How many vertical asymptotes does the equation \(y=\frac{x-1}{x^2+6x-7}\) have?
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. .
+23245 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.
