Problem 12 - Precalculus : Building Concepts and Connections

Question

Find the domain and the vertical and horizontal asymptotes (if any). $$h(x)=\frac{-3 x^{2}+12}{x^{2}-9}$$

Step-by-Step Solution

Answer

The domain of the function is all real numbers except 3 and -3. The vertical asymptotes are at \( x=-3 \) and \( x=3 \), and the horizontal asymptote is \( y=-3 \).

Step 1: Determine the domain

The domain of a function consists of all the real number values of \( x \) for which the function is defined. In the given function, the denominator cannot be equal to zero (as it would make the function undefined). Setting the denominator equal to zero, we solve the following equation to find the values \( x \) can't take: \(x^{2}-9=0\). Factoring gives \( (x-3)(x+3)=0 \), so \( x \) can't be 3 or -3. Therefore, the domain of \( h(x) \) is all real numbers except 3 and -3.

Step 2: Find the vertical asymptotes

Vertical asymptotes occur at the values of \( x \) that are not included in the domain. From Step 1, these are at \( x=-3 \) and \( x=3 \). Therefore, the vertical asymptotes of \( h(x) \) are at \( x=-3 \) and \( x=3 \).

Step 3: Find the horizontal asymptote

To find the horizontal asymptote, you compare the degrees of the polynomials in the numerator and the denominator. Here, both are of degree 2, so you take the ratio of the leading coefficients, which in this case is \(-3/1 = -3\). So, the horizontal asymptote is \( y=-3 \).

Key Concepts

Domain of a Function

The domain of a function is critical because it tells us the set of all possible inputs (or real number values of \( x \)) for which the function is defined. Put simply, these are the values that won't cause the function to "blow up," like dividing by zero. In the function \( h(x)=\frac{-3x^2+12}{x^2-9} \), we need to ensure the denominator is never zero because division by zero is undefined in mathematics.

For this function, set the denominator equal to zero and solve: \( x^2 - 9 = 0 \). Factoring gives \((x-3)(x+3)=0\), leading to the solutions \( x = 3 \) and \( x = -3 \). Thus, \( x \) cannot be 3 or -3. The domain of \( h(x) \) is therefore all real numbers except 3 and -3. Visually, you can think of this as a number line where there are "gaps" at 3 and -3.

• Domain: All real numbers \( x \), except \( x = 3 \) and \( x = -3 \).

Vertical Asymptotes

Vertical asymptotes are the lines \( x = a \) where the function grows infinitely large or drops down past any limit as it approaches the line \( x = a \) from either side. This often happens where the function's denominator equals zero and flips the function's value to infinity or negative infinity. In our function \( h(x)=\frac{-3x^2+12}{x^2-9} \), we found that the denominator becomes zero at \( x = 3 \) and \( x = -3 \).

Thus, these are the points where vertical asymptotes occur, as the function cannot be defined at these \( x \)-values. So when you approach \( x = 3 \) or \( x = -3 \) from the left or right, \( h(x) \) tends to ±∞.

• Vertical Asymptotes: \( x = 3 \) and \( x = -3 \).

Horizontal Asymptotes

Horizontal asymptotes tell us the behavior of a function as \( x \) approaches infinity or negative infinity. Unlike vertical asymptotes, horizontal asymptotes are about long-term behavior. For rational functions like \( h(x)=\frac{-3x^2+12}{x^2-9} \), they can be found by comparing the degrees of the polynomial in the numerator and the denominator.

Here, the degrees are both 2, so we take the ratio of the leading coefficients (the highest-degree coefficients), which are -3 from the numerator and 1 from the denominator, giving us -3. This tells us that as \( x \) goes to infinity or negative infinity, \( h(x) \) approaches \( y = -3 \).

• Horizontal Asymptote: \( y = -3 \).