Problem 5 - Algebra and Trigonometry

Question

The following questions are about the rational function $$ I(x)=\frac{(x+1)(x-2)}{(x+2)(x-3)} $$ The function \(r\) has vertical asymptotes \(x=\) ________ and \(\boldsymbol{X}=\) ______

Step-by-Step Solution

Answer

The function has vertical asymptotes at \(x = -2\) and \(x = 3\).

Step 1: Understanding Vertical Asymptotes

Vertical asymptotes occur where the function is undefined due to division by zero in the rational function. This means you need to look for values of \(x\) that make the denominator equal to zero.

Step 2: Identify the Denominator

For the given function \(I(x) = \frac{(x+1)(x-2)}{(x+2)(x-3)}\), the denominator is \((x+2)(x-3)\). We need to find the values of \(x\) that make \((x+2)(x-3) = 0\).

Step 3: Setting the Denominator to Zero

Set each factor of the denominator equal to zero to find the vertical asymptotes. Solve the equations \(x+2 = 0\) and \(x-3 = 0\) for \(x\).

Step 4: Solve for Vertical Asymptotes

Solving \(x+2=0\) gives \(x = -2\). Solving \(x-3=0\) gives \(x = 3\). These are the points where the function has vertical asymptotes.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are a fundamental concept in rational functions. They occur at specific values of \(x\) where the function approaches infinity or negative infinity. In simple terms, a vertical asymptote is a vertical line at a particular \(x\)-value where the function cannot produce a real-number output. For a rational function like \(I(x) = \frac{(x+1)(x-2)}{(x+2)(x-3)}\), vertical asymptotes exist where the denominator equals zero, provided these points do not also cancel out the numerator. This results in the function being undefined at those \(x\)-values. The process involves finding values of \(x\) which make the denominator zero, while ensuring these points are not cancelled by corresponding zero points in the numerator.

Denominator Analysis

The denominator of a rational function is key to understanding when and where vertical asymptotes occur. In the given function \(I(x) = \frac{(x+1)(x-2)}{(x+2)(x-3)}\), our primary focus is on the expression \((x+2)(x-3)\).

• Look at each factor within the denominator.
• Set each factor equal to zero separately.
• This will help locate the \(x\)-values that make the function undefined.

For \((x+2)\), the function is undefined when \(x = -2\).For \((x-3)\), the function is undefined when \(x = 3\).These \(x\)-values \((-2\) and \(3\)) are where the vertical asymptotes appear in the function. It is crucial to remember these steps because a different function would require fresh analysis, considering its unique denominator factors.

Division by Zero

Division by zero is a common pitfall in algebra that leads to undefined expressions. In rational functions, this occurs when the denominator becomes zero while the numerator remains non-zero, indicating a potential vertical asymptote. In \(I(x) = \frac{(x+1)(x-2)}{(x+2)(x-3)}\), when the denominator \((x+2)(x-3)\) equals zero, the function cannot compute a result for that particular \(x\)-value.This means:

• If \((x+2) = 0\), then \(x = -2\). Here, division by zero makes the function undefined at \(x = -2\), creating a vertical asymptote.
• Similarly, \((x-3) = 0\) leads to \(x = 3\), another point of division by zero, and thus a vertical asymptote exists at \(x = 3\).

Understanding these points helps prevent errors in calculations, as recognizing when and why a function is undefined due to division by zero is essential in higher-level math topics.