Problem 16 - A Graphical Approach to College Algebra

Question

Give the equations of any vertical, horizontal, or oblique asymptotes for the graph of each rational function. State the domain of \(f .\) $$f(x)=\frac{3 x^{2}-6 x-24}{5 x^{2}-26 x+5}$$

Step-by-Step Solution

Answer

Vertical asymptotes: \(x = 5\) and \(x = 0.2\). Horizontal asymptote: \(y = \frac{3}{5}\). Domain: \(x \in \mathbb{R}, x \neq 5, x \neq 0.2\).

Step 1: Find Vertical Asymptotes

To determine the vertical asymptotes, set the denominator equal to zero and solve for \(x\). The denominator of the function is \[5x^2 - 26x + 5 = 0\]Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) for the equation:\[a = 5, \; b = -26, \; c = 5\]Substituting in the values, we have:\[x = \frac{26 \pm \sqrt{(-26)^2 - 4 \, \times \, 5 \, \times \, 5}}{2 \, \times \, 5}\]\[x = \frac{26 \pm \sqrt{676 - 100}}{10}\]\[x = \frac{26 \pm \sqrt{576}}{10}\]\[x = \frac{26 \pm 24}{10}\]So the roots are:\[x = \frac{50}{10} = 5 \quad \text{and} \quad x = \frac{2}{10} = 0.2\]Thus, the vertical asymptotes are \(x = 5\) and \(x = 0.2\).

Step 2: Find Horizontal or Oblique Asymptotes

To find the horizontal or oblique asymptotes, compare the degrees of the numerator and denominator.- The degree of the numerator is 2 (because of \(3x^2\)).- The degree of the denominator is 2 (because of \(5x^2\)).Since the degrees are equal, the horizontal asymptote is determined by dividing the leading coefficients of the numerator and denominator.The leading coefficients are 3 and 5, respectively. Therefore:\[y = \frac{3}{5}\]This means the horizontal asymptote is \(y = \frac{3}{5}\).

Step 3: Identify the Domain

The domain of \(f(x)\) is all real numbers except where the denominator is zero (the locations of the vertical asymptotes). Since the denominator is zero at \(x = 5\) and \(x = 0.2\), the domain of \(f(x)\) is:\[x \in \mathbb{R}, \; x eq 5, \; x eq 0.2\]

Key Concepts

Vertical Asymptotes

Vertical asymptotes tell us where the function is undefined, and they usually appear where the denominator of a rational function equals zero. To find them, we simply set the denominator equal to zero and solve for the variable.
For our function, the denominator is given by \[5x^2 - 26x + 5 = 0\]. Using the quadratic formula, we input the coefficients: \(a = 5\), \(b = -26\), and \(c = 5\) into the formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
Solving this gives us the solutions \(x = 5\) and \(x = 0.2\). These solutions are the vertical asymptotes because they make the denominator zero, and the function thus becomes undefined at these x-values.

• Vertical asymptotes mark the points of discontinuity on the graph.
• They indicate the locations where the function spikes to infinity or negative infinity.

Horizontal Asymptotes

Horizontal asymptotes illustrate the behavior of a rational function as \(x\) approaches infinity or negative infinity. To find the horizontal asymptote, we compare the degrees of the polynomial in the numerator and the polynomial in the denominator.
If the degrees are the same, as in our function, the horizontal asymptote is determined by the ratio of the leading coefficients.
In our case, both the numerator and the denominator have a degree of 2. Thus, we divide the leading coefficient of the numerator, 3, by the leading coefficient of the denominator, 5, to find the horizontal asymptote:
\[y = \frac{3}{5}\]
This means the graph of the function approaches \(y = \frac{3}{5}\) as \(x\) becomes very large or very small.

• Horizontal asymptotes provide a "guideline" towards which the function's graph tends for extreme values of \(x\).
• They hint at the long-term behavior of the function.

Domain of a Function

The domain of a function is the complete set of all possible input values (usually \(x\) values) that allow the function to be defined and produce a valid output. For rational functions, the domain includes all real numbers except those that make the denominator zero.
From our vertical asymptote analysis, we know the denominator is zero at \(x = 5\) and \(x = 0.2\). Therefore, these values must be excluded from the domain.
Thus, the domain of our given function is expressed as:

• All real numbers except \(x = 5\) and \(x = 0.2\).
• Mathematically, this can be represented as: \(x \in \mathbb{R}, \; x eq 5, \; x eq 0.2\).

The domain is crucial because it tells us where the function can be utilized and predicts potential undefined points in the graph.