Answer:
a) The domain of the function is
.
,
, b) The range of the function is
.
,
, c) The ball is 73 meters off of the ground at x = 3 seconds.
Step-by-step explanation:
The complete statement is: A ball is thrown upward off of a 100 meter cliff with an initial velocity of 6 m/s. The function
represents this situation where x is time and y is the distance off of the ground.
a) What domain does the function make sense?
b) What range does the function make sense
?
c) How far off the ground is the ball at time x = 3 seconds?
a) Let
and
be the time, measured in seconds, and the distance of the ground, measured in meters, respectively. Time is a positive variable, so domain corresponds to the interval when
and
. That is:
![-5\cdot x^{2} + 6\cdot x + 100 \geq 0](https://tex.z-dn.net/?f=-5%5Ccdot%20x%5E%7B2%7D%20%2B%206%5Ccdot%20x%20%2B%20100%20%5Cgeq%200)
![-(x-5.112\,s)\cdot (x+3.912\,s) \geq 0](https://tex.z-dn.net/?f=-%28x-5.112%5C%2Cs%29%5Ccdot%20%28x%2B3.912%5C%2Cs%29%20%5Cgeq%200)
Therefore, the domain of the function is
.
, ![\forall x \in \mathbb{R}](https://tex.z-dn.net/?f=%5Cforall%20x%20%5Cin%20%5Cmathbb%7BR%7D)
b) The distance off of the ground is also a positive variable, where ball is thrown upward at a height of 100 meters and hits the ground at a height of 0 meters. Hence, the range of the function is
.
, ![\forall y\in \mathbb{R}](https://tex.z-dn.net/?f=%5Cforall%20y%5Cin%20%5Cmathbb%7BR%7D)
c) The distance of the ball off of the ground at x = 3 seconds is found by evaluating the function:
![f(3\,s) = -5\cdot (3\,s)^{2} + 6\cdot (3\,s) + 100](https://tex.z-dn.net/?f=f%283%5C%2Cs%29%20%3D%20-5%5Ccdot%20%283%5C%2Cs%29%5E%7B2%7D%20%2B%206%5Ccdot%20%283%5C%2Cs%29%20%2B%20100)
![f(3\,s) = 73\,m](https://tex.z-dn.net/?f=f%283%5C%2Cs%29%20%3D%2073%5C%2Cm)
The ball is 73 meters off of the ground at x = 3 seconds.