Answer:
x = 3 seconds
Step-by-step explanation:
The time (x) at which the function [tex]-(x-3)^2+23[/tex] reaches its maximum can be determined in couple of ways. If you know calculus, you differentiate the quadratic, set to 0 and solve: [tex]y'= -2(x-3)=0 \implies x=3[/tex]. If you do not yet know how to do that, you can just apply the following argument:
The expression [tex]-(x-3)^2[/tex] will be non-positive for any x. The largest it can be is 0. So the entire function [tex]-(x-3)^2+23[/tex] can be at most 23, or less. It is 23 only when [tex]-(x-3)^2=0[/tex] and that happens exactly for x=3 (seconds).
As a third option, you can also use a formula for the maximum point of a parabola, but the formula is non-intuitive so I prefer to skip that.
