## College Algebra (11th Edition)

Domain: $(-\infty, +\infty)$ Range: $(-\infty, +\infty)$ Refer to the image below for the graph.
$f(x)=\frac{2}{3}x+2$ The x-intercept can be found by letting f(x)=0. Here, $0=\frac{2}{3}x+2$ $-2=\frac{2}{3}x$ $-6=2x$ $-3=x$ Hence, -3 is the x-intercept, therefore we plot (-3,0). The y-intercept can be found by letting x=0. Here, $f(x)=-0+2$ $f(x)=2$ Hence, 2 is the y-intercept, therefore we plot (0, 2). By connecting these two points by a straight line, we get the graph. The domain is $(-\infty, \infty)$ as the function will "work" for all real x-values. (We can see the domain on the graph too. As for every x-value we will find a corresponding point on the graph.) The range is $(-\infty, \infty)$ as we will get all real y-values after substituting all real x-values in the function. (We can see the domain on the graph too. As for every y-value we will find a corresponding point on the graph.)
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