## University Calculus: Early Transcendentals (3rd Edition)

a) $y=-x+2\ \$ on $\ \ (0,2]$ $y=-\frac{1}{3}x+\frac{5}{3}\ \$ on $\ \ (2,5]$ b) $y=-3x-3\ \$ on $\ \ (-1,0]$ $y=-2x+3\ \$ on $\ \ (0,2]$
$a).$ Find the equation of line that goes through the points $\ (0,2)\ \ \mathrm{and}\ \ (2,0)\$ and the line that goes through $\ (2,1)\ \ \mathrm{and}\ \ (5,0).$ Secondly, write down the domain restrictions. $y-2=\frac{0-2}{2-0}(x-0)$ $y=-x+2\ \$ on $\ \ (0,2]$ $y-1=\frac{0-1}{5-2}(x-2)$ $y=-\frac{1}{3}x+\frac{5}{3}\ \$ on $\ \ (2,5]$ $b).$ Find the equation of line that goes through the points $\ (-1,0)\ \ \mathrm{and}\ \ (0,-3)\$ and the line that goes through $\ (0,3)\ \ \mathrm{and}\ \ (2,-1).$ Secondly, write down the domain restrictions. $y-0=\frac{-3-0}{0+1}(x+1)$ $y=-3x-3\ \$ on $\ \ (-1,0]$ $y-3=\frac{-1-3}{2-0}(x-0)$ $y=-2x+3\ \$ on $\ \ (0,2]$