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I can use y = 1/100x as a linear transition between x:0, y:0 and x100, y1 enter image description here

I'm trying to understand how to create a "curved" transition thats "easing in" and "easing out" in a symmetrical fashion?

The function y = x^2 is curved but its not symmetrical within x:0, y:0 and x100, y1

How can I create "symmetrically" curved transitions in the style of the ones below? enter image description here

Ideally the function would contain a variable that controls the amount of curve, (from curved the one way to a straight line and then to curved again the other way). Please excuse my lack of vocabulary and understanding in the field, I hope the question makes sense. Thanks

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    What about $\left(dfrac x{100}\right)^2=\dfrac{x^2}{100^2}$? – 2017-01-16
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    [Graph of the above](https://i.imgur.com/cyHh16n.jpg) – 2017-01-16
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    Thank you! It doesn't look symmetrical though? It's changing more quickly in the beginning and slower towards the end (by beginning I mean x0, y0 and by end i mean x100, y1) . In other words it's "easing out" more than it's "easing in" – 2017-01-16
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    So do you want a [semicircle](https://i.imgur.com/8kFAZEo.jpg)? – 2017-01-16
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    Wow cool! That's the closest thing so far!! But I have a feeling it would be difficult to change it's "curviness" because then it's not a semi circle anymore? – 2017-01-16
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    How about [a hyperbola](https://i.imgur.com/wfSifeS.jpg)? Note that the denominator is equal to $3(100-x)+100$. You can change the $3$ to other things to change "curviness". – 2017-01-16
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    Yes yes yes!!! You're amazing!! This looks absolutely right for what I'm trying to do! But I don't think I fully understand all the concepts I should understand because when I try to use 3(100−x)+100, this is what I get [link](http://imgur.com/4PPfwQB) – 2017-01-16
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    Sorry, that's just the denominator. The whole thing is $\dfrac x{3(100-x)+100}$ (or, equivalently, $\dfrac x{400-3x}$ like it says in the image) – 2017-01-16
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    It's perfect! Perfectly answers my imperfect question! Thank you – 2017-01-16
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    @AkivaWeinberger Please write it as an answer so I can accept it as the correct one, and also, If you have any suggestion to improve the question that would be appreciated!! Again, thank you so so much, ive been struggling with this for days, youre the best! – 2017-01-16

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$$\frac x{a(100-x)+x}$$ works for $1

Example for $a=3$:x/(3(100-x)+x)

If $a=1$, it's just a line; if $0

Example for $a=\frac13$:x/((100-x)/3+x)