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Chapter 3: Problem 26

Find the vertical asymptotes, if any, and the values of \(x\) corresponding to holes, if any, of the graph of each rational function. $$h(x)=\frac{x}{x(x-3)}$$

Short Answer

Expert verified
For the function \(h(x)=\frac{x}{x(x-3)}\), there is one hole at \(x=0\) and one vertical asymptote at \(x=3\).

Step by step solution

01

Factor the Denominator

Start by factoring out the equation. \(h(x) = \frac{x}{x(x-3)}\). This leaves us with the denominator in the factored form \(x(x-3)\).
02

Identify the Undefined Points

Set each factor in the denominator equal to zero and solve for \(x\). This will give the values of \(x\) for which the function is undefined. For \(x\), it is 0. For \(x-3\), by adding 3 to both sides it can be found that \(x=3\). The function is therefore undefined at \(x=0\) and \(x=3\).
03

Identify Asymptotes or Holes

A value will provide a hole in the function if, once factored, it cancels out with a factor in the numerator. If it does not cancel out, it will provide a vertical asymptote. Here, the factor \(x\) in the denominator cancels with the \(x\) in the numerator, hence \(x=0\) is a hole, not an asymptote. However, the factor \(x-3\) does not cancel out, thus \(x=3\) is a vertical asymptote.

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