Exercise 9 Page 585

Consider the given rational function. f(x)=6/x+2-5 Let's start by identifying any x-values for which f(x) is undefined. To do this, we must recall that a denominator cannot be zero. x+2≠ 0 ⇔ x≠ - 2 The function is not defined when x=- 2, so there is a vertical asymptote at x=- 2. Let's now consider the given graph.

We can see that from x=- 2, as the x-values decrease, f(x) values approach - 5. Similarly, as the x-values increase, f(x) values approach - 5. This means there is a horizontal asymptote at f(x)=- 5. Let's draw the asymptotes on the given coordinate plane.

In the graph we see that the domain is all real numbers except - 2, and the range is all real numbers except - 5. Domain:& { x | x ≠ - 2 } Range:& { f(x) | f(x) ≠ - 5 }