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Unformatted text preview: e however, managed to
find a vertical asymptote already.
Now, let’s see if we’ve got xintercepts. x2  4 = 0 Þ x = ±2 So, we’ve got two of them. © 2007 Paul Dawkins 242 http://tutorial.math.lamar.edu/terms.aspx College Algebra We’ve got one vertical asymptote, but there may be more so let’s go through the process and see. x2  4 x = x ( x  4) = 0 Þ x = 0, x = 4
So, we’ve got two again and the three regions that we’ve got are x < 0 , 0 < x < 4 and x > 4 .
Next, the largest exponent in both the numerator and denominator is 2 so by the fact there will be
a horizontal asymptote at the line, 1
y = =1
1
Now, one of the xintercepts is in the far left region so we don’t need any points there. The other
xintercept is in the middle region. So, we’ll need a point in the far right region and as noted in
the previous example we will want to get a couple more points in the middle region to completely
determine its behavior. f (1) = 1 f ( 3) = f ( 5) = (1,1) 5ö
æ
ç 3,  ÷
3ø
è
æ 21 ö
ç 5, ÷
è...
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 Spring '12
 MrVinh

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