Answer
$f(t)=-6.25t^{2}-100t+1200$
In 2008, the model predicts ${{\$}} 1425$ billion, ${{\$}} 25$ billion lower than the actual value.
Work Step by Step
$ y_{1}\sim ax_{1}^{2}+bx+c$
Result:
$f(t)=-6.25t^{2}-100t+1200$
The year 2008 is 18 years after 1990,
$f(18)=1425$ billion,
The chart in exercise 19 has this value to be ${{\$}} 1450$ billion,
so the model predicts ${{\$}} 25$ billion lower than the actual value.