Problem 61 - Algebra and Trigonometry

Question

Find the vertical asymptotes, if any, of the graph of $$ f(x)=\frac{3 x^{2}}{(x-3)(x+1)} $$

Step-by-Step Solution

Answer

The vertical asymptotes are at \(x = 3\) and \(x = -1\).

Step 1: Identify Denominator

To find the vertical asymptotes of the function, start by identifying where the denominator equals zero. The function given is \[ f(x) = \frac{3x^2}{(x-3)(x+1)} \] The vertical asymptotes occur where the denominator is zero, but the numerator is not zero.

Step 2: Set Denominator to Zero

Set the denominator equal to zero to find the potential vertical asymptotes: \[ (x-3)(x+1) = 0 \] Solve for x: \[ x - 3 = 0 \quad \text{or} \quad x + 1 = 0 \] \[ x = 3 \quad \text{or} \quad x = -1 \]

Step 3: Confirm Asymptotes

Make sure the numerator does not cancel out these values. Since the numerator \(3x^2\) is never zero for \(x = 3\) and \(x = -1\), both these values confirm vertical asymptotes.

Key Concepts

rational functions

Rational functions are mathematical expressions that involve the ratio of two polynomials. They are written in the form \[ f(x) = \frac{P(x)}{Q(x)} \], where \(P(x)\) and \(Q(x)\) are polynomials. In the given exercise, the function provided is \[f(x) = \frac{3x^{2}}{(x-3)(x+1)} \]. The numerator here is \[3x^2 \] and the denominator is \[ (x-3)(x+1) \]. Rational functions are important in algebra and calculus because they often exhibit interesting properties, such as asymptotes and discontinuities. Understanding how to manipulate and analyze these functions can provide deeper insights into their behavior.

function asymptotes

Asymptotes are lines that a graph of a function approaches but never actually touches. In this exercise, the focus is on vertical asymptotes. Vertical asymptotes occur when the function approaches infinity or negative infinity as \(x\) approaches a certain value. For the given rational function, the vertical asymptotes are determined by examining the values that make the denominator zero. Vertical asymptotes indicate a point where the function is undefined and tends to become extremely large or small. They provide crucial information about the behavior of the function near certain values of \(x\).

denominator zero

To find the vertical asymptotes of a rational function, you need to identify where the denominator equals zero. This is because the function is undefined at these points. For the given function \[ f(x) = \frac{3x^2}{(x-3)(x+1)} \], the values of \(x\) that make the denominator zero are determined by solving the equation \[ (x-3)(x+1) = 0 \]. Solving this, we get \[ x = 3 \] and \[ x = -1 \]. Since the numerator does not cancel out these values, these points confirm the presence of vertical asymptotes. Always remember that the vertical asymptotes are found where the denominator is zero, but the numerator is not zero.