Question

If an object is dropped from a height of 55 feet, the function d=-16t^2+55 gives the height of the object after t seconds. Graph this function. Approximately how long does it take the object to reach the ground (d=0)

Answer 1

notice, it hits the ground when d = 0, thus 

$\bf \stackrel{d}{0}=-16t^2+55\implies 16t^2=55\implies t^2=\cfrac{55}{16}\implies t=\sqrt{\cfrac{55}{16}}$

that's when.

as far as its graph, is more or less like the picture below, now, notice that if we put the quadratic in "vertex form", it'd look like 

$\bf d=-16t^2+55\implies d=-16(t-\stackrel{h}{0})^2+\stackrel{k}{55}\qquad vertex~(h,k)$

so its vertex is at (0, 55), so just pick one point to the left of the vertex, and one to the right of the vertex, and draw it away