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# HELP ASAP IN LATEXX

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How many vertical asymptotes does the equation $$y=\frac{x-1}{x^2+6x-7}$$ have?

Jul 29, 2020

#1
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Whenever the denominator equals zero.

x2 + 6x – 7 factors to (x+7)(x–1) so ordinarily I'd say that asymptotes are at x = –7 and x = +1

However, when x = +1 both the numerator and denominator equal zero..

Maybe the (x–1) in the numerator and denominator cancel and you're left with y = 1/8.  .

Jul 29, 2020
edited by Guest  Jul 29, 2020
#2
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First, factor the denominator:  x2 + 6x - 7  =  (x + 7)(x - 1).

Now, the function becomes:  (x - 1) / (x2 + 6x - 7)   --->   (x - 1) / [ (x + 7)(x - 1) ]

The "x - 1" term in the denominator cancels the "x - 1" term in the numerator; this means that there

will be a "hole" in the graph at the point where x = 1.

The function has become:  1 / (x + 7)

There will be a vertical asymptote at the point where x = -7.

There is one asymptote.

Jul 29, 2020