Answer if u only know or will be deleted by mod pls
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Let's first factor the numerator and the denominator so we can find the input values that make them equal to zero.
the function is  undefined  for inputs that make the denominator equal to zero. 
inputs where the function is defined and the numerator is equal to zero are  zeros  of the function. these are simply the  xxx-intercepts  of the graph of the function. 
now we can simplify the function expression by canceling out common factors. any undefined input that no longer makes the denominator equal to zero is a  removable discontinuity. the remaining undefined inputs are  vertical asymptotes. 
   
                      factoring the numerator and the denominator
 
we see that the denominator is equal to zero when the input is x=-4  this means that  the function isn't defined for this input. it may either have a vertical asymptote or a removable discontinuity at this 
x -value. 
we still need to determine the exact type of discontinuity at this point, but let's determine the zeros of the function first.

we see that the numerator is equal to zero when the input is either x=-4 or x=-2. 

however, the function  doesn't   have a zero at  x=-4, because it isn't defined for that value.

therefore,  the function has a zero at  x=-2.
the function can be simplified by canceling out a factor of  (x+4).
the simplified expression,  (x+2)  is no longer undefined for  x=-4.

therefore,  the function has a removable discontinuity at  x=-4. 

after simplifying the function expression, no values make the denominator equal to zero.

therefore,  the function has no vertical asymptotes.

in conclusion,  q   has:  
a zero at x=-2 

a removable discontinuity at x=-4