I'm working on vertical asymptotes when factoring 6x^2 - 6 (denominator). I came up with 6(x^2-1). Then I was trying to find the zeros of the denominator but 6 does not = 0 and x^2-1=0 then I got x^2=1 and x = plus or minus sqrt 1. Can you tell me what I am doing wrong?
(6x²-6)
=6(x²-1)
=6(x+1)(x-1)
If the above expression is the denominator of a given function, there will be two vertical asymptotes, namely at the zeroes of the given expression, at x=1 and x=-1.
The expression equals zero at these two places since 6*0=0.
Based on your explanation, it seems like you might have made a small error in your understanding of finding the zeros of the denominator. Let me explain the correct process to find the zeros in this case.
To find the zeros of the denominator, which would indicate the potential vertical asymptotes, you need to solve the equation x^2 - 1 = 0.
To solve x^2 - 1 = 0, you can use the factoring method as you correctly mentioned earlier.
By factoring, you would have (x - 1)(x + 1) = 0.
Now, to find the values of x, you need to set each factor equal to zero and solve for x separately.
Setting x - 1 = 0, you would get x = 1.
Setting x + 1 = 0, you would get x = -1.
Therefore, the zeros of the denominator are x = 1 and x = -1.
These values indicate that there are potential vertical asymptotes at x = 1 and x = -1.
The mistake you made was when you said x^2 = 1, implying that x = ±√1. However, the correct approach is to factor the expression further and solve for x by setting each factor equal to zero.
I hope this clarifies the issue. If you have any further questions, feel free to ask!