Question

State the characteristics of the graph of the function \(y=\frac{x}{x+2}+\frac{x-4}{x-2}\)

Short answer

Domain: all reals except \(x = -2\) and \(x = 2\). Vertical asymptotes at \(x = -2\) and \(x = 2\). No horizontal asymptotes. No critical points and no points of inflection.

Step-by-step solution

Identify the Domain

First, find the domain of the function by identifying values of x that make the denominator zero. Solve for when the denominators of both terms are equal to zero: - For the first term: \(x + 2 = 0 \implies x = -2\) - For the second term: \(x - 2 = 0 \implies x = 2\). Thus, the domain is all real numbers except \(x = -2\) and \(x = 2\).

Find Vertical Asymptotes

Vertical asymptotes occur where the function is undefined within its domain. From the previous step, these values are \(x = -2\) and \(x = 2\). Therefore, the function has vertical asymptotes at \(x = -2\) and \(x = 2\).

Determine Horizontal Asymptotes

To find horizontal asymptotes, analyze the behavior of the function as x goes to \(+\text{\textbackslash}infty\) or \( -\text{\textbackslash}infty\). Simplify the fractions: \( \lim_{{x \to \text{\textbackslash}infty}} \frac{x}{x+2} = \lim_{{x \to \text{\textbackslash}infty}} \frac{1}{1 + \frac{2}{x}} = 1\) and \( \lim_{{x \to -\text{\textbackslash}infty}} \frac{x}{x+2} = \lim_{{x \to -\text{\textbackslash}infty}} \frac{1}{1 + \frac{2}{x}} = 1\). Similarly, for the second fraction: \( \lim_{{x \to \text{\textbackslash}infty}} \frac{x-4}{x-2} = \lim_{{x \to \text{\textbackslash}infty}} \frac{1 - \frac{4}{x}}{1 - \frac{2}{x}} = 1\) and \( \lim_{{x \to -\text{\textbackslash}infty}} \frac{x-4}{x-2} = \lim_{{x \to -\text{\textbackslash}infty}} \frac{1 - \frac{4}{x}}{1 - \frac{2}{x}} = 1\). Hence, there are no horizontal asymptotes, and instead, the function values approach infinity.

Determine Critical Points

Set the first derivative of the function to zero to find critical points. Derivative calculation: \(y' = \left(\frac{d}{dx} \left(\frac{x}{x+2}\right) + \frac{d}{dx} \left(\frac{x-4}{x-2}\right)\right)\). Use the quotient rule to find: For \(f(x) =\frac{x}{x+2}\), \(f'(x) = \frac{(x + 2)(1) - x(1)}{(x+2)^2} = \frac{2}{(x+2)^2}\). For \(g(x) = \frac{x-4}{x-2}\), \(g'(x) = \frac{(x-2)(1) - (x-4)(1)}{(x-2)^2} = \frac{2}{(x-2)^2}\). Therefore, \(y' = \frac{2}{(x+2)^2} + \frac{2}{(x-2)^2}\). There are no critical points where the derivative equals zero.

Identify Points of Inflection

Set the second derivative to zero to find points of inflection. The process involves taking the second derivative: For \(\frac{d^2}{dx^2}\left(\frac{2}{(x+2)^2}\right)\) and \(\frac{d^2}{dx^2}\left(\frac{2}{(x-2)^2}\right)\), which simplifies to \(\frac{-4}{(x+2)^3} + \frac{-4}{(x-2)^3}\). Solve \( \frac{-4}{(x+2)^3} + \frac{-4}{(x-2)^3} = 0 \) to find no real points of inflection.

Key concepts

Domain of a Function

To find the domain of the function, we need to identify values of x that make the function undefined. These are values that make any denominator zero. For the given function, the denominators are zero when:

• For \ \( \frac{x}{x+2} \): \ \( x + 2 = 0 \implies x = -2 \).
• For \ \( \frac{x-4}{x-2} \): \ \( x - 2 = 0 \implies x = 2 \).

Thus, the domain of the function is all real numbers except \ \( x = -2 \) and \ \( x = 2 \). Make sure to avoid these values while graphing.

Vertical Asymptotes

Vertical asymptotes occur where the function becomes undefined within its domain. From our previous step, we identified such points:

• The first term, \ \( x + 2 = 0 \), gives \ \( x = -2\).
• The second term, \ \( x - 2 = 0 \), gives \ \( x = 2\).

Therefore, the function has vertical asymptotes at \ \( x = -2 \) and \ \( x = 2 \). These asymptotes are important to know because the graph will grow infinitely large as it approaches these lines both from the left and right.

Horizontal Asymptotes

Horizontal asymptotes describe the behavior of the function as \ \( x \) approaches \ \( \infty \) or \ \( -\infty \). Let's analyze the behavior of each fraction:

• \ \( \lim_ {x \to \infty} \frac{x}{x + 2} = \lim_ {x \to \infty} \frac{1}{1 + \frac{2}{x}} = 1 \) and \ \( \lim_ {x \to -\infty} \frac{x}{x + 2} = \lim_ {x \to -\infty} \frac{1}{1 + \frac{2}{x}} = 1 \).
• For the second fraction: \ \( \lim_ {x \to \infty} \frac{x - 4}{x - 2} = \lim_ {x \to \infty} \frac{1 - \frac{4}{x}}{1 - \frac{2}{x}} = 1 \) and \ \( \lim_ {x \to -\infty} \frac{x - 4}{x - 2} = \lim_ {x \to -\infty} \frac{1 - \frac{4}{x}}{1 - \frac{2}{x}} = 1 \).

In both cases, the fractions approach 1 as \ \( x \) approaches \ \( \pm \infty \), suggesting there are no horizontal asymptotes and the function values approach infinity.

Critical Points

Critical points are found where the first derivative of the function is zero or undefined. We find the first derivative using the quotient rule for each fraction:

• For \ \( f(x) = \frac{x}{x+2} \), \ \( f'(x) = \frac{(x+ 2)(1) - x(1)}{(x+2)^2} = \frac{2}{(x+2)^2} \).
• For \ \( g(x) = \frac{x-4}{x-2} \), \ \( g'(x) = \frac{(x-2)(1) - (x-4)(1)}{(x-2)^2} = \frac{2}{(x-2)^2} \).

Combining the derivatives: \ \( y' = \frac{2}{(x+2)^2} + \frac{2}{(x-2)^2} \).Since the sum is always positive and never equals zero, there are no critical points.

Points of Inflection

Points of inflection occur where the second derivative changes sign. We find the second derivative of each fraction:

• For \ \( \frac{2}{(x+2)^2} \), \ \( \frac{d^2}{dx^2} \left( \frac{2}{(x+2)^2} \right) = \frac{-4}{(x+2)^3} \).
• For \ \( \frac{2}{(x-2)^2} \), \ \( \frac{d^2}{dx^2} \left( \frac{2}{(x-2)^2} \right) = \frac{-4}{(x-2)^3} \).

Combine the second derivatives: \ \( \frac{-4}{(x + 2)^3} + \frac{-4}{(x - 2)^3} = 0 \).Solving the equation, we see no real solutions, indicating there are no points of inflection.