## Calculus with Applications (10th Edition)

$y=33.7t+68.3$
Graphing the approximately linear function, the points on the graph are (t, y(t)). Let y be in billions of dollars, t is the number of years after 2000. Given points: (1, 102) and (8, 338) Slope: $m=\displaystyle \frac{338-102}{8-1}=\frac{236}{7}\approx 33.7$ Point-slope equation: $y-102=33.7(t-1)$ $y-102=33.7t-33.7\qquad/+102$ $y=33.7t+68.3$