91Ó°ÊÓ

Vertical asymptotes in rational functions are lines that the graph approaches but never actually touches. They occur specifically at values of \( x \) where the function's denominator is zero, resulting in an undefined expression. For our function \( f(x) = \frac{2(x+4)}{5(x-1)} \), the simplified denominator is \( 5(x-1) \).
To pinpoint the vertical asymptote, set the denominator equal to zero and solve for \( x \): \( 5(x-1) = 0 \). This simplifies to \( x = 1 \).
At \( x = 1 \), the function does not exist as the denominator becomes zero, creating a vertical asymptote. Graphically, as \( x \) approaches this value, the function's output will shoot up towards positive infinity or down towards negative infinity. This dramatic behavior results in the function's graph having a visible "gap" or break at \( x = 1 \).
It's key to remember that for vertical asymptotes, canceled terms don't count towards their determination. Even if a factor in the denominator cancels with a similar term in the numerator, it leads instead to a hole in the function, as we'll discuss later.

Horizontal asymptotes describe the behavior of a rational function as \( x \) tends to infinity or negative infinity. They provide a sense of what value the function may approach but not necessarily reach.
For our given function, after simplification, both the numerator \( 2(x+4) \) and the denominator \( 5(x-1) \) are linear, being degree one. When the degrees of both the numerator and the denominator are equal, the horizontal asymptote is determined by dividing the leading coefficients of each.
In this case, the leading coefficient of the numerator is 2, and for the denominator, it is 5. Thus, the horizontal asymptote is given by the line \( y = \frac{2}{5} \). This line indicates that as the values of \( x \) increase or decrease without bound, the value of \( f(x) \) approaches \( \frac{2}{5} \).
It's important to understand that horizontal asymptotes don't allow a function to strictly approach just a particular \( x \) from one side as in vertical asymptotes; instead, they convey end behavior, meaning what happens to the function's values as \( x \) becomes very large or very small.

Holes in a rational function occur where specific values of \( x \) lead to a zero that is cancelled out in the function's simplification process. In simpler terms, while the original function is undefined at this point, due to division by zero, the simplification omits this discontinuity.
To identify holes in our function \( f(x) = \frac{2(x+4)}{5(x-1)} \), we look at the factors that were cancelled during simplification. The factor \( (x+2) \) in both the numerator and the denominator was removed, leaving the function undefined at \( x = -2 \), despite this term no longer appearing in the simplified expression.
At \( x = -2 \), the original function would have had a zero in the denominator, which means the graph of the function has a "hole" or a point of discontinuity there. In a graph, this shows up as a little circle or an omitted point at that \( x \) value. Although the function doesn't exist exactly at \( x = -2 \), everywhere else behaves normally as anticipated from the simplified function.
Recognizing holes requires checking places where cancellations have taken place. If a term cancels, it indicates a missing point rather than infinite growth or decay seen in asymptotes.