Problem 24 - A Graphical Approach to Precalculus with Limits

Question

Explain how the graph of \(f\) can be obtained from the graph of \(y=\frac{1}{x}\) or \(y=\frac{1}{x^{2}} .\) Draw a sketch of the graph of \(f\) by hand. Then generate an accurate depiction of the graph with a graphing calculator. Finally, give the domain and range. $$f(x)=\frac{2}{x+2}-1$$

Step-by-Step Solution

Answer

Transformations: left 2, stretch by 2, down 1. Domain is \(x \neq -2\), range is \(y \neq -1\).

Step 1: Identify the Base Graph

The base graph for the function is derived from the parent function \(y = \frac{1}{x}\). We will use transformations to modify this graph to obtain \(f(x) = \frac{2}{x+2} - 1\).

Step 2: Horizontal Shift

The term \(x + 2\) in the denominator indicates a horizontal shift of the base graph. The graph of \(y = \frac{1}{x}\) is shifted 2 units to the left.

Step 3: Vertical Stretch and Reflection

The coefficient \(2\) in the numerator indicates a vertical stretch by a factor of 2. Since there is no negative sign, we do not reflect the graph.

Step 4: Vertical Shift

The \(-1\) at the end of the function signifies a downward shift of the entire graph by 1 unit.

Step 5: Sketch the Graph

Begin by sketching the asymptotes. The vertical asymptote is at \(x = -2\) due to the horizontal shift, and the horizontal asymptote is at \(y = -1\) due to the vertical shift. Draw the hyperbola with these transformations.

Step 6: Depict with Graphing Calculator

Use a graphing calculator to input the function \(f(x) = \frac{2}{x+2} - 1\). Confirm that the calculator's graph matches the hand-drawn sketch.

Step 7: Domain and Range

Since the graph has a vertical asymptote at \(x = -2\), the domain is \(x eq -2\). Due to the horizontal asymptote at \(y = -1\), the range is \(y eq -1\).

Key Concepts

Horizontal Shift

When dealing with graph transformations, understanding horizontal shifts is key to locating your graph accurately on the x-axis. In this exercise, the function given is \[ f(x) = \frac{2}{x+2} - 1 \]The term \(x+2\) inside the denominator indicates a horizontal shift. In simpler terms, this means we're moving the entire graph of the base function \(y = \frac{1}{x}\) left or right.

• If we see \(x - c\), it shifts right by \(c\) units.
• If we see \(x + c\), it shifts left by \(c\) units.

For this function, \(x + 2\) tells us to shift the graph 2 units to the left. To visualize, imagine sliding the graph so every point steps 2 units in the negative x direction. This movement affects the vertical asymptote, placing it at \(x = -2\).

Vertical Stretch

Vertical stretching (or compression) modifies how the graph rises or falls. The impact of this transformation is determined by the coefficient multiplied by the function.Our function is again:\[ f(x) = \frac{2}{x+2} - 1 \]Notice the "2" in the numerator? It's multiplying \(\frac{1}{x}\), indicating a vertical stretch. Here’s what happens:

• a coefficient greater than 1 stretches the graph vertically.
• a coefficient between 0 and 1 compresses it.

In this case, the graph is stretched vertically by a factor of 2. This means that any point \((x, y)\) on the base graph \(y = \frac{1}{x}\) is vertically stretched to \((x, 2y)\) on the new graph, making the graph appear steeper.

Asymptotes

Understanding asymptotes helps us recognize where the graph will approach but never actually touch. The transformed function, \[ f(x) = \frac{2}{x+2} - 1 \]exhibits both a vertical and horizontal asymptote, crucial in shaping the graph.**Vertical Asymptote:**- Created by setting the denominator equal to zero: \(x + 2 = 0\). Thus, we have a vertical asymptote at \(x = -2\).**Horizontal Asymptote:**- Determined by looking at when \(x\) grows very large or very small. Here, the asymptote is \(y = -1\) because the function simplifies towards \(y = -1\) as \(x\) approaches infinity. These asymptotes are the invisible boundaries that guide the hyperbolic shape of the graph and signify where it will hover but never cross.

Domain and Range

Knowing the domain and range of a function is essential for identifying its usable x and y values.For the function \[ f(x) = \frac{2}{x+2} - 1 \]let’s determine these values:**Domain:**- This asks: "What \(x\) values can we use without breaking the function (e.g., division by zero)?"- Because the denominator \(x + 2\) cannot be zero, \(x = -2\) is off-limits.- Thus, the domain is \(x eq -2\).**Range:**- This answers: "What \(y\) values are possible outcomes?"- With a horizontal asymptote at \(y = -1\), \(y\) never actually equals -1.- Consequently, the range is \(y eq -1\).These boundaries help us map where the graph exists and exclude those points where it does not.