Posted by Sydney on Saturday, July 25, 2009 at 8:59pm.
I'm working on vertical asymptotes when factoring 6x^2  6 (denominator). I came up with 6(x^21). Then I was trying to find the zeros of the denominator but 6 does not = 0 and x^21=0 then I got x^2=1 and x = plus or minus sqrt 1. Can you tell me what I am doing wrong?

College Algebra  MathMate, Saturday, July 25, 2009 at 11:10pm
(6x²6)
=6(x²1)
=6(x+1)(x1)
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.
Answer This Question
Related Questions
 pre cal  g(x) = x^2  x 12 / x + 1 I know the division brings it to x 2 + 10...
 College Algebra  Using the 7 steps outlined in section 4.3 of your book, ...
 Math  State domain, range, period, vertical asymptotes, zeros, symmetry and y...
 algebra  Create a rational function with a linear binomial in both the ...
 ALGEBRA  Create a rational function with a linear binomial in both the ...
 math  the sum of numerator and denominator is 3 less than twice the denominator...
 math  1)The area A of a triangle varies jointly as the lengths of its base b ...
 math,correction please..  (x)/(x2)  (x+1)/(x) = (8)/(x^22x) I keep getting x...
 Calculus  Ok so I have to find a vertical asymptote with denominator 3x^3x1. ...
 Calculus/College Algebra  I'm so lost. We're working on limits for this ...
More Related Questions