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

Rewrite f(x) = [3x(x−1)] / [x^2(x+3)(x+1)]. Zeros of the numerator: x = 0 and x = 1. But x = 0 (and x = −3, −1) makes the denominator zero, so x = 0 is not in the domain.

Cancel one x to see the simplified rule:
f(x) = 3(x−1) / [x(x+3)(x+1)] with domain x ≠ 0, −3, −1.
At x = 1 the denominator ≠ 0, so f(1)=0 and the graph has the x‑intercept (1, 0). At x = 0 the function is undefined (vertical asymptote), so (0,0) is not an x‑intercept.

Thus the correct description is: the graph has a single x‑intercept at (1,0).