Construct several parabolas passing through two given points I have two coordinates (x,y) =(1,3.5) and (3,0.5) and
I want to have 3 different "parabola looking" curves that go through these points that
are progressively falling steeper. The "belly" should be downward -convex.
I solved
$$3.5=1^2\cdot a_1 +a_0 $$
$$0.5=3^2\cdot a_1 + a_0 $$
giving $a_1=-0.375$ and $a_0=3.875$ but that one is belly up.
Edit: Sorry, forgot to add that (3,0.5) is the minimum-point - $y_{min}=0.5$
 A: Let the parabola be given by the equation $y=ax^2+bx+c$. Subbing in the points we want, we get the system of linear equations
$$\begin{cases}\frac{7}{2}=a+b+c \\ \frac{1}{2}=9a+3b+c.\end{cases}$$
Now you can solve this system for $a,b,c\in\mathbb R$, and get infinitely many parabolas that pass through your two points. Choosing $a<0$ will make the parabola "downward facing".
A: Select the minimum point of the parabola $(x_2, y_2)$ where $(x_1, y_1) = (1, 3.5) $ and $(x_3, y_3) = (3, 0.5) $.  $ x_2 $ should be between $x_1 $ and $x_3$, i.e. $ 1 \lt x_2 \lt 3 $ and $ y_2 $ is yet to be determined.  Now the equation of the parabola having $(x_2, y_2)$ as it minimum point is
$ y = A (x - x_2)^2  + y_2 $
And you want to determine $A$ and $y_2$ by using the two given points, i.e.
$ y_1 = A (x_1 - x_2)^2 + y_2 $
$ y_3 = A (x_3 - x_2)^2 + y_2 $
Solving this $2 \times 2 $ system gives you the desired parabola.
You have (using Cramer's rule)
$ A = \dfrac{  (y_1 - y_3) }{ (x_1 - x_2)^2 - (x_3 - x_2)^2 } $
$ y_2 = \dfrac{ y_3  (x_1 - x_2)^2  - y_1  (x_3 - x_2)^2 }{(x_1 - x_2)^2  - (x_3 - x_2)^2 } $
For example, if you select $x_2 = 2.5$ , then
$ A = \dfrac{ (3.5 - 0.5) }{ 1.5^2 - 0.5^2 } = 1.5 $
$ y_2 = \dfrac{ 0.5( 1.5)^2 - 3.5 (0.5)^2 }{ 2 } = 0.125 $
Hence, the parabola will be,
$ y = 1.5 (x - 2.5)^2 + 0.125 $
Note that we want $A \gt 0$
From the equation of $A$,
$ A = \dfrac{  (y_1 - y_3) }{ (x_1 - x_2)^2 - (x_3 - x_2)^2 } $
If $y_1 \gt y_3 $ , then we want $ (x_1 - x_2)^2 \gt (x_3 - x_2)^2 $, i.e. we should choose $x_2$ closer to $x_3$ than $x_1$, and if $y_1 \lt y_3 $ then we want $ (x_1 - x_2)^2 \lt (x_3 - x_2)^2 $, i.e. we should choose $x_2$ closer to $x_1$ than $x_3$.
If $ y_1 = y_3$ then we can choose $A$ to be any positive number, but $x_2$ should be chosen half way between $x_1 $ and $x_3$.
