Answer:
See below
Step-by-step explanation:
The given rational function is;
[tex]y=\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}[/tex]
The given function is not continuous where the denominator is equal to zero.
[tex](x+4)(x-4)(x+2)=0[/tex]
The function is discontinuous at [tex]x=-4,x=4,x=-2[/tex]
b) The point at x=-2 is a removable discontinuity(hole) because (x+2) is common to both the numerator and the denominator.
The point at x=-4 and x=4 are non-removable discontinuities(vertical asymptotes)
c) The equation of the vertical asymptotes are x=-4 and x=4
To find the equation of the horizontal asymptote, we take limit to infinity.
[tex]\lim_{x\to \infty}\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}=0[/tex]
The horizontal asymptote is y=0