Parabola where the $x$ intersect is always $0$ and $1$

Question

I am working on creating a parabola where the $x$ intersect is always $0$ and $1$ for basic needs the one below works great!

$$ y=-d\cdot a^{2}\left(x\ -.5\right)^{a}\ +d $$

but I am attempting to change the angle which, I am not even sure if that is the correct term, but it is my $a$ above, only $2$ and $4$ seem to actually work with this equation, I kind of understand why, but I cannot determine what I need to change for allow any even value for $a$.

I am very bad at math (compared to pretty much everyone on here) and I really get by on testing and doing and seeing how it affects the curve. I would like to have it as simple as possible and getting "close" is good enough, for me, I just need it between 0 and 1 for a percentage.

edit Based upon the suggestions, I realized I forgot an aspect to the question, the d variable is required as I want it set from $0$ to $1$ to where if $d$ is $1$ and $x$ is $.5$ then $y$ is $1$.

My goal is to find the distance from 0 (y) based upon the location (x) where (d) is the max distance to travel.

Example of what I want: https://desmos.com/calculator/vld0cp7zs0 and how a when anything other than 2 or 4 breaks the x intercept requirement.

edit 2 after working with this a bit more I ended up with: $$ y=-d\cdot2^{a\cdot.4}\left(x\ -.5\right)^{a\cdot.4}\ +d $$

which is simplified to:

$$ y=d\left(1-(2x-1)^{0.4a}\right) $$

a now has to be a whole number and it gets me pretty much exactly what I want. Here is a link to both of my new equations. https://www.desmos.com/calculator/jicyg0sa7h

Thank you!

Answer 1

Perhaps try this: $$ y = a x(x-1) $$ where $a$ can vary however you like. Every parabola passing through $0$ and $1$ on the $x$-axis has this form, so it's a complete solution to your problem.

If you'd like a little more adjustability, you could try

$$ y = a x^k(x-1)^k $$ where $k = 2, 4, 6, $ should give you parabolic-like things with pointier or flatter bottoms.

Answer 2

For a parabola to always intersect with the $x$ axis at $0$ and $1$, it must have the factors $(x-0)$ and $(x-1)$. The product of these already gives you a parabola, and all possible parabolas satisfying your condition are: $$y = A(x-0)(x-1)=Ax(x-1)$$ where $A\neq 0$ is any scalar (real number).

Answer 3

The general form of a parabola with roots at $0$ and $1$ is $y = a x (x-1)$.

Because parabolas are symmetrical, the maximum height it will obtain will be in the middle of two roots (so in this case the maximum $y$ value will be obtained when $x = \frac{1}{2}$).

So $y_\max = a \frac{1}{2} (\frac{1}{2} - 1) = -\frac{1}{4}a$.

In other words $a = -4 y_\max = -4d$, where I'm defining $d = y_\max$.

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Putting $y = ax(x-1)$ and $a = -4d$ together gives us:

$$y = -4dx(x-1)$$