Question 745041
y=-16t^2+100

since co-efficient of x^2 is negative the parabola opens downwards

find the two values of x for which y=0

t= +2.5 and -2.5

the two points are (-2.5,0) and (2.5,0)

when t=0, y=100 you get another point (0,100) this is the vertex of the parabola.

take t=+/-1, y1=84 (+/-1,84)

take t=+/-2,y2=68(+/-2,36)

The parabola is symmetric about the axis of symmetry. 

Plot the points and trace a symmetric curve.

{{{drawing(500,500,-10,10,-10,100,grid(1),circle(-2.5,0,0.2),circle(2.5,0,0.2),circle(1,84,0.2),circle(-1,84,0.2),circle(2,36,0.2),circle(-2,36,0.2),graph(500,500,-10,10,-10,100,y=-16x^2+100))}}}