vertical asymptotes

roger22797

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Feb 8, 2013
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I use apex, but when i gave me this problem: 2/ 3x(x-1)(x+5). The 2 is over the whole equation. Apex just has you use the opposites of the numbers in the parenthesis, and x equals 0, but it doesn't say what to do the 3...... Help!?
 
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I use apex, but when i gave me this problem: 2/ 3x(x-1)(x+5). The 2 is over the whole equation. Apex just has you use the opposites of the numbers in the parenthesis, and x equals 0, but it doesn't say what to do the 3...... Help!?
The factor of 3 does not affect the roots, because 3×0 is also 0. (And 3 itself is never 0.)
 
I use apex, but when i gave me this problem: 2/ 3x(x-1)(x+5). Note that you have no "f(x) =" or "y =" in front of it.


The 2 is over the > > whole equation < < . Regardless, you must use grouping symbols. f(x) = 2/(3x(x - 1)(x + 5)) or f(x) = 2/[3x(x - 1)(x + 5)], for instance.


Apex just has you use the opposites of the numbers in the parenthesis, and x equals 0, but it doesn't say what to do the 3...... Help!?

What you typed is an expression, not an equation.
 
"Vertical asymptotes" occure when the denominator (but not the numerator) is 0. Here the denominator is 3x(x- 1)(x- 5). That product will be 0 if and only if one of the factors is 0 (basic rule ab= 0 if and only if one or both of a and b is 0). As Dr Phil said, 3 is never 0 so you wind up looking at x= 0, x- 1= 0, x- 5= 0.
 
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