Problem 1 - Calculus Early Transcendentals

Question

Give a graph of the rational function and label the coordinates of the stationary points and inflection points. Show the horizontal and vertical asymptotes and label them with their equations. Label point(s), if any, where the graph crosses a horizontal asymptote. Check your work with a graphing utility. $$ \frac{2 x-6}{4-x} $$

Step-by-Step Solution

Answer

Vertical asymptote at \(x = 4\), horizontal asymptote at \(y = -2\), no stationary or inflection points.

Step 1: Identify Vertical Asymptotes

The vertical asymptotes occur where the denominator of the function is zero. For the function \( f(x) = \frac{2x - 6}{4 - x}\), set the denominator equal to zero: \(4 - x = 0\). Solve for \(x\), giving the vertical asymptote at \(x = 4\).

Step 2: Identify Horizontal Asymptotes

To find the horizontal asymptote, evaluate the degrees of the numerator and the denominator. Both are linear (degree 1), thus take the ratio of the leading coefficients. The leading coefficient in the numerator is 2, and the leading coefficient in the denominator is -1. So, the horizontal asymptote is \(y = -2\).

Step 3: Find Stationary Points

Stationary points are found where the derivative is zero. First, find the derivative \( f'(x) \) using the quotient rule: \[ f'(x) = \frac{(4-x)(2) - (2x-6)(-1)}{(4-x)^2} \] Simplify to find the critical points: \[ f'(x) = \frac{8 - 2x + 2x - 6}{(4-x)^2} = \frac{2}{(4-x)^2} \]Setting \(f'(x) = 0\) gives no solution, as the numerator (2) is never zero. Thus, there are no stationary points.

Step 4: Find Inflection Points

Inflection points occur where the second derivative changes sign. Differentiate \(f'(x)\) to find \(f''(x)\): \[ f''(x) = \frac{d}{dx}\left( \frac{2}{(4-x)^2} \right) = \frac{d}{dx}[2(4-x)^{-2}]\]Using the power rule, find: \[ f''(x) = 2\cdot(-2)(4-x)^{-3}\cdot(-1) = \frac{4}{(4-x)^3} \] This is always positive, indicating no change in concavity and thus no inflection points.

Step 5: Intersections with Horizontal Asymptote

Check if and where the curve intersects the horizontal asymptote \(y = -2\). Set \(\frac{2x - 6}{4-x} = -2\) and solve: \[ 2x - 6 = -2(4-x) \]\[ 2x - 6 = -8 + 2x \]\[ -6 = -8 \] This is incorrect, so the curve never intersects the horizontal asymptote.

Step 6: Graph the Function

Graph \(f(x) = \frac{2x - 6}{4-x}\). Label the vertical asymptote \(x = 4\) and the horizontal asymptote \(y = -2\). Indicate that there are no stationary or inflection points. Verify using a graphing utility.

Key Concepts

Vertical Asymptotes

In the context of rational functions, vertical asymptotes are lines that the graph of the function approaches but never touches or crosses. These occur at the values of the variable that cause the denominator of the function to be zero. For the given function \( f(x) = \frac{2x - 6}{4 - x} \), we find the vertical asymptote by setting the denominator equal to zero:

• Solve \( 4 - x = 0 \)
• Find \( x = 4 \)

This tells us that the vertical asymptote is at \( x = 4 \). This means that as \( x \) approaches 4, the value of \( f(x) \) increases or decreases without bound, which is graphically represented by a vertical line on the graph of the function.

Horizontal Asymptotes

Horizontal asymptotes indicate the behavior of the function as \( x \) approaches infinity or negative infinity. For a rational function like \( f(x) = \frac{2x - 6}{4 - x} \), calculating the horizontal asymptote involves comparing the degrees of the numerator and the denominator. In this case:

• The numerator \( 2x - 6 \) is degree 1
• The denominator \( 4 - x \) is also degree 1

Since the degrees are equal, the horizontal asymptote is determined by the ratio of the leading coefficients:

• The leading coefficient of the numerator is 2
• The leading coefficient of the denominator is \(-1\)
• The horizontal asymptote is therefore \( y = \frac{2}{-1} = -2 \).

This means that as \( x \) goes to infinity or negative infinity, the function approaches the line \( y = -2 \). The graph gets closer to this line but never actually meets it.

Stationary Points

Stationary points are where the slope of the function is zero, i.e., where the derivative equals zero. These points often indicate local maximums, minimums, or points of inflection on the graph. To find stationary points for \( f(x) = \frac{2x - 6}{4 - x} \), we calculate the derivative using the quotient rule:
The derivative is given by \( f'(x) = \frac{2}{(4-x)^2} \). Setting this equal to zero gives:

• \( \frac{2}{(4-x)^2} = 0 \)
• The numerator is constant and not zero, hence no solution.

This tells us that there are no real stationary points because the derivative does not equal zero for any real \( x \). Thus, the graph does not have any peak or trough where the slope becomes zero.

Inflection Points

Inflection points occur where the graph changes concavity. To find these points, we take the second derivative and find where it changes sign. For our function, starting from \( f'(x) = \frac{2}{(4-x)^2} \), we calculate the second derivative:
Using the power rule and chain rule, we find \( f''(x) = \frac{4}{(4-x)^3} \).

• This derivative \( f''(x) \) is always positive for all real values of \( x \).
• As it does not change sign, it indicates a constant concavity, without a switch from concave up to concave down or vice versa.

This means there are no inflection points in the function \( f(x) = \frac{2x - 6}{4-x} \), as the graph maintains the same curvature throughout its domain.