How do you find a hole in a rational function graph?

Question:

How do you find a hole in a rational function graph?

Rational Functions:

A rational function is a function that is a ratio of two polynomials, i.e. of the form

{eq}\begin{align*} f (x) ****= \frac{P (x)}{Q (x)} \end{align*} {/eq}

where {eq}P {/eq} and {eq}Q {/eq} are both polynomials. Note that this function is undefined whenever {eq}Q = 0 {/eq} and in those special places where part of the denominator cancels part of the numerator, we will find holes in the graph.

Answer and Explanation:

It is important to know that when we graph a function, holes remain hidden! This is because the hole is infinitesimally small, and is easily glossed over by our graphing utilities. For this reason it is important to check for holes. To do so for a rational function we need to factor the numerator and denominator see if it anything can "cancel out" (we use quotes because we need to be very careful when we simplify). If it does, then we have a hole in the function. The easiest function to see this in is probably {eq}f (x) = \frac{x^2}x {/eq}. Note that {eq}x \neq 0 {/eq}. And if it does not, then {eq}f (x) = x {/eq}. And so the function {eq}f (x) = \frac{x^2}x {/eq} is the line {eq}f (x) = x {/eq} with a hole in it at {eq}x = 0 {/eq}.

More generally if we have a rational function of the form

{eq}\begin{align*} f (x) &= \frac{ax^2 + bx + c}{x+r} \end{align*} {/eq}

where we can write {eq}ax^2 + bx + c = (x+r)(x+s) {/eq}, then

{eq}\begin{align*} f (x) &= \frac{ax^2 + bx + c}{x+r} \end{align*} {/eq}

is actually the line

{eq}\begin{align*} f (x) &= \frac{ax^2 + bx + c}{x+r} \\ &= \frac{ (x+r)(x+s) }{x+r} \\ &= x+s \end{align*} {/eq}

with a hole at {eq}x = r {/eq}.

This can basic idea can be extended to any degree polynomial to find all holes.


Learn more about this topic:

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Rational Function: Definition, Equation & Examples

from GMAT Prep: Help and Review

Chapter 10 / Lesson 11
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