Rational function graphs are defined by (and you get marks for) \u00a0the locations of the asymptotes (if any), as well as the x and y intercepts (once again, provided they exist) and holes (if any).\u00a0 Today, we’ll look at two of these features:\u00a0 horizontal asymptotes and holes.<\/p>\n
First, to holes:\u00a0 consider the following rational function:<\/p>\n
f(x)=((x-1)(x+2))\/((x+2)(x-3)) eqn 1<\/i><\/p>\n
You can see that, since (x+2) is both in the top and the bottom, the function simplifies to<\/p>\n
f(x)=(x-1)\/(x-3) eqn 2<\/i><\/p>\n
However,\u00a0when you cancel, you are really dividing.\u00a0 Since you can’t divide by zero, you can’t cancel x+2 when x=-2.\u00a0 At x=-2, the equation\u00a0remains undefined.\u00a0 Therefore, you will get a hole there.\u00a0 The graph of eqn 1<\/i>, above, will follow the graph of eqn 2<\/i> identically, except for a hole at x=-2.<\/p>\n
Now, to horizontal asymptotes:\u00a0 you get them when the degree on top matches the degree on the bottom or is less than the degree on the bottom.<\/p>\n
Case 1:\u00a0 the degrees on top and bottom are the same.\u00a0 Consider<\/p>\n
f(x)=(2x^2 – 2x -3)\/(x^2 +17x+11)<\/p>\n
To get the horizontal asymptote, divide the coefficient of the highest exponent term on top by the coefficient of the highest exponent term on the bottom.\u00a0 The horizontal asymptote will be y=2\/1, or just y=2.<\/p>\n
Case 2:\u00a0 the degree on the bottom is greater than that on top.<\/p>\n
Simple:\u00a0 the horizontal asymptote is\u00a0y=0.<\/p>\n