Answer
a. $y=1.6x +35$
b. $y=-1.3x +57$
c. $8$ years after 2001, or in year 2009.
Work Step by Step
a. Based on the given graph, with $x=year-2001$, we can identify two points on the line as $(0,35)$ and $(14,57)$. We can find the slope as $m=\frac{57-35}{14-0}=\frac{22}{14}\approx1.6$. We can write the line equation as $y-35=1.6(x-0)$ or in slope-intercept form as $y=1.6x +35$
b. Based on the given graph, with $x=year-2001$, we can identify two points on this line as $(0,57)$ and $(14,39)$. We can find the slope as $m=\frac{39-57}{14-0}=\frac{-18}{14}\approx-1.3$. We can write the line equation as $y-57=-1.3(x-0)$ or in slope-intercept form as $y=-1.3x +57$
c. Let $1.6x +35=-1.3x +57$; we have $2.9x=22$ and $x\approx8$ years after 2001 or in year 2009.
