3. The distance in feet that an object has fallen t seconds after being dropped: g(t)=16t2

Question

3. The distance in feet that an object has fallen t seconds after being dropped:  g(t)=16t2

a. Domain:

b. Modify graph:

c. Vertex:

d. Zeros:

Answer 1

good answer but the graph is a parabola. I get tongue tied too...

Answer 2

Your function of how far the object has fallen is y = 16t² where t is the number of seconds. As you can see, this is a quadratic equation so its graph is a hyperbola.

The domain is all the values that make sense for t. Since you can't have a negative number of seconds, you know that t >= 0.

If you are told that the object was dropped from a specific height, then the time t would have an upper bound too. For example, if it was dropped from 1600 feet, then t would also have to satisfy 16t² <= 1600, so t <= 10.

The vertex of this graph is the lowest point, so you can see that g(0) = 0 is the minimum because it will never be negative.

Your zeroes are where the graph meets the axes, so this vertex is also the only zero.

The part asking you to modify the graph needs more information. Modify it how?