Question

How would you describe the relationship between the real zero(s) and x-intercept(s) of the function

3x(x-1)
f(x) =
x+3)(x + 1)
x²(x+
41:07
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).
O When you substitute x = 0 into the function, there is no solution; therefore, the graph will not have any x-intercepts.
O 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

The real zeros of the function correspond to the x-intercepts of the graph.

Step-by-step explanation:

The relationship between the real zero(s) and x-intercept(s) of the function f(x) = \(\frac{3x(x-1)}{(x+3)(x + 1)}\) can be described as follows:

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). This means that the real zeros of the function are 0 and 1, and they correspond to the x-intercepts of the graph.

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