## Intermediate Algebra for College Students (7th Edition)

if $x=-2$, $y=-\frac{1}{2}$ --> $(-2,-\frac{1}{2})$ if $x=-1$, $y=-\frac{1}{1}$ --> $(-1,-1)$ if $x=-\frac{1}{2}$, $y=-2$ --> $(-\frac{1}{2},-2)$ if $x=2$, $y=\frac{1}{2}$ --> $(2,\frac{1}{2})$ if $x=-\frac{1}{3}$, $y=-3$ --> $(-\frac{1}{3},-3)$ if $x=\frac{1}{3}$, $y=3$ --> $(\frac{1}{3},3)$ if $x=\frac{1}{2}$, $y=2$ --> $(\frac{1}{2},2)$ if $x=1$, $y=1$ --> $(1,1)$ if $x=2$, $y=\frac{1}{2}$ --> $(2,\frac{1}{2})$
$$y = \frac{1}{x}$$ (Let $x = -2, -1, -\frac{1}{2}, 2, -\frac{1}{3}, \frac{1}{3}, \frac{1}{2}, 1,$ and $2$.) Substitute the values of $x$ to the equation: if $x=-2$, $y=-\frac{1}{2}$ --> $(-2,-\frac{1}{2})$ if $x=-1$, $y=-\frac{1}{1}$ --> $(-1,-1)$ if $x=-\frac{1}{2}$, $y=-2$ --> $(-\frac{1}{2},-2)$ if $x=2$, $y=\frac{1}{2}$ --> $(2,\frac{1}{2})$ if $x=-\frac{1}{3}$, $y=-3$ --> $(-\frac{1}{3},-3)$ if $x=\frac{1}{3}$, $y=3$ --> $(\frac{1}{3},3)$ if $x=\frac{1}{2}$, $y=2$ --> $(\frac{1}{2},2)$ if $x=1$, $y=1$ --> $(1,1)$ if $x=2$, $y=\frac{1}{2}$ --> $(2,\frac{1}{2})$