The height, h, of a ball that is tossed into the air is a function of the time, t, it is in the air. The height in feet for t seconds is given by the function h(t)=βˆ’16t2+96tWhat is the domain of the function?Question 8 options:a)[0, [infinity] )b)(-[infinity], [infinity])c)(0, [infinity])d)(0, 5)e)none

Accepted Solution

Answer:e) noneStep-by-step explanation:The first thing that we need to realise is that this question is set to a practical example of a ball being tossed into the air for a period of time. Thus, we know that:a) time cannot be negativeb) a ball cannot travel a negative distance in height from where it was tossed it into the air (given that this starting position is also the ground)In the context of what the problem asks us to achieve, which is to find the domain, we now know that:a) t β‰₯ 0b) the function h(t) is only practical when h(t) β‰₯ 0In other words, we know that the domain starts at t = 0, however we need to find when the ball drops back onto the ground. We can do this by solving h(t) = 0:h(t) = -16t^(2) + 96t0 = -16t^(2) + 96t0 = -16t(t - 6) (Factorise -16t^(2) + 96t)So, now we get:-16t = 0, therefor t = 0ort - 6 = 0, therefor t = 6Thus, the ball is on the ground at t = 0 seconds and t = 6 seconds; it is between these two values of t that the domain exists and that the problem is practical.Now, looking at the multiple choice options, it seems as though none of them are correct, therefor the answer would be e) none.The other way to work through this question (particularly if it is just multiple choice) is that, after realising that the ball is tossed into the air at t = 0 seconds and then drops at some point later, we could already discount a), b) and c) as answers since they include infinity in the domain, which is not practical for this problem. Then, we could substitute t = 5 into the equation to get:h(t) = -16(5)^2 + 96*5h(t) = -16*5*5 + 96*5h(t) = -80*5 + 96*5Since we need h(t) to be 0 at the end value of the domain, this is not the correct answer. Thus, the answer is e) none.