## Intermediate Algebra (6th Edition)

$f(x)=x-\dfrac{3}{4}$
Using $y-y_1=\dfrac{y_1-y_2}{x_1-x_2}(x-x_1)$ or the Two-Point Form of linear equations, then the equation of the line passing through $\left( \dfrac{1}{2},-\dfrac{1}{4} \right) \text{ and } \left( \dfrac{3}{2},\dfrac{3}{4} \right) ,$ is \begin{array}{l}\require{cancel} y-\left( -\dfrac{1}{4} \right)=\dfrac{-\dfrac{1}{4}-\dfrac{3}{4}}{\dfrac{1}{2}-\dfrac{3}{2}}\left(x-\dfrac{1}{2} \right) \\\\ y+\dfrac{1}{4}=\dfrac{-\dfrac{4}{4}}{-\dfrac{2}{2}}\left(x-\dfrac{1}{2} \right) \\\\ y+\dfrac{1}{4}=\dfrac{-1}{-1}\left(x-\dfrac{1}{2} \right) \\\\ y+\dfrac{1}{4}=x-\dfrac{1}{2} \\\\ y=x-\dfrac{1}{2}-\dfrac{1}{4} \\\\ y=x-\dfrac{2}{4}-\dfrac{1}{4} \\\\ y=x-\dfrac{3}{4} .\end{array} In function notation, this is equivalent to $f(x)=x-\dfrac{3}{4} .$