Answer:
0 ≤ t ≤ 4
Step-by-step explanation:
Given function:
[tex]a(t)=-16t^2+48t+64[/tex]
where:
The event starts when t = 0. This is the first endpoint of the domain.
If the event ends when there are no attendees, the other endpoint of the domain will be the greater value of t when a(t) = 0.
Set the function to zero and factor it:
[tex]\implies -16t^2+48t+64=0[/tex]
[tex]\implies -16(t^2-3t-4)=0[/tex]
[tex]\implies t^2-3t-4=0[/tex]
[tex]\implies t^2-4t+t-4=0[/tex]
[tex]\implies t(t-4)+1(t-4)=0[/tex]
[tex]\implies (t+1)(t-4)=0[/tex]
Apply the zero-product property:
[tex]\implies t+1=0 \implies t=-1[/tex]
[tex]\implies t-4=0 \implies t=4[/tex]
Therefore, the domain is 0 ≤ t ≤ 4.
When graphing inequalities on a number line:
Therefore, to represent the found domain on the number line: