## Calculus: Early Transcendentals (2nd Edition)

The linear function is $$f(x)=-\frac{2}{3}x-1.$$
The equation of the linear function passing through points $P(x_1,y_1)$ and $Q(x_2,y_2)$ is given by $$y=\frac{y_2-y_1}{x_2-x_1}(x-x_1)+y_1.$$ Using the value from the problem $x_1=0,\quad y_1=-1,\quad x_2=3,\quad y_2=-3$ we get $$y=\frac{-3-(-1)}{3-0}(x-0)+(-1)$$ which gives $$y=-\frac{2}{3}x-1,$$ so the linear functon is $$f(x)=-\frac{2}{3}x-1.$$