Given:
The function is

To find:
The asymptotes and zero of the function.
Solution:
We have,

For zeroes, f(x)=0.



Therefore, zero of the function is 0.
For vertical asymptote equate the denominator of the function equal to 0.


Taking square root on both sides, we get


So, vertical asymptotes are x=-4 and x=4.
Since degree of denominator is greater than degree of numerator, therefore, the horizontal asymptote is y=0.
Answer:answer is b
Step-by-step explanation: Just makes since
Answer:
(3+p)
Step-by-step explanation:
solution
first expression=9-P^2
=3^2 - P^2
=(3+P) (3-p)
second expression=p^2 + 3p
= p(3+p)
Therefore lCM= (3+p)