq 0$. Then show that this is also the equation for the parabola with its focus at the point $(h,k+\frac{1}{4a})$ and directrix at the horizontal line $y=k-\frac{1}{4a}$. These are well-defined since $a\neq 0$ and clearly the focus is not on the directrix. I'll leave this second part to you--you can use the first part above to help.