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irspow
I will assume that you meant:

(x+4)/(-3x^2+12x+36)  factor the denominator...

(x+4)/(-3(x^2-4x-12))

To factor a quadratic of the form ax^2+bx+c you need to find two values, j and k, which satisfy two conditions...

jk=ac=-12 and j+k=b=-4 so j and k must be 2 and -6 and the factors are then:

(x+2)(x-6) so we are now left with:

(x+4)/(-3(x+2)(x-6))

The only restriction on this function is that division by zero is undefined, so x cannot equal -2 or 6
boffeemadrid

The excluded values of [tex]x[/tex] are required which make the function undefined.

The excluded points are [tex]-2,6[/tex].

The given equation is

[tex]\dfrac{x+4}{-3x^2+12x+36}[/tex]

Since, [tex]x[/tex] is present in the denominator.The denominator must not become zero as it will become undefined if is zero.

So, the zeros of the denominator will be the excluded points.

Finding the zeros

[tex]-3x^2+12x+36=0\\\Rightarrow x=\dfrac{-12\pm \sqrt{12^2-4\left(-3\right)\times 36}}{2\left(-3\right)}\\\Rightarrow x=-2,6[/tex]

So, the excluded points are [tex]-2,6[/tex].

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