Answer:
you didn't feature any graphs for me to choose from
Answer:

Step-by-step explanation:
We have been given that an arrow is shot straight up from a cliff 58.8 meters above the ground with an initial velocity of 49 meters per second. Let up be the positive direction. Because gravity is the force pulling the arrow down, the initial acceleration of the arrow is −9.8 meters per second squared.
We know that equation of an object's height t seconds after the launch is in form
, where
g = Force of gravity,
= Initial velocity,
= Initial height.
For our given scenario
,
and
. Upon substituting these values in object's height function, we will get:

Therefore, the function for the height of the arrow would be
.
Answer:
4(x - 9)^2 - 17
Step-by-step explanation:
A horizontal shift to the right by 9 changes x^2 to (x - 9)^2.
A vertical shift down by 1 changes - 16 to -17.
Therefore, the new equation is:
4(x - 9)^2 - 17
K well uh i don't know what u meant but the 4th one i think so u
It means the ball did not move forward or backward. It stayed on the same line where it was