Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 4: Problem 41
Simplify \(f(x)\), and sketch the graph of \(f\). $$ f(x)=\frac{x^{2}+x-2}{x+2} $$
Short Answer
Expert verified
The simplified function is \(f(x) = x - 1\) with a hole at \(x = -2\).
Step by step solution
01
Factor the Numerator
The first step is to factor the numerator of the function \(f(x) = \frac{x^2 + x - 2}{x + 2}\). The quadratic expression \(x^2 + x - 2\) can be factored into \((x - 1)(x + 2)\).
02
Cancel Common Factors
After factoring the numerator, \(f(x) = \frac{(x - 1)(x + 2)}{x + 2}\), we can cancel the common factor \((x + 2)\) from both the numerator and the denominator, assuming \(x eq -2\) to avoid division by zero. This simplifies \(f(x)\) to \(x - 1\).
03
State the Simplified Function
The simplified function is \(f(x) = x - 1\), with the important condition that \(x eq -2\) because the original expression is undefined at this point.
04
Sketch the Graph
Since \(f(x) = x - 1\), the graph is a straight line with a slope of 1 and a y-intercept at -1. However, since \(x eq -2\), there is a hole in the graph at \(x = -2\).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Factoring Quadratics
Factoring quadratics involves rewriting a quadratic expression as a product of its factors. This process plays a crucial role in simplifying expressions, especially when dealing with rational functions. In the given exercise, the quadratic expression in the numerator is \(x^2 + x - 2\). To factor it, we look for two numbers whose product is -2 (the constant term) and whose sum is 1 (the coefficient of the linear term). These numbers are -1 and 2, as \((-1) \times 2 = -2\) and \((-1) + 2 = 1\).
This allows us to factor the expression into \((x - 1)(x + 2)\).
This allows us to factor the expression into \((x - 1)(x + 2)\).
- Helps identify and cancel common factors in rational expressions.
- Essential for simplifying expressions and finding undefined points.
Graphing Functions
Graphing functions gives a visual representation of how a function behaves. In the case of the simplified function \(f(x) = x - 1\), this is a linear equation.
To graph this function:
Remember, there's a special case at \(x = -2\), which should be highlighted as a hole, indicating the function is not defined there.
To graph this function:
- Identify the slope, which is 1. This means the function rises by 1 unit vertically for every 1 unit it moves horizontally.
- Find the y-intercept, which is -1. This is where the line crosses the y-axis.
Remember, there's a special case at \(x = -2\), which should be highlighted as a hole, indicating the function is not defined there.
Simplifying Expressions
Simplifying expressions is a key step in algebra, particularly for rational functions. This involves reducing the function to its simplest form. In the given problem, we simplified \(f(x) = \frac{(x - 1)(x + 2)}{x + 2}\).
By canceling the common factor \((x + 2)\) from both the numerator and the denominator, we get the simpler expression \(x - 1\).
By canceling the common factor \((x + 2)\) from both the numerator and the denominator, we get the simpler expression \(x - 1\).
- Avoids complexity when analyzing the function.
- Makes it easier to understand and graph the function behavior.
Undefined Points
Undefined points occur where a function does not have a value. These typically arise when the denominator of a rational expression equals zero.
In our exercise, \(f(x) = \frac{x^2 + x - 2}{x + 2}\) becomes undefined when \(x + 2 = 0\), or \(x = -2\). At this point, any calculation involving the function would result in division by zero, which is impossible.
In our exercise, \(f(x) = \frac{x^2 + x - 2}{x + 2}\) becomes undefined when \(x + 2 = 0\), or \(x = -2\). At this point, any calculation involving the function would result in division by zero, which is impossible.
- Always identify undefined points before simplifying.
- Essential for understanding any discontinuities in the graph.
Holes in Graphs
Holes are unique points on a graph where the function is not defined due to cancellation of factors in rational functions. In our exercise, after canceling the factor \((x + 2)\), we obtain \(f(x) = x - 1\).
However, since we initially had a common factor \((x + 2)\) in both numerator and denominator, \(x = -2\) is a point where the function has a hole.
However, since we initially had a common factor \((x + 2)\) in both numerator and denominator, \(x = -2\) is a point where the function has a hole.
- Holes are not visible in the function's equation unless considered.
- Appear as skipped points on the graph.
- Important for accurate graphing and understanding function behavior.
