Precalculus

Hello,

So I'm determining the non permissible values of the variables & for this one:

(x^2+7x+12)/(x^2-x-12)

I know the restrictions are 4<x<-3
but I can't seem to get the math to support x<-3...

I factored the denominator into
(x-4)(x+3)
x+3>0
x>-3

What am I doing wrong?

Thanks for your time! :)

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  1. You've factored correctly, but each factor is restricted to not being equal to 0. In other words, x + 3 ≠ 0, giving x ≠ -3, and x - 4 ≠ 0, giving x ≠ 4.

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  2. A little bit more tricky than that.

    (x^2+7x+12)/(x^2-x-12)
    = (x+3)(x+4)/( (x-4)(x+3) )
    = (x+4)/(x-4) , where x ≠ -3

    From what I did above, it is clear that x ≠ 4, or else I am dividing by 0 in my
    final line if x = 4 , in both forms of the expression

    If we graph y = (x^2+7x+12)/(x^2-x-12)
    we could just as well graph y = (x+4)/(x-4)

    Both graphs will have a vertical asymptote at x = 4
    but we don't see a vertical asymptote at x = -3

    https://www.wolframalpha.com/input/?i=Plot+y+%3D+(x%5E2%2B7x%2B12)%2F(x%5E2-x-12)

    Why not??
    Notice when x = -3, y = 0/0, which is indeterminate, that is,
    the graph actually has a " hole" in it, namely at (-3, -1/7)
    In the simplified equation the point (-3, -1/7) actually exists
    https://www.wolframalpha.com/input/?i=Plot+y+%3D+(x%2B4)%2F(x-4)

    You stated that the restrictions are 4<x<-3
    That is false, the restrictions are x = 4 and x = -3
    By saying 4<x<-3 , you are also cutting out values like x = 1.5, x = 2, x = -2 etc

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  3. 4 < x < -3.
    Do you mean -3 < x < 4?.

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