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When working with a rational function like \(h(x) = \frac{x^2 + x - 2}{x^2 - 5x}\), it's important to accurately calculate the numerator. The numerator in this particular function is \(x^2 + x - 2\). Let's break this down:

- **Quadratic Expression**: The term \(x^2\) represents a quadratic expression where the variable \(x\) is squared.- **Linear Expression**: The term \(x\) is a linear component, contributing directly to the value when \(x\) is substituted.- **Constant Term**: Finally, \(-2\) is a constant that shifts the entire expression up or down the y-axis.
To find the value of the numerator for any given \(x\), simply substitute the \(x\) value into the expression and perform basic arithmetic operations. For instance, substituting \(x = 5\) gives:
\[5^2 + 5 - 2 = 25 + 5 - 2 = 28\]Substituting \(x = -2\) results in:
\[(-2)^2 + (-2) - 2 = 4 - 2 - 2 = 0\]The evaluation of these expressions is straightforward once you understand the role of each part of the numerator.

The denominator of a rational function like \(x^2 - 5x\) is equally important. It determines where the function is undefined (i.e., where the denominator equals zero). This denominator can be broken down into its components:

- **Quadratic Term**: The \(x^2\) part represents a quadratic component which rises rapidly as \(x\) increases.- **Linear Term**: The \(-5x\) linearly subtracts from the quadratic part, influencing the function's shape significantly.
To evaluate the denominator, substitute the specific \(x\) value into the expression and calculate. When substituting \(x = 5\):
\[5^2 - 5 \cdot 5 = 25 - 25 = 0\]This results in a zero denominator, making the function undefined at this point. Conversely, for \(x = -2\):
\[(-2)^2 - 5 \cdot (-2) = 4 + 10 = 14\]Here the denominator does not become zero, allowing the function to have a defined value. Remember, a non-zero denominator is key to ensuring the function is valid.

Function evaluation involves substituting a particular value of \(x\) into a function and calculating the resulting value. This basic process is crucial for understanding the behavior of functions, especially rational ones.

Consider the process for evaluating \(h(5)\) which involves substituting \(x = 5\). Calculate the numerator and denominator separately:
- **Numerator**: \(5^2 + 5 - 2 = 28\).- **Denominator**: \(5^2 - 5 \cdot 5 = 0\).
Since the denominator is zero, \(h(5)\) is undefined. This can often be a critical point indicating restrictions or discontinuities.

Next, let's evaluate \(h(-2)\). Substitute \(x = -2\) into the function:
- **Numerator**: \((-2)^2 - 2 - 2 = 0\).- **Denominator**: \((-2)^2 - 5 \cdot (-2) = 14\).
With a zero numerator and non-zero denominator, the function evaluates to zero, i.e., \(h(-2) = 0\). This illustrates where the function crosses the x-axis.