mathhelp1a
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x^2 - 4 / x -2
does this equation have any vertical asymptotes. i think it does at x = 2?
does this equation have any vertical asymptotes. i think it does at x = 2?
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asymptotes question
- Thread starter mathhelp1a
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No, it does not. It has a HOLE at x=2.
That is because \(\displaystyle \frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2\)
See why?. A good calculator may show it. If you work a TI-83 just right, it will display the hole in the graph.
Whenever you have a rational expression that has the same term in the numerator and denominator that cancel one another, it more than likely is a hole.
Now, if that were, say \(\displaystyle \frac{x+3}{x-2}\), then there would be a VA at x=2. The x-2 does not cancel. See?.
That is because \(\displaystyle \frac{x^{2}-4}{x-2}=\frac{(x-2)(x+2)}{x-2}=x+2\)
See why?. A good calculator may show it. If you work a TI-83 just right, it will display the hole in the graph.
Whenever you have a rational expression that has the same term in the numerator and denominator that cancel one another, it more than likely is a hole.
Now, if that were, say \(\displaystyle \frac{x+3}{x-2}\), then there would be a VA at x=2. The x-2 does not cancel. See?.
mathhelp1a
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but it does have a slant assymptote at x+2
D
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mathhelp1a said:but it does have a slant assymptote at x+2
Not really - the graph of your function is the straight line( y = x + 2) with a hole at x = 2. It does not satisfy the definition of asymptote.