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In rational functions, vertical asymptotes represent values of \( x \) where the function takes on infinite behavior—essentially points at which the graph of the function shoots up to positive or negative infinity. Vertical asymptotes occur wherever the denominator of a rational function becomes zero, as these points are not defined for real number outputs.
To find these asymptotes, you simply set the denominator of the function equal to zero and solve for \( x \). Take the function \( y = \frac{x^5}{x^3 - 1} \) for example. To find the vertical asymptotes, solve \( x^3 - 1 = 0 \), resulting in \( x = 1 \). This means the graph has a vertical asymptote at \( x = 1 \).
When graphing, vertical asymptotes are important as they guide the viewer in understanding where the rational function will not be defined.

Intercepts are critical points on the graph of a function where it crosses the axes. In the context of rational functions, finding intercepts helps in understanding the function's graph. There are two types of intercepts: \( x \)-intercepts and \( y \)-intercepts.

• \( x \)-Intercepts: These occur where the graph crosses the \( x \)-axis. To find \( x \)-intercepts, you set the numerator equal to zero because the output (\( y \)-value) needs to be zero for the graph to touch the \( x \)-axis. For the function \( y = \frac{x^5}{x^3 - 1} \), this gives \( x^5 = 0 \), so the \( x \)-intercept is at \( (0,0) \).
• \( y \)-Intercepts: These occur where the graph crosses the \( y \)-axis (i.e., when \( x = 0 \)). Substituting \( x = 0 \) into the function yields \( y = \frac{0^5}{0^3 - 1} = 0 \), so the \( y \)-intercept is also at \( (0,0) \).

Understanding the intercepts is crucial for sketching an accurate graph of the function.

Local extrema refer to the local highest and lowest points on the graph of a function, commonly known as local maxima and minima. To find these points in a rational function, calculus and specifically derivatives, are used.
Begin by finding the derivative of the rational function using the quotient rule, which states that for a function \( y = \frac{u}{v} \), the derivative is \( y' = \frac{v u' - u v'}{v^2} \). Applying this to our function \( y = \frac{x^5}{x^3 - 1} \), we find the derivative: \[ y' = \frac{(x^3-1)5x^4 - x^5(3x^2)}{(x^3-1)^2} = \frac{-2x^7 - 5x^4}{(x^3-1)^2} \].
Set \( y' = 0 \) to find local extrema. Solving leads to \( x^4(-2x^3 - 5)=0 \), which provides an extremum at \( x = 0 \). A second derivative test confirms this is a point of local extremum.

Polynomial division, specifically long division, is a process that divides a polynomial by another polynomial. It's particularly useful for rational functions like \( y = \frac{x^5}{x^3 - 1} \) to find a polynomial that approximates their behavior at extreme values of \( x \).
The main goal is to simplify the rational function into something more manageable, typically capturing the end behavior of the original function. For our function, perform long division by dividing \( x^5 \) by \( x^3 - 1 \). Start with dividing the leading term \( x^5 \) by \( x^3 \), resulting in \( x^2 \). Multiply \( x^2 \) by \( x^3 - 1 \) to get \( x^5 - x^2 \), subtract this from the original to find the remainder, which is \( x^2 \).
This division reveals that as \( x \) grows very large, the rational function resembles \( x^2 \), guiding us in graphing and understanding the end behavior of the original function.