Problem 62 - A Graphical Approach to College Algebra

Question

Sketch a graph of each rational function. Your graph should include all asymptotes. Do not use a calculator. $$f(x)=\frac{x^{2}-5}{x-3}$$

Step-by-Step Solution

Answer

Domain: all \(x \neq 3\); Vertical Asymptote: \(x = 3\); Oblique Asymptote: \(y = x + 3\); Intercepts: (0, \(\frac{5}{3}\)), (\(\sqrt{5}\), 0), and (\(-\sqrt{5}\), 0).

Step 1: Determine the Domain

The function \(f(x) = \frac{x^2 - 5}{x - 3}\) is undefined where the denominator equals zero. Set \(x - 3 = 0\) and solve for \(x\). Thus, \(x = 3\) is not in the domain. So, the domain of \(f(x)\) is all real numbers except \(x = 3\).

Step 2: Find the Vertical Asymptote

Vertical asymptotes occur where the denominator is zero, provided the numerator is not zero at that point. Since \(x = 3\) makes the denominator zero and the numerator \((3^2 - 5 = 4)\), \(x = 3\) is a vertical asymptote.

Step 3: Identify Horizontal Asymptote

For large values of \(x\), the behavior of \(f(x)\) depends on the leading term of the numerator and denominator. The degree of the numerator \(x^2\) is greater than the degree of the denominator \(x\), hence, there is no horizontal asymptote.

Step 4: Find Oblique Asymptote

Since the degree of the numerator is one more than the degree of the denominator, perform polynomial long division to find the oblique asymptote. Dividing \(x^2 - 5\) by \(x - 3\), we get \(x + 3\) as the quotient. Thus, the oblique asymptote is the line \(y = x + 3\).

Step 5: Find Intercepts

Calculate the y-intercept by substituting \(x = 0\): \(f(0) = \frac{0^2 - 5}{0 - 3} = \frac{-5}{-3} = \frac{5}{3}\). For x-intercepts, set the numerator equal to zero: \(x^2 - 5 = 0\). Solving gives \(x = \pm\sqrt{5}\). Thus, intercepts are (0, \(\frac{5}{3}\)), (\(\sqrt{5}\), 0), and (\(-\sqrt{5}\), 0).

Step 6: Sketch the Graph

Plot the vertical asymptote at \(x = 3\), the oblique asymptote \(y = x + 3\), and the intercepts. Sketch the behavior of the function near the asymptotes, ensuring that it approaches the asymptotes as \(x\) tends to positive or negative infinity.

Key Concepts

Domain

The domain of a function describes all the possible input values (x-values) that will yield a valid output (y-value). In rational functions, which are ratios of polynomials, the domain is determined by identifying the values of x that make the denominator zero, as division by zero is undefined.

For the rational function \(f(x) = \frac{x^2 - 5}{x - 3}\), we set the denominator \(x - 3\) equal to zero: \(x - 3 = 0\). Solving this gives \(x = 3\). This means that the function is not defined at \(x = 3\). Thus, the domain consists of all real numbers except \(x = 3\).

If you were to graph this, there would be a gap or an "undefined point" at \(x = 3\), so no part of the graph can cross this line.

Vertical Asymptotes

Vertical asymptotes give us information on where the function increases or decreases without bound (heads to infinity or negative infinity) as it approaches a specific x-value.

To find vertical asymptotes in rational functions, we look for the values of x that make the denominator zero, without also making the numerator zero. This is because if both are zero, it indicates a different behavior, potentially a hole in the graph.

In our function \(f(x) = \frac{x^2 - 5}{x - 3}\), the denominator \(x - 3\) results in zero at \(x = 3\). Since the numerator \(x^2 - 5\) does not become zero when \(x = 3\) \((3^2 - 5 = 4)\), we can say confidently that \(x = 3\) is a vertical asymptote.

This means, on the graph, there will be a vertical line at \(x = 3\) which the function will approach, but never touch or cross, instead moving towards infinity or negative infinity as it nears this line.

Oblique Asymptotes

Oblique (or slant) asymptotes occur in rational functions where the degree of the polynomial in the numerator is exactly one more than the degree of the polynomial in the denominator.

For the function \(f(x) = \frac{x^2 - 5}{x - 3}\), the degree of the numerator \(x^2\) is two, and the degree of the denominator \(x\) is one. Because the numerator's degree is one more, we need to use polynomial long division to find the asymptote.

Carrying out the division of \(x^2 - 5\) by \(x - 3\) gives a quotient of \(x + 3\). Hence, the oblique asymptote is the line \(y = x + 3\). This line shows the direction in which the function progresses as the x-values grow large in the positive or negative direction.

Hence, on the graph, as x moves towards infinity or negative infinity, the curve of the rational function will closely approach the line \(y = x + 3\), but will never actually reach or overlap it.

Intercepts

Intercepts are points where the graph of the function crosses the x-axis and y-axis. These points help piece together the overall picture of the graph.

• **Y-intercept:** This occurs where the graph crosses the y-axis (where \(x=0\)). For \(f(x) = \frac{x^2 - 5}{x - 3}\), substitute \(x = 0\): \(f(0) = \frac{0^2 - 5}{0 - 3} = \frac{-5}{-3} = \frac{5}{3}\). So, the y-intercept is (0, \(\frac{5}{3}\)).


• **X-intercepts:** These happen where the graph crosses the x-axis (where \(f(x) = 0\)), which means we set the numerator equal to zero. Solving \(x^2 - 5 = 0\) gives \(x = \pm\sqrt{5}\). Thus, the x-intercepts are at the points (\(\sqrt{5}\), 0) and (\(-\sqrt{5}\), 0).

These intercepts are crucial as they indicate the exact points where the graph makes contact with the axis, helping to sketch the overall path and shape of the rational function's graph.