Chapter 2: Problem 12
Find a single function defined implicitly by the given equation. Graph the function and give its domain. Use a graphing utility if necessary.] $$ x y+2 y+3=x $$
Short Answer
Expert verified
The function is \( y = \frac{x - 3}{x + 2} \) with domain \( x \neq -2 \).
Step by step solution
01
Rearrange the Equation
Start with the given equation: \( x y + 2 y + 3 = x \). Move all terms involving \( y \) to one side to prepare for solving for \( y \), resulting in the equation: \( xy + 2y - x + 3 = 0 \).
02
Factor Out y
Factor \( y \) from the terms on the left-hand side: \( y(x + 2) = x - 3 \). This step simplifies the expression so that we can isolate \( y \).
03
Solve for y
Solve for \( y \) by dividing both sides by the expression \( (x + 2) \): \( y = \frac{x - 3}{x + 2} \). This gives us a single implicit function that relates \( x \) and \( y \).
04
Identify the Domain
Determine the domain of the function by identifying values of \( x \) that do not make the denominator zero. Since the denominator is \( x + 2 \), \( x eq -2 \) to avoid division by zero. Thus, the domain is all real numbers except \( x = -2 \).
05
Graph the Function
To graph the function \( y = \frac{x - 3}{x + 2} \), note that it is a rational function with a vertical asymptote at \( x = -2 \) and it crosses the x-axis when \( x - 3 = 0 \), i.e., \( x = 3 \). The y-intercept happens at \( x = 0 \), giving \( y = -\frac{3}{2} \). Use a graphing utility if needed.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Domain of a Function
The domain of a function is the complete set of possible input values (usually represented by the variable \( x \)) that a function can accept without resulting in errors like division by zero or taking the square root of a negative number. For rational functions, which are ratios of polynomials, the domain is all real numbers except for the values that make the denominator zero.
In the function \( y = \frac{x - 3}{x + 2} \), to find the domain, look at the denominator \( x + 2 \). Set the denominator equal to zero and solve for \( x \):
In the function \( y = \frac{x - 3}{x + 2} \), to find the domain, look at the denominator \( x + 2 \). Set the denominator equal to zero and solve for \( x \):
- \( x + 2 = 0 \)
- \( x = -2 \)
Rational Function Graphing
Graphing a rational function involves identifying and plotting key points such as intercepts, holes, and asymptotes. For the rational function \( y = \frac{x - 3}{x + 2} \), start by finding the intercepts and asymptotes.
X-intercept: Set \( y = 0 \) and solve for \( x \):
X-intercept: Set \( y = 0 \) and solve for \( x \):
- \( \frac{x - 3}{x + 2} = 0 \) implies \( x - 3 = 0 \)
- Thus, \( x = 3 \) is the x-intercept.
- \( y = \frac{0 - 3}{0 + 2} = -\frac{3}{2} \)
- Thus, \( y = -\frac{3}{2} \) is the y-intercept.
Function Asymptotes
Asymptotes are lines that a graph approaches but never actually touches or crosses as it extends into infinity. These can be horizontal, vertical, or even oblique. For the function \( y = \frac{x - 3}{x + 2} \), focus on vertical and horizontal asymptotes.
Vertical Asymptote:
Vertical Asymptote:
- Occurs where the denominator is zero and the numerator is not zero.
- For \( x + 2 = 0 \), \( x = -2 \) is a vertical asymptote.
- Compare degrees of the polynomials in the numerator and denominator.
- Since the degrees of both are 1 (linear), take the ratio of leading coefficients: \( \frac{1}{1} = 1 \).
- This gives a horizontal asymptote at \( y = 1 \).
Solving Equations for y
Solving for \( y \) is about manipulating an equation so that \( y \) is on one side of the equation, usually to express \( y \) explicitly in terms of \( x \). Begin with an equation that may involve both variables. The goal is to isolate \( y \).
In our exercise \( xy + 2y = x - 3 \), to solve for \( y \):
In our exercise \( xy + 2y = x - 3 \), to solve for \( y \):
- First, factor \( y \) from the left-hand side: \( y(x + 2) = x - 3 \).
- Next, divide both sides by \( x + 2 \): \( y = \frac{x - 3}{x + 2} \).