Help on one quick Math 2 question!!

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1

<p>Hey guys, I was just reviewing for tomorrow and found something I don't understand. Can someone explain?</p>

<p>So the question asks for the horizontal and vertical asymptotes of this equation:
(3x^2 - 13x - 10) / (x^2 - 4x - 5)</p>

<p>So I factor both out and get this
(3x+2)(x-5) / (x+1)(x-5)</p>

<p>so I understand there is a horizontal asymptote y=3, and that there is a vertical asymptote at x=-1. What I dont understand is that shouldnt there be an asymptote at x=5 as well? I understand it gets cancelled out, but when I enter the equation in my graphing calculator, It still shows the value x=5 as error. If anyone could explain this to me, it'd help a lot.</p>

2

<p>No asymptote at x = 5. For all points in the neighborhood of 5 that are not equal to 5 (x = 5 ± ε for 0 < ε < 0.1 for example), the function is defined and is equal to (3x+2)/(x+1). So the graph looks the same as f(x) = (3x+2)/(x+1), except it is not defined at x = 5. This is sometimes referred to as a “hole” in the graph.</p>

3

<p>Thanks!</p>

4

<p>No problem - good luck!</p>