Problem 49
Question
Graph the functions in Exercises \(35-54\) $$ y=\frac{1}{x}+2 $$
Step-by-Step Solution
Verified Answer
The graph of \( y = \frac{1}{x} + 2 \) is a hyperbola vertically shifted up by 2 units with a horizontal asymptote at \( y = 2 \).
1Step 1: Identify the Function Type and Basic Graph
The given function is of the form \( y = \frac{1}{x} + 2 \). This is a transformation of the parent function \( y = \frac{1}{x} \), which represents a hyperbola. The parent hyperbola is centered at the origin.
2Step 2: Apply Vertical Translation
The function \( y = \frac{1}{x} + 2 \) implies a vertical shift of the hyperbola upwards by 2 units. The horizontal asymptote changes from \( y = 0 \) to \( y = 2 \).
3Step 3: Horizontal Asymptote Identification
Identify the horizontal asymptote of the function. Since the transformation only affects the vertical position, the horizontal asymptote is now \( y = 2 \).
4Step 4: Sketch the Parent Hyperbola
Begin by sketching the graph of \( y = \frac{1}{x} \), which has a vertical asymptote at \( x = 0 \) and horizontal asymptote at \( y = 0 \). It consists of two branches, located in the first and third quadrants.
5Step 5: Apply the Vertical Shift to the Graph
Adjust the sketch of the hyperbola by shifting the entire graph upward by 2 units to account for the \(+2\) in the equation. This results in the hyperbola's horizontal asymptote moving from \( y = 0 \) to \( y = 2 \).
6Step 6: Label Final Asymptotes and Intersections
The vertical asymptote remains at \( x = 0 \). The horizontal asymptote is \( y = 2 \). Find points by plugging values for \( x \) to determine some key points. For example, \( x = 1 \) gives \( y = 3 \) and \( x = -1 \) gives \( y = 1 \), which aids in sketching the correct positioning of the graph.
Key Concepts
Vertical AsymptoteHorizontal AsymptoteFunction Transformations
Vertical Asymptote
In the realm of rational functions, a vertical asymptote represents a line where the function approaches infinity, meaning the function grows larger and larger positively or negatively. For the parent function \( y = \frac{1}{x} \), this behavior is exhibited at \( x = 0 \). This is because as \( x \) gets closer and closer to zero from the positive or negative direction, the expression \( \frac{1}{x} \) tends to infinity or negative infinity.
In our exercise, the function \( y = \frac{1}{x} + 2 \) also has a vertical asymptote at \( x = 0 \). No transformation has been applied to affect the values of \( x \), so the vertical asymptote remains unchanged.
To identify vertical asymptotes easily:
In our exercise, the function \( y = \frac{1}{x} + 2 \) also has a vertical asymptote at \( x = 0 \). No transformation has been applied to affect the values of \( x \), so the vertical asymptote remains unchanged.
To identify vertical asymptotes easily:
- Look for values of \( x \) that make the denominator zero. In \( \frac{1}{x} \), \( x = 0 \) makes it undefined, indicating a vertical asymptote.
- Vertical asymptotes are denoted as vertical dashed lines on graphs as \( x \) approaches these critical points, but the function never actually touches the line.
Horizontal Asymptote
Horizontal asymptotes describe the behavior of a graph as \( x \) approaches plus or minus infinity. They show where the graph levels out on a horizontal plane. For the base function \( y = \frac{1}{x} \), the horizontal asymptote is \( y = 0 \), as the function values get closer to zero as \( x \) becomes very large in the positive or negative direction.
When transformations are applied, like in \( y = \frac{1}{x} + 2 \), the horizontal asymptote shifts accordingly. The `+2` in this equation indicates every point is shifted upwards by two units, moving the horizontal asymptote from \( y = 0 \) to \( y = 2 \).
Key points to consider when identifying horizontal asymptotes:
When transformations are applied, like in \( y = \frac{1}{x} + 2 \), the horizontal asymptote shifts accordingly. The `+2` in this equation indicates every point is shifted upwards by two units, moving the horizontal asymptote from \( y = 0 \) to \( y = 2 \).
Key points to consider when identifying horizontal asymptotes:
- The horizontal asymptote provides the end behavior of the function, which helps in understanding its long-term behavior.
- In transformations, if a constant \( k \) is added or subtracted, the horizontal asymptote changes to \( y = k \).
- Graphs never actually touch the horizontal asymptote; instead, they approach it infinitely as \( x \) becomes very large or small.
Function Transformations
Function transformations involve changing the base graph of a function in various ways, such as shifting, stretching, or reflecting. In the equation \( y = \frac{1}{x} + 2 \), we see a basic example of vertical transformation. Let's delve into this aspect further.
A vertical transformation occurs when a constant is added or subtracted, adjusting the graph upwards or downwards on the y-axis. In our example, adding `+2` creates a vertical translation upwards by two units.
Other types of transformations can include:
A vertical transformation occurs when a constant is added or subtracted, adjusting the graph upwards or downwards on the y-axis. In our example, adding `+2` creates a vertical translation upwards by two units.
Other types of transformations can include:
- Horizontal Translation: Shifting the graph to the left or right, achieved by adding or subtracting a constant inside the function's argument.
- Reflections: Flipping the graph over an axis, accomplished by multiplying by \(-1\).
- Stretching or Compressing: Scaling the graph along an axis, achieved by multiplying the function by a constant factor.
Other exercises in this chapter
Problem 48
Graph the functions in Exercises \(35-54\) $$ y=\frac{1}{x}-2 $$
View solution Problem 48
In Exercises \(47-58,\) say whether the function is even, odd, or neither. Give reasons for your answer. $$ f(x)=x^{-5} $$
View solution Problem 49
In Exercises \(47-58,\) say whether the function is even, odd, or neither. Give reasons for your answer. $$ f(x)=x^{2}+1 $$
View solution Problem 50
Graph the functions in Exercises \(35-54\) $$ y=\frac{1}{x+2} $$
View solution