Chapter 1: Problem 10
Sketch the graph of each function. Do not use a graphing calculator. (Assume the largest possible domain.) $$ y=1+1 /(x+2)^{2} $$
Short Answer
Expert verified
The graph has a vertical asymptote at \(x = -2\), a horizontal asymptote at \(y = 1\), and a y-intercept at \((0, 1.25)\).
Step by step solution
01
Identify Domain
To find the domain of the function \(y = 1 + \frac{1}{(x+2)^2}\), we need to identify any values of \(x\) that would cause division by zero. The expression \((x+2)^2\) is zero when \(x = -2\). Therefore, \(x = -2\) is not in the domain. The domain of the function is all real numbers except \(x=-2\).
02
Determine Asymptotes
Vertical asymptotes occur where the function is undefined. The function has a vertical asymptote at \(x = -2\) where \((x+2)^2 = 0\). As \(x\) approaches \(-2\), the expression \((x+2)^2\) approaches zero, and \(\frac{1}{(x+2)^2}\) increases towards infinity. There is a horizontal asymptote at \(y = 1\), because as \(x\) goes to infinity or negative infinity, \(\frac{1}{(x+2)^2}\) tends to zero.
03
Determine Intercepts
To find the y-intercept, set \(x = 0\): \(y = 1 + \frac{1}{(0+2)^2} = 1 + \frac{1}{4} = 1.25\). Thus, the y-intercept is \((0, 1.25)\). There is no x-intercept, as the equation \(1 + \frac{1}{(x+2)^2} = 0\) has no solution.
04
Analyze Behavior Around Critical Points
On either side of the vertical asymptote (\(x = -2\)), the function approaches positive infinity. As \(x \to -2^+\), \(y \to +\infty\), and as \(x \to -2^-\), \(y \to +\infty\). To the left of \(x = -2\), \(y\) decreases as \(x\) moves away from \(-2\) and heads towards the horizontal asymptote \(y = 1\). To the right of \(x = -2\), \(y\) decreases and approaches \(y = 1\) as \(x\) increases.
05
Sketch the Graph
Plot the vertical asymptote at \(x = -2\) as a dotted line. Draw the horizontal asymptote at \(y = 1\). Plot the y-intercept at \((0, 1.25)\). Sketch the curve starting higher on the left approaching \(x = -2\), shooting up as \(x \to -2^-\), then from above as \(x \to -2^+\), the curve comes down and moves closer to the \(y = 1\) line as \(x\) goes to positive infinity.
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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 refers to all the possible input values (or x-values) that you can plug into a function without causing any problems like division by zero or taking the square root of a negative number. For the function \( y = 1 + \frac{1}{(x+2)^2} \), the denominator \((x+2)^2\) cannot be zero, as division by zero is undefined in mathematics. The expression \((x+2)^2\) equals zero when \(x = -2\). Therefore, this particular x-value is excluded from the domain of the function. For this function, the domain includes all real numbers except \(x = -2\).
To summarize:
To summarize:
- The domain is all real numbers except \(x = -2\).
Asymptotes
Asymptotes are lines that a graph approaches but never actually touches. They help you understand the behavior of functions as they move toward infinity. For this function, there are two types of asymptotes: vertical and horizontal.
**Vertical Asymptotes**
The vertical asymptote of this function is at \(x = -2\), determined by where the function is undefined. As \(x\) gets closer to \(-2\), without actually touching it, the values of \(y\) shoot up towards infinity. This behavior is typical near vertical asymptotes, where the graph of a function experiences a sort of "break."
**Horizontal Asymptotes**
A horizontal asymptote looks at the behavior of the function as \(x\) goes towards positive or negative infinity. For this function, the horizontal asymptote is \(y = 1\). As \(x\) moves towards very large positive or negative numbers, \(\frac{1}{(x+2)^2}\) tends toward zero, making \(y\) approach 1. So, the graph will get closer and closer to this y-value but never actually reach it.
**Vertical Asymptotes**
The vertical asymptote of this function is at \(x = -2\), determined by where the function is undefined. As \(x\) gets closer to \(-2\), without actually touching it, the values of \(y\) shoot up towards infinity. This behavior is typical near vertical asymptotes, where the graph of a function experiences a sort of "break."
**Horizontal Asymptotes**
A horizontal asymptote looks at the behavior of the function as \(x\) goes towards positive or negative infinity. For this function, the horizontal asymptote is \(y = 1\). As \(x\) moves towards very large positive or negative numbers, \(\frac{1}{(x+2)^2}\) tends toward zero, making \(y\) approach 1. So, the graph will get closer and closer to this y-value but never actually reach it.
Intercepts
Intercepts are the points where the graph of a function crosses the x-axis or y-axis. These points give us a snapshot of key intersections in the graph, which are crucial for sketching it.
**Y-intercept**
To find the y-intercept, we set \(x = 0\) in the function, giving us \(y = 1 + \frac{1}{4} = 1.25\). Thus, the graph crosses the y-axis at the point \((0, 1.25)\). This point helps in positioning the overall graph on the coordinate plane.
**X-intercept**
An x-intercept occurs when \(y = 0\). Solving \(1 + \frac{1}{(x+2)^2} = 0\), we find there is no solution, meaning the graph never crosses the x-axis. The absence of an x-intercept is important because it confirms the graph doesn't dip below the x-axis.
**Y-intercept**
To find the y-intercept, we set \(x = 0\) in the function, giving us \(y = 1 + \frac{1}{4} = 1.25\). Thus, the graph crosses the y-axis at the point \((0, 1.25)\). This point helps in positioning the overall graph on the coordinate plane.
**X-intercept**
An x-intercept occurs when \(y = 0\). Solving \(1 + \frac{1}{(x+2)^2} = 0\), we find there is no solution, meaning the graph never crosses the x-axis. The absence of an x-intercept is important because it confirms the graph doesn't dip below the x-axis.
Behavior at Critical Points
Critical points in a function's graph are where you might experience abrupt changes or an interesting behavior, such as the graph shooting toward infinity or settling into a flat line.
**Near the Vertical Asymptote**
The behavior of the function \(y = 1 + \frac{1}{(x+2)^2}\) around the vertical asymptote \(x = -2\) is telling. As \(x\) approaches \(-2\) from the left \((x \to -2^-)\), the values of \(y\) increase sharply toward positive infinity. Similarly, from the right \((x \to -2^+)\), \(y\) also rises rapidly towards positive infinity. This simultaneous rise on both sides reflects the distinct nature of a vertical asymptote.
**Approaching the Horizontal Asymptote**
As \(x\) moves away in either direction from the asymptote \(x = -2\), the graph stabilizes and approaches the horizontal asymptote \(y=1\). Whether \(x\) is much larger or smaller than \(-2\), \(y\) gently decreases and approaches \(y=1\), highlighting the role of horizontal asymptotes in anchoring the long-term behavior of a function's graph.
**Near the Vertical Asymptote**
The behavior of the function \(y = 1 + \frac{1}{(x+2)^2}\) around the vertical asymptote \(x = -2\) is telling. As \(x\) approaches \(-2\) from the left \((x \to -2^-)\), the values of \(y\) increase sharply toward positive infinity. Similarly, from the right \((x \to -2^+)\), \(y\) also rises rapidly towards positive infinity. This simultaneous rise on both sides reflects the distinct nature of a vertical asymptote.
**Approaching the Horizontal Asymptote**
As \(x\) moves away in either direction from the asymptote \(x = -2\), the graph stabilizes and approaches the horizontal asymptote \(y=1\). Whether \(x\) is much larger or smaller than \(-2\), \(y\) gently decreases and approaches \(y=1\), highlighting the role of horizontal asymptotes in anchoring the long-term behavior of a function's graph.