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Magazine Chapter 1: Problem 50
Graph the functions. $$y=\frac{1}{x+2}.$$
Short Answer
Expert verified
Graph the hyperbola with a vertical asymptote at \(x=-2\) and a horizontal asymptote at \(y=0\), using points like \((-1,1)\) and \((0,0.5)\).
Step by step solution
01
Identify the Type of Function
The given function is \( y = \frac{1}{x+2} \), which is a reciprocal function. Reciprocal functions have the general form \( y = \frac{1}{x} \) and involve a hyperbola shape when graphed.
02
Identify Vertical and Horizontal Asymptotes
For the function \( y = \frac{1}{x+2} \), a vertical asymptote occurs where the denominator is zero. Setting \( x+2 = 0 \) gives \( x = -2 \). There is no horizontal asymptote directly in the form \( y = 0 \) for this type of function. Instead, note that as \( x \to \pm \infty \), \( y \approx 0 \), indicating a horizontal asymptote at \( y = 0 \).
03
Determine Critical Points
Find some critical points to understand function behavior around the asymptotes. Substitute values into the function, like \( x = -1, 0, -3 \): - For \( x = -1 \), \( y = \frac{1}{-1+2} = 1 \).- For \( x = 0 \), \( y = \frac{1}{0+2} = \frac{1}{2} \).- For \( x = -3 \), \( y = \frac{1}{-3+2} = -1 \).
04
Sketch the Graph
Begin by drawing the vertical asymptote at \( x = -2 \) and horizontal asymptote on the x-axis at \( y=0 \). Plot the critical points found: \(( -1, 1) \), \((0, 0.5) \), and \(( -3, -1) \). Draw two branches of the hyperbola, one in the second quadrant approaching the asymptotes, and another in the first quadrant, also approaching both asymptotes.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Asymptotes
Asymptotes are lines that a graph approaches but never actually touches or crosses, giving us significant information about the behavior of the graph around them. For the function \( y = \frac{1}{x+2} \), we have vertical and horizontal asymptotes to consider.
- Vertical Asymptote: This occurs where the denominator of the reciprocal function is zero, causing the function to be undefined. Setting the denominator equal to zero, \( x+2 = 0 \), we find the vertical asymptote at \( x = -2 \). As \( x \) approaches \( -2 \), the value of \( y \) will approach infinity or negative infinity, depending on the direction.
- Horizontal Asymptote: For \( y = \frac{1}{x+2} \), the horizontal asymptote is at \( y = 0 \), which can be deduced as \( x \) tends to positive or negative infinity. The value of \( y \) dips closer to zero, never quite reaching it, indicating the graph approaches but never hits the x-axis.
Critical Points
Critical points in a function's graph are specific points that help us understand its behavior and outline its structure. In our function \( y = \frac{1}{x+2} \), calculating critical points around the asymptotes provides a clearer picture of how the graph behaves. These points were calculated for values of \( x \) like \( -1 \), \( 0 \), and \( -3 \).
- When \( x = -1 \), we find \( y = 1 \). This point \((-1, 1)\) is located in the first quadrant.
- At \( x = 0 \), \( y = \frac{1}{2} \), this gives the point \((0, 0.5)\) and shows the graph nearing the horizontal asymptote.
- For \( x = -3 \), we get \( y = -1 \). This places the point \((-3, -1)\) in the third quadrant.
Hyperbola
The graph of the function \( y = \frac{1}{x+2} \) takes the shape of a hyperbola, which is a type of curve found commonly in reciprocal functions. A hyperbola consists of two separate branches that mirror each other diagonally across the axes or asymptotes. Here is a closer look at the characteristics:
- Branches: Each branch of the hyperbola is asymptotic, heading off towards infinity and adhering closely to the asymptotes without touching them.
- Quadrants: One branch lies in the second quadrant, curving from the top left towards the negative side of the x-axis, while the second branch is in the first quadrant, moving from the bottom right towards the positive side of the x-axis.
- Symmetry: The hyperbola has rotational symmetry around the origin when seen through a transformation of its centroid.
