Using the general formula from the Brachistochrone Problem on different functions. When solving the Brachiostone Problem we reach the "general formula":
$$T[y]=\frac{1}{\sqrt{2g}}\int^{\bar{x}}_{0}\sqrt{\frac{1+(\frac{dy}{dx})^2}{-y(x)}}dx$$
I am unsure about how to apply this formula to different types of curves, e.g. a quadratic.
Take a quadratic equation $-x^2$ in domain $[0, 1]$:
What I assume to do is below:

The $\frac{dy}{dx}$ of this curve is $-2x$. Moreover, $-y(x)=-(-1)=1$.
$$T[y]=\frac{1}{\sqrt{2g}}\int^{1}_{0}\sqrt{1+(-2x)^2}$$
$$T[y]=\frac{1}{\sqrt{2g}}\int^1_0\sqrt{1+4x^2}=0.33...$$ (https://www.wolframalpha.com/input?i=1%2F4+%282+sqrt%285%29+%2B+sinh%5E%28-1%29%282%29%29&assumption=%22ClashPrefs%22+-%3E+%7B%22Math%22%7D)
Would this be correct? And obviously I haven't evaluated the integral by hand because of complexity, but are there graphs of functions yielded from my "general formula" which are more easily differentiable?
Edit: another question which I have is how can I determine the maximum velocity when going down a curve?
 A: I think you have the right idea, but you've made a mistake in evaluating the integral. If  $\ y(x)=-x^2\ $ then $\ {-}y(x)\ $ is not $\ 1\ $, but $\ x^2\ $. Your integral should be
$$
\frac{1}{\sqrt{2g}}\int_0^1\sqrt{\frac{1+4x^2}{x^2}}\,dx
$$
You can evaluate this integral by making the substitution $\ x=\frac{w-w^{-1}}{4}\ $, which reduces it to
\begin{align}
&\frac{1}{2\sqrt{2g}}\int_1^{2+\sqrt{5}}\frac{\big(w^2+1\big)^2}{w^2\big(w^2-1\big)}dw\\
=&\frac{1}{2\sqrt{2g}}\int_1^{2+\sqrt{5}}\left(1+\frac{2}{w-1}-\frac{2}{w+1}-\frac{1}{w^2}\right)dw\ .
\end{align}
This integral diverges over any interval starting at $\ w=1\ $ (corresponding to $\ x=0\ $).  While it's finite over any interval whose lower limit is strictly greater than $\ 1\ $, it nevertheless grows without bound as that lower limit tends to $\ 1\ $ (that is, as the starting ordinate of the curve tends to $\ 0\ $).
You can see why this must be the case from the physics of the situation.  The integral is the time it would take a particle sliding frictionlessly down the curve under gravity to move from the point $\ (0,0)\ $ to the point $\ (1,-1)\ $.  But at $\ x=0\ $ the curve is absolutely horizontal (that is, it's slope is $\ 0\ $). Therefore a particle at rest on the curve at $\ x=0\ $ will be in equilibrium.  Although the equilibrium is unstable, the particle is nevertheless not going to move unless it's subjected to some disturbance.  This will be the case for any (differentiable) curve whose slope is $\ 0\ $ at $\ x=0\ $.  For the particle to start moving in the direction of positive $\ x\ $, the curve must have a strictly negative slope at $\ x=0\ $.
There's one other minor fault in the presentation of your answer. For physical problems you should always specify the units of the physical quantities appearing in your analysis.  Judging from the numerical value you give for your answer,  I presume it's $\ 0.33...\ \color{red}{\text{seconds}}\ $, and the units of your $\ x$-$y\ $ coordinates are metres, but that's not something your readers should have to guess at.
One quadratic curve between the same two points, $\ (0,0)\ $ and $\ (1,-1)\ $ as yours, but which the particle will slide down in finite time is $\ y=$$x^2-2x\ $. Another is $\ y=\frac{-(x^2+x)}{2}\ $.  For the first of these, you'll have $\ \frac{dy}{dx}=2(x-1)\ $ and
\begin{align}
T[y]&=\frac{1}{\sqrt{2g}}\int_0^1\sqrt{\frac{2-8x+4x^2}{2x-x^2}}\,dx\ .
\end{align}
The maplesoft calculator app confirms my suspicion that an indefinite integral with the above integrand can only be expressed in terms of elliptic integral functions, not in terms of elementary functions. Wolfram alpha gives
$$
\int_0^1\sqrt{\frac{2-8x+4x^2}{2x-x^2}}\,dx\approx2.63518538\ .
$$
So if $\ x\ $ and $\ y\ $ are measured in metres, and we take $\ g=9.80665\,\text{m/s$^2$}\ $ we'll have $\ T[y]\approx0.595025\,$s for this curve.
For $\ y=\frac{-(x^2+x)}{2}\ $, you'll have $\ \frac{dy}{dx}=-x-\frac{1}{2}\ $ and
\begin{align}
T[y]&=\frac{1}{\sqrt{2g}}\int_0^1\sqrt{\frac{5+4x+4x^2}{2(x^2+x)}}\,dx\ .
\end{align}
Again, elliptic integral functions are needed to evaluate this integral, for which Wolframalpha gives
$$
\int_0^1\sqrt{\frac{5+4x+4x^2}{2(x^2+x)}}\approx3.24104\ .
$$
So if $\ x\ $ and $\ y\ $ are measured in metres, and we take $\ g=9.80665\,\text{m/s$^2$}\ $ we'll have $\ T[y]\approx0.731828\,$s for this curve.
I expect there are very few functions $\ y\ $ for which the integral
$$
{\Large\int}\sqrt{\frac{1+\left(\frac{dy}{dx}\right)^2}{-y}}\,dx\\
$$
is expressible in terms of elementary functions. Your choice of $\ y=-x^2\ $ happens to be one of them.
From the law of conservation of energy it follows that the speed $\ v\ $ of a particle falling frictionlessly down a curve under gravity is completely determined by the distance $\ d\ $ that it has fallen from rest:
$$
v=\sqrt{2dg}\ .
$$
Its maximum speed will therefore always occur at the lowest point of the curve.
