Let's say we want our acceleration to be an affine function of time, meaning we have a constant jerk. That makes our velocity vs time graph take the form of a parabola, with its directrix parallel to the time axis.
The general equation for such a curve is:
$$v(t) = a \cdot t^2 + b \cdot t + c$$
Since your parabola goes through the point \$(t, v(t)) = (0, 0)\$, we can conclude that \$c = 0\$.
Next, the derivative of this equation is:
$$\frac {d v(t)} {d t} = 2 a \cdot t + b$$
So if you want the initial acceleration to be \$a_0\$ then we can conclude \$b = a_0\$
Then if we want \$v(T_{max}) = v_{max}\$, we can solve for \$a\$:
$$\begin{align}
v_{max} &= a \cdot T_{max}^2 + a_0 \cdot T_{max}\\
v_{max} - a_0 \cdot T_{max} &= a \cdot T_{max}^2\\
\frac {v_{max}} {T_{max}^2} - \frac {a_0} {T_{max}} &= a
\end{align}$$
So we end up with the equation for speed:
$$v(t) = \left( \frac {v_{max}} {T_{max}^2} - \frac {a_0} {T_{max}} \right) \cdot t^2 + a_0 \cdot t$$
...which you can solve for \$t\$ using the quadratic formula (taking the positive root):
$$t = \frac {-a_0 + \sqrt{a_0^2 + 4 v \left( \frac {v_{max}} {T_{max}^2} - \frac {a_0} {T_{max}} \right) }} {2 \left( \frac {v_{max}} {T_{max}^2} - \frac {a_0} {T_{max}} \right) }$$
Note that with this construction, you need to keep \$a_0 \cdot T_{max} <= 2 v_{max}\$ if you want the velocity function to be non-decreasing over the interval \$t \in [0, T_{max}]\$. If your initial acceleration is too steep or your time horizon is too short, your velocity will have to exceed your max, then slow down again to hit your target. If you need curves in this range, then we'll need to fall back on a different class of function with a more aggressive braking behaviour than the constant jerk we use here.
Here's another approach that works better for higher values of the initial acceleration, based on a kind of hyperbolic function:
$$x = \frac t {T_{max}}\\
v(t) = v_{max}\cdot x \cdot \frac {k + 1} {x + k}$$
Where
$$k = \frac 1 {T_{max} \frac {a_0} {v_{max}} - 1}$$
This one can handle arbitrarily high values of \$a_0\$, but it blows up for the constant acceleration case \$a_0 = \frac {v_{max}} {T_{max}}\$, so you'd need to special-case values in that vicinity. It also tends to make a sharper "corner" than the quadratic approach I showed first:
(Quadratic / constant jerk curve in blue, new hyperbolic function in orange)
The hyperbolic function is reasonably straightforward to invert too:
$$vx + vk = v_{max} x k + v_{max}x\\
vk = x \left(v_{max}(k + 1) - v\right)\\
x = \frac {vk} {v_{max}(k + 1) - v}\\
t = T_{max}\cdot x =T_{max}\frac {vk} {v_{max}(k + 1) - v}$$
