## Calculus: Early Transcendentals (2nd Edition)

The required function is $$f(x)=-\frac{4}{5}x+5.$$
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 data given in the problem: $$x_1=0,\quad y_1=5,\quad x_2=5,\quad y_2=1$$ we have $$y=\frac{1-5}{5-0}(x-0)+5$$ giving $$y=-\frac{4}{5}x+5,$$ so our linear function is $$f(x)=-\frac{4}{5}x+5.$$