Problem 39 - Algebra 2

Question

Graph each rational function. $$ f(x)=\frac{3}{(x-1)(x+5)} $$

Step-by-Step Solution

Answer

Vertical asymptotes at \( x = 1 \) and \( x = -5 \); horizontal asymptote at \( y = 0 \).

Step 1: Identify the Vertical Asymptotes

To find the vertical asymptotes of the function \( f(x) = \frac{3}{(x-1)(x+5)} \), set the denominator equal to zero: \( (x-1)(x+5) = 0 \). This gives us \( x = 1 \) and \( x = -5 \). Thus, the vertical asymptotes are at \( x = 1 \) and \( x = -5 \).

Step 2: Determine the Horizontal Asymptote

The function \( f(x) = \frac{3}{(x-1)(x+5)} \) is of the form \( \frac{p(x)}{q(x)} \). Since the degree of the numerator (0, constant 3) is less than the degree of the denominator (2), the horizontal asymptote is \( y = 0 \).

Step 3: Identify the Y-Intercept

The y-intercept of the function is found by evaluating \( f(0) \). Substitute \( x = 0 \) into the function: \[ f(0) = \frac{3}{(0-1)(0+5)} = \frac{3}{-5} = -\frac{3}{5}. \] This means the y-intercept is \( (0, -\frac{3}{5}) \).

Step 4: Find Critical Points and Behavior

Examine the behavior near the asymptotes. Calculate additional points around \( x = 1 \) and \( x = -5 \) to see how the graph behaves. For example, evaluate \( f(2) \), \( f(-6) \), \( f(-4) \), etc., to assist in sketching the graph.

Step 5: Graph the Function

Now, plot the vertical asymptotes at \( x = 1 \) and \( x = -5 \), the horizontal asymptote at \( y = 0 \), and the y-intercept at \( (0, -\frac{3}{5}) \). Use the additional points to sketch the curve, ensuring the graph approaches the asymptotes.

Key Concepts

Vertical Asymptotes

Vertical asymptotes are vertical lines on a graph that a function approaches but never actually reaches. They occur where the denominator of a rational function is zero and the numerator is not zero. For the given function \( f(x) = \frac{3}{(x-1)(x+5)} \), set the denominator equal to zero to find vertical asymptotes:

• Find the zeros of the denominator: \((x-1)(x+5) = 0\).
• Solve for \( x \) to get \( x = 1 \) and \( x = -5 \).

These are the points where the vertical asymptotes occur, at \( x = 1 \) and \( x = -5 \). As \( x \) approaches these values, \( f(x) \) will increase or decrease without bound. Watch how the function behaves moving toward these lines.

Horizontal Asymptote

Horizontal asymptotes describe the behavior of a function as \( x \) approaches infinity or negative infinity. They tell us the value that the function approaches but does not reach as \( x \) becomes extremely large or small.
With \( f(x) = \frac{3}{(x-1)(x+5)} \), we compare the degrees of the numerator and denominator:

• The numerator, a constant \( 3 \), has degree 0.
• The denominator \((x-1)(x+5)\), expanded to \(x^2 + 4x - 5\), has degree 2.

Since the degree of the numerator is less than that of the denominator, the horizontal asymptote is \( y = 0 \). This means as \( x \) goes to infinity or negative infinity, the value of \( f(x) \) approaches 0.

Y-Intercept

The y-intercept of a function is the point where the graph crosses the y-axis. To find the y-intercept, set \( x = 0 \) in the function and solve for \( f(x) \).
For \( f(x) = \frac{3}{(x-1)(x+5)} \), substitute \( x = 0 \):

• Calculate \( f(0) = \frac{3}{(0-1)(0+5)} = \frac{3}{-5} = -\frac{3}{5} \).

Thus, the y-intercept of the function is at \( (0, -\frac{3}{5}) \). This provides a starting point for sketching the graph, along with the asymptotes.

Critical Points

Critical points of a function are where the derivative is zero or undefined. They are crucial for understanding the behavior of the function through its increasing and decreasing intervals and local maxima and minima.
While finding critical points typically involves calculus, an intuition-based approach can help here without derivatives. Examine the function's behavior near the vertical asymptotes and y-intercept:

• Calculate \( f(x) \) at points around the asymptotes, like \( x = 2 \), \( x = -6 \), and \( x = -4 \), to see how the graph behaves.

This helps predict the curve's approach towards the asymptotes and interconnection with known points. Critical values are crucial for sketching a reasonable graph of the function.