Question

Use the equation to answer the following question y=(x-3(x+2)/(x+4)(x-4)(x+2)

Answer 1

Answer:

See below

Step-by-step explanation:

The given rational function is;

$y=\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}$

The given function is not continuous where the denominator is equal to zero.

$(x+4)(x-4)(x+2)=0$

The function is discontinuous at $x=-4,x=4,x=-2$

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.

$\lim_{x\to \infty}\frac{(x-3)(x+2)}{(x+4)(x-4)(x+2)}=0$

The horizontal asymptote is y=0