The function is a parabola in inverted U form.
The form of the equation tells us that the vertex is at t=4, h=12.8, because this point is uniquely the point where there is only one value of t, instead of two, for each value of h in range.
The vertex at (4,12.8) sits between the zeroes of the function given by putting h=0.
0.8(t-4)^2=12.8, so (t-4)^2=16 and t-4=±4. The zeroes are at (0,0) (the origin, where the h (vertical) and t (horizontal) axes intersect) and (8,0).
This information should be sufficient to sketch the graph.
We can tabulate some values: