Question 1076744
You have to calculate derivatives.
{{{df/dt=t^2-4t-45=(t-9)(t+5)}}}
That derivative function is
negative for {{{-5<t<9}}}
(meaning f(t) decreases in that interval);
it is zero at {{{t=-5}}} and {{{t=9}}}
(indicating local extreme values of the function),
and the derivative is positive for any other {{{t}}} value
(indicating that f(t) is increasing).
So, f(t) increases with t for {{{t<-5}}} ,
reaches a local maximum at {{{t=-5}}} ,
decreases until reaching a local minimum at {{{t=9}}} ,
and then increases for {{{t>9}}} .
There is no absolute minimum or maximum.
{{{lim(t->infinity,f(t))}}}{{{"="}}}{{{infinity}}} and
{{{lim(t->-infinity,f(t))=-infinity}}} .
 
For inflection points,
we need the second derivative.
{{{d^2f/dt^2=2t-4=2(t-2)}}}
shows you that the inflection point is at
{{{t=2}}} .