Problem 489 - Precalculus

Question

For the following exercises, use the given transformation to graph the function. Note the vertical and horizontal asymptotes. The reciprocal function shifted up two units.

Step-by-Step Solution

Answer

The new vertical asymptote is at \( x=0 \) and the horizontal asymptote is at \( y=2 \).

Step 1: Understanding the Reciprocal Function

The reciprocal function is given by \( f(x) = \frac{1}{x} \). Its graph has a vertical asymptote at \( x = 0 \) where the function is undefined, and a horizontal asymptote at \( y = 0 \) as \( x \to \pm \infty \).

Step 2: Identifying the Transformation

The problem states that the reciprocal function is shifted up by two units. This means the entire graph of the function is moved up on the y-axis by two units. The new function is \( g(x) = \frac{1}{x} + 2 \).

Step 3: Graphing the Transformed Function

The transformation results in each point of the function \( f(x) = \frac{1}{x} \) being moved up by 2 units. Thus, the horizontal asymptote, originally at \( y = 0 \), shifts to \( y = 2 \). The vertical asymptote remains unchanged at \( x = 0 \). Plot the new function \( g(x) = \frac{1}{x} + 2 \) with these asymptotes.

Step 4: Concluding the Graph

After graphing, the transformed function \( g(x) = \frac{1}{x} + 2 \) has a vertical asymptote at \( x = 0 \) and a horizontal asymptote at \( y = 2 \). The function will appear as two curves, one in each quadrant of the Cartesian plane, similar to its precursor but shifted upward by 2 units.

Key Concepts

Reciprocal Function

The reciprocal function is an interesting mathematical concept. It is defined by the expression \( f(x) = \frac{1}{x} \). This means that for each input value \( x \), the output is the reciprocal of \( x \). Often, students encounter this function as it showcases critical properties of function transformation such as asymptotes.

One key feature of the reciprocal function is its graph, which typically consists of two separate branches. These branches are located in quadrants one and three of a Cartesian plane. Why? Because:\

• In quadrant one, both \( x \) and \( y = \frac{1}{x} \) are positive.
• In quadrant three, both \( x \) and \( y = \frac{1}{x} \) are negative.

This arrangement results in the characteristic hyperbolic shape of the graph of \( f(x) = \frac{1}{x} \). Understanding the basic shape helps when you need to graph transformations of this function.

Vertical Asymptote

A vertical asymptote is a line that a graph approaches but never touches or crosses. For the reciprocal function \( f(x) = \frac{1}{x} \), the vertical asymptote is found at \( x = 0 \). This is because the function is undefined when \( x \) equals zero. As \( x \) gets close to this value, the outputs of the function become extremely large or extremely small, but they do not settle on any finite result.

When you plot \( f(x) = \frac{1}{x} \), you will notice that as \( x \to 0^+ \) (approaching zero from the right), \( f(x) \to +∞ \), and as \( x \to 0^- \) (approaching zero from the left), \( f(x) \to -∞ \). This behavior confirms the existence of a vertical asymptote at \( x = 0 \). Regardless of transformation, the vertical asymptote remains anchored at its original position unless there's a horizontal shift.

Horizontal Asymptote

Horizontal asymptotes are another important feature of the reciprocal function. For the basic function \( f(x) = \frac{1}{x} \), the horizontal asymptote appears at \( y = 0 \). This occurs because as \( x \to \, \pm \, \infty \), the value of \( f(x) = \frac{1}{x} \) trends closer to zero without ever actually reaching it.

In transformations, a horizontal asymptote can shift. When the function is shifted up by units, such as in the function \( g(x) = \frac{1}{x} + 2 \), the horizontal asymptote moves from \( y = 0 \) to \( y = 2 \). This shift indicates that as \( x \to \, \pm \, \infty \), the function values approach 2 instead of zero. Typically, to determine a horizontal asymptote after a transformation, you observe the constant added to or subtracted from \( f(x) \). In this exercise, the addition of 2 directly relocates the horizontal asymptote upwards by 2 units.