Answered

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

$f(x)=\frac{3 x(x-1)}{x^2(x+3)(x+1)}$?


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

Verified Answer :

  • The domain of the function excludes \(x = 0\), \(x = -3\), and \(x = -1\) because these values make the denominator zero.
  • Setting the numerator to zero gives candidates \(x = 0\) and \(x = 1\).
  • Since \(x = 0\) is outside the domain, only \(x = 1\) is a valid zero.
  • Consequently, the only x-intercept is \((1,0)\).
  • Final answer: The graph has an x-intercept of \((1,0)\).

Explanation

  1. Determining the Domain of the Function

To understand the relationship between the real zeros and x-intercepts of the function \(f(x) = \frac{3x(x-1)}{x^2(x+3)(x+1)}\), we first identify the function's domain. The function is undefined when the denominator is equal to zero:
\[x^2(x+3)(x+1) = 0\]
This occurs when \(x = 0\), \(x = -3\), or \(x = -1\). Thus, the domain of the function is all real numbers except \(x \in \{0, -3, -1\}\).

  1. Finding Potential Zeros

Next, we find the potential zeros by setting the numerator of the function equal to zero:
\[3x(x-1) = 0\]
This equation holds when \(x = 0\) or \(x = 1\). These are the candidates for the real zeros of the function. However, a zero must also be within the domain of the function to qualify as an x-intercept. We previously determined that \(x=0\) is excluded from the domain because it results in a division by zero in the original expression. Therefore, \(x=0\) is not a zero of the function.

  1. Verifying Zeros and Identifying X-Intercepts

We check the candidate \(x=1\). Since \(1\) is not in the set of excluded values \(\{0, -3, -1\}\), it is within the domain of \(f(x)\). Evaluating the function at \(x=1\):
\[f(1) = \frac{3(1)(1-1)}{1^2(1+3)(1+1)} = \frac{3(1)(0)}{1(4)(2)} = \frac{0}{8} = 0\]
Because \(f(1) = 0\), the value \(x=1\) is a valid real zero of the function, and consequently, the point \((1, 0)\) is an x-intercept of the graph.

  1. Conclusion

Comparing our findings with the given options, we conclude that setting the function equal to zero yields the solution \(x=1\), confirming that the graph has an x-intercept at \((1,0)\). The other candidates do not satisfy the condition of being in the domain, so they do not produce x-intercepts. Thus, the first statement provided is the correct one.

Examples

Understanding the domain of a function is crucial in real-world modeling. For example, if a function represents the concentration of a chemical in a reaction over time, certain values of time might lead to division by zero or negative time, which are physically impossible. By identifying the domain, scientists ensure that their mathematical models accurately reflect physical reality, correctly identifying valid points where concentrations reach zero without being misled by algebraic artifacts.

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