x/abs(x^2 - 4) Would there be a vertical asymptote at only x = -2?
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Originally Posted by explodingtoenails x/abs(x^2 - 4) Would there be a vertical asymptote at only x = -2? There are vertical asymptotes at x = 2 and x = -2, and the function is not defined at only these two values of x.
Why x = 2? I thought the absolute value bars would negate the (x - 2)?
Originally Posted by explodingtoenails Why x = 2? I thought the absolute value bars would negate the (x - 2)? Why did you think that? Note that .
I thought if you set abs(x - 2) to 0, it would be like this: abs(x - 2) = 0 abs(x) = -abs(2) x = -2
Originally Posted by explodingtoenails I thought if you set abs(x - 2) to 0, it would be like this: abs(x - 2) = 0 abs(x) = -abs(2) x = -2
Why do the absolute value bars not do anything?
Originally Posted by explodingtoenails Why do the absolute value bars not do anything? The only number whose absolute value is 0, is 0.
So when there are absolute value bars around a binomial, it does nothing?
Originally Posted by explodingtoenails So when there are absolute value bars around a binomial, it does nothing? Do you know what absolute value means?
Yes. It's the value of the number, except it's positive no matter what. Correct?
Originally Posted by explodingtoenails Yes. It's the value of the number, except it's positive no matter what. Correct? Correct.
abs(x^2 - 4) = abs[(x - 2) (x + 2)] For (x - 2), I keep thinking the absolute value bars would make the - 2 into a + 2. So I keep doing this: abs(x - 2) = 5 abs(x) + abs(-2) = 5 abs(x) = 5 - 2 Why would the abs bars not doing anything to (x - 2)?
Originally Posted by explodingtoenails abs(x^2 - 4) = abs[(x - 2) (x + 2)] For (x - 2), I keep thinking the absolute value bars would make the - 2 into a + 2. So I keep doing this: abs(x - 2) = 5 abs(x) + abs(-2) = 5 abs(x) = 5 - 2 Why would the abs bars not doing anything to (x - 2)? For example, , whereas .
So when would abs bars actually affect the value of a binomial?
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