F(x)=x+2 If you know the domain is {4,12}, then the range must be { , }
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The given expression is
And since we have x+2 common in numerator and denominator, there is a hole at x+2=0, that is at x=-2
And a function is undefined when denominator is zero. And at x=-6, denominator become zero.
So, at x=-6, the function is undefined, or there is a vertical asymptote at x=-6 and hole at x=-2 .