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Rational functions are one of the foundational concepts in algebra, particularly involving fractions where both the numerator and the denominator are polynomials. These functions are represented as a ratio of two polynomial expressions. For example, \(f(x) = \frac{P(x)}{Q(x)}\), where \(P(x)\) and \(Q(x)\) are polynomials and \(Q(x) \eq 0\). In the exercise provided, we have \(h(x) = \frac{7x}{(x-1)(x+5)}\), making it a rational function since it is a ratio of the polynomial \(7x\) to the polynomial \( (x-1)(x+5)\). Understanding how rational functions behave, including their zeros and restrictions due to the denominator, is crucial for solving and graphing them effectively.

One key step in solving rational functions is setting the function equal to zero. This step often helps in finding the values of \('x'\) where the function intersects the x-axis. To determine the solutions to \(h(x) = 0\), set the entire rational expression equal to zero: \(\frac{7x}{(x-1)(x+5)} = 0\). To solve this, it's vital to remember that a fraction equals zero only when its numerator is zero while the denominator doesn’t equal zero (since division by zero is undefined). Therefore, we solve for \(7x = 0\). By simplifying this, we get \(x = 0\). This means the function \(h(x)\) intersects the x-axis at \(x = 0\).

In rational functions, understanding the roles of the numerator and denominator is essential. The numerator is the top part of the fraction, while the denominator is the bottom part. In the given exercise, the numerator is \(7x\) and the denominator is \( (x-1)(x+5)\).
When solving rational functions, setting the numerator equal to zero helps find the zeros of the function. For the function to be zero, \(7x = 0\), because a fraction equals zero only if its numerator is zero, irrespective of the denominator.
Furthermore, the values that make the denominator zero must be excluded from the domain because division by zero is undefined. In this example, the critical values from the denominator would be \(x - 1 = 0\) and \(x +5 = 0\), which solve to \(x = 1\) and \(x = -5\) respectively. These values are where the function is undefined.