Problem 470 - Precalculus

Question

For the following exercises, find the domain, vertical asymptotes, and horizontal asymptotes of the functions. $$f(x)=\frac{x^{2}-1}{x^{3}+9 x^{2}+14 x}$$

Step-by-Step Solution

Answer

Domain: All real numbers except \(x=0, -2, -7\). Vertical asymptotes: \(x=0, -2, -7\). Horizontal asymptote: \(y=0\).

Step 1: Find the Domain

The domain of a function is all the values of \(x\) for which the function is defined. In this case, \(f(x)=\frac{x^{2}-1}{x^{3}+9x^{2}+14x}\) is defined except where the denominator is zero. So, set the denominator \(x^{3}+9x^{2}+14x=0\) to find these points. Factor the denominator: \(x(x^{2}+9x+14)=0\). Solving gives \(x=0\) or \(x^{2}+9x+14=0\). Solving \(x^{2}+9x+14=0\) using the quadratic formula, we find the roots \(x=-2\) and \(x=-7\). Thus, the domain is all real numbers except \(x=0, -2, -7\).

Step 2: Identify Vertical Asymptotes

Vertical asymptotes occur where the denominator is zero (i.e., the function is undefined) but the numerator is not zero. We already found the critical points where the denominator is zero: \(x=0, -2, -7\). Check if the numerator \(x^{2}-1\) is zero at these points. \(x^{2}-1=0\) implies \(x=\pm1\). Since none of these occur at \(x=0, -2, -7\), there are vertical asymptotes at \(x=0, -2, -7\).

Step 3: Determine Horizontal Asymptotes

Horizontal asymptotes are found by comparing the degrees of the polynomial in the numerator and denominator. The degree of the numerator \(x^2-1\) is 2, and the degree of the denominator \(x^3+9x^2+14x\) is 3. Since the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y=0\).

Key Concepts

Domain of a Function

The domain of a function refers to all possible input values (typically denoted as \(x\)) for which the function is defined. In the case of rational functions, such as \(f(x) = \frac{x^2 - 1}{x^3 + 9x^2 + 14x}\), the function can become undefined if the denominator equals zero. Rational functions are typically undefined at points where the denominator is zero because division by zero is impossible in mathematics.

To find these points, set the denominator equal to zero and solve the resulting equation:

• Factor: \(x(x^2 + 9x + 14) = 0\).
• Solve: This gives the equation \(x = 0\) or \(x^2 + 9x + 14 = 0\).
• Quadratic Solution: Solving \(x^2 + 9x + 14 = 0\) using the quadratic formula yields roots \(x = -2\) and \(x = -7\).

Thus, the domain of the function excludes these points, so it is all real numbers except \(x = 0, -2, -7\). These excluded values indicate that the function is not defined at these points.

Vertical Asymptotes

Vertical asymptotes in a rational function occur at specific points where the function goes to infinity. These are places where the denominator equals zero but the numerator does not, causing the function's value to grow infinitely large or small.

From the factorization of the denominator \(x(x^2 + 9x + 14)\), potential vertical asymptotes are at \(x = 0, -2,\) and \(-7\). We need to ascertain that the numerator \(x^2 - 1\) does not zero out at these points:

• Substitute \(x = 0\) into the numerator: \(0^2 - 1 = -1\) (not zero).
• Substitute \(x = -2\) into the numerator: \((-2)^2 - 1 = 3\) (not zero).
• Substitute \(x = -7\) into the numerator: \((-7)^2 - 1 = 48\) (not zero).

Since the numerator equals zero only at \(x = \pm 1\) and not at \(x = 0, -2, -7\), these values of \(x\) are indeed vertical asymptotes of the function. This means the graph of the function will shoot up to positive or negative infinity as \(x\) approaches these values.

Horizontal Asymptotes

Horizontal asymptotes occur in the long-run behavior of a function as \(x\) approaches infinity or negative infinity. To find horizontal asymptotes of rational functions, we often compare the degrees of the numerator and the denominator.

For \(f(x) = \frac{x^2 - 1}{x^3 + 9x^2 + 14x}\), the numerator has a degree of 2 (\(x^2\)) while the denominator has a degree of 3 (\(x^3\)). Here are the typical scenarios and what they mean:

• If the degree of the numerator is less than the degree of the denominator, the horizontal asymptote is \(y = 0\).
• If the degrees are equal, the horizontal asymptote would be at \(y = \frac{leading\ coefficient\ of\ numerator}{leading\ coefficient\ of\ denominator}\).
• If the numerator has a higher degree, there is no horizontal asymptote.

In our function, since the degree of the numerator (2) is less than that of the denominator (3), the horizontal asymptote is \(y = 0\). This signifies that as \(x\) moves towards positive or negative infinity, the function value \(f(x)\) will approach 0.