Question 1101603
the first derivative is 0 at t=8
Here is the graph.
{{{graph(300,300,-5,15,-50,40,x^2-16x+28,2x-16)}}}
The particle is moving left from [0, 8). and it is moving right from (8, oo). ANSWER
At t=8 the instantaneous velocity is 0.
Can visualize moving left as down in this instance and moving right as up.  The x-intercepts are only where the particle is at 0, which is not a consideration here.