Question

How would you describe the relationship between the real zero(s) and x-intercept(s) of the function f(x)= 3x(x-1)/ x^2(x+3)(x+1)

When you set the function equal to zero, the solution is x = 1; therefore, the graph has an x-intercept of (1, 0).
When you set the function equal to zero, the solutions are x = 0 or x = 1; therefore, the graph has x-intercepts at (0, 0) and (1, 0).
When you substitute x = 0 into the function, there is no solution; therefore, the graph will not have any x-intercepts.
Since there are asymptotes at x = –3, x = –1, and x = 0, the graph has no x-intercepts and, therefore, no real zeros.

Answer 1

Set f(x)=0 by solving numerator = 0, but exclude any x that makes the denominator 0.

Numerator 3x(x−1)=0 → x=0 or x=1.
Denominator x^2(x+3)(x+1)=0 → x=0, −3, −1 are NOT allowed.

x=0 is ruled out because it makes the denominator 0 (actually x=0 is a vertical asymptote here — the denominator has multiplicity 2 while the numerator has multiplicity 1, so one x remains in the denominator after cancellation). x=1 is allowed, so the only real zero and x-intercept is (1, 0).

Therefore the first statement is correct. (There are vertical asymptotes at x = −3, −1, and x = 0.)