## Calculus: Early Transcendentals (2nd Edition)

The function is given by $$y=\Bigg\{_{-\frac{1}{2}x+3,\quad x>0.}^{x+3,\quad x\leq0;}$$
This is an example of a part by part linear function. We will use the fact that the equation of the linear function passing through points $(x_1,y_1)$ and $(x_2,y_2)$ is given by $$y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1$$ For $x\leq0$ we will choose the points $(-3,0)$ and $(0,3)$ giving $$y=\frac{3-0}{0-(-3)}(x-(-3))+0=\frac{3}{3}(x+3)$$ which gives finally $$y=x+3.$$ For $x\geq0$ we will chose $(0,3)$ and $(4,1)$ $$y=\frac{1-3}{4-0}(x-0)+3=\frac{-2}{4}x+3$$ finally giving $$y=-\frac{1}{2}x+3.$$ Now we can write $$y=\Bigg\{_{-\frac{1}{2}x+3,\quad x>0.}^{x+3,\quad x\leq0;}$$