Answer:
D. (4,0) and (-4,0)
C. there is one discontinuity at x=3
Explanation:
Roots can be found by setting f(x)=0:
f(x)=(x^2-16)/(x^2-2x-3)
0=(x^2-16)/(x^2-2x-3), multiply both sides by x^2-2x-3 to get rid of the denominator
0=x^2-16, use difference of squares to factor it
0=(x+4)(x-4)
x=4,-4
Therefore the roots are (4,0) and (-4,0)
Discontinuities are when f(x) is undefined, and f(x) is undefined when something is divided by 0
Set the denominator equal to 0:
0=x^2-6x+9, use the form a^2 - 2ab + b^2 to factor
0=(x-3)^2
x=3
Therefore, there is a discontinuity at x=3