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Vertical asymptotes play a significant role in the graph of a rational function as they represent lines that the graph approaches but never crosses. They occur at values of \( x \) that make the denominator of a rational function equal to zero, provided these values do not also make the numerator zero.
In our given rational function \( f(x) = \frac{x}{x^2 + x - 12} \), the denominator is \( x^2 + x - 12 \), which can be factored as \((x + 4)(x - 3)\). Solving \( x + 4 = 0 \) gives \( x = -4 \), and solving \( x - 3 = 0 \) gives \( x = 3 \).
Thus, the vertical asymptotes are located at:

• \( x = -4 \)
• \( x = 3 \)

These lines should be drawn as dashed lines on the graph to indicate that the function never actually reaches these \( x \) values.

Horizontal asymptotes describe the behavior of a rational function as \( x \) approaches infinity or negative infinity. They provide insight into the end behavior of the function.
To determine the horizontal asymptote, compare the degrees of the numerator and denominator of the rational function. The rational function \( f(x) = \frac{x}{x^2 + x - 12} \) has a numerator of degree 1 and a denominator of degree 2. Whenever the degree of the numerator is less than the degree of the denominator, the horizontal asymptote of the function is \( y = 0 \).
This means that as \( x \) tends to either positive or negative infinity, \( f(x) \) approaches 0. This asymptote should be represented as a dashed line along the \( y \)-axis at \( y = 0 \) on the graph.

Intercepts are points where the graph crosses the axes. In a rational function, we often find both the x-intercept and y-intercept to help locate the function's position on the graph.

Y-intercept:

The y-intercept occurs when \( x = 0 \). Inserting 0 into the function, \( f(0) = \frac{0}{0^2 + 0 - 12} = 0 \), indicates that the y-intercept is at the point \((0, 0)\).

X-intercept:

The x-intercept occurs when the numerator equals zero, which means setting \( x = 0 \). This also results in an x-intercept at \((0, 0)\).
Both intercepts for the given function are the same, found at the origin point \((0, 0)\) on the graph.

The domain of a rational function includes all real numbers except those that cause the denominator to be zero (as this would make the function undefined).
For the function \( f(x) = \frac{x}{x^2 + x - 12} \), we factor the denominator as \((x + 4)(x - 3)\). The values \( x = -4 \) and \( x = 3 \) make this denominator zero.
Therefore, the domain of the function is all real numbers except for these values:

• \( x eq -4 \)
• \( x eq 3 \)

This means that \( f(x) \) can take any real number except \( -4 \) and \( 3 \), ensuring that the function remains defined throughout its domain.