Problem 29 - 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. $$f(x)=\frac{1}{(x-3)^{2}}$$

Step-by-Step Solution

Answer

Shift \(y=\frac{1}{x^2}\) right by 3 units. Domain: \((- fty, 3) \cup (3, fty)\); Range: \((0, fty)\).

Step 1: Identify the Parent Function

First, identify which parent function \(f(x) = \frac{1}{(x-3)^{2}}\) is derived from. It's a transformation of the graph \(y = \frac{1}{x^2}\).

Step 2: Horizontal Shift

The expression \((x-3)^2\) indicates a horizontal shift to the right by 3 units. This means the graph of \(y = \frac{1}{x^2}\) is shifted rightwards so that its vertical asymptote moves from \(x = 0\) to \(x = 3\).

Step 3: Sketch the Transformed Graph

Draw the graph of \(y = \frac{1}{(x-3)^2}\) by moving the basic shape of \(y = \frac{1}{x^2}\) to the right by 3 units. This graph will have a vertical asymptote at \(x = 3\) and will curve upwards on both sides of this asymptote.

Step 4: Verify with Graphing Calculator

Use a graphing calculator to plot \(f(x) = \frac{1}{(x-3)^2}\) and confirm the position of the asymptote and the shape observed in your sketch.

Step 5: Determine the Domain and Range

For the function \(f(x) = \frac{1}{(x-3)^2}\), the domain is all real numbers except where the function is undefined due to the denominator being zero. So, the domain is all real numbers except \(x = 3\), or in interval notation: \((-fty, 3) \cup (3, fty)\). Since the function approaches infinity as \(x\) approaches \(3\) from either direction, and is positive for all other \(x\), the range is \((0, fty)\).

Key Concepts

Horizontal Shift

When discussing the concept of a horizontal shift, it's crucial to understand how changes in the equation of a function affect its graph. A horizontal shift involves moving the graph of a function left or right along the x-axis. This shift occurs in response to changes inside the function's variable expression. For the function in this exercise, we have

• The expression \[ (x-3)^2 \] indicates that the graph of the original function, \[ y = \frac{1}{x^2} \], is shifted horizontally.
• Specifically, the \(-3\) in the expression signals a shift to the right by 3 units because any expression of the form \( (x-h) \) shifts the graph to the right by \( h \) units.

This links directly to the idea that modifying the input of a function directly affects where the graph will appear along the x-axis. Shifting by changing what the variable \(x\) is compared to is a basic technique for adjusting the position of graphs without altering their shape.

Vertical Asymptote

Understanding vertical asymptotes is key to mastering graph transformations, especially for rational functions or those involving division by a variable expression. A vertical asymptote is a line where the graph of a function approaches but never touches or crosses. In functions such as

• \( y = \frac{1}{(x-3)^2} \), the denominator \( (x-3)^2 \) becomes zero when \( x = 3 \) .
• This indicates a vertical asymptote at \( x = 3 \), meaning as you approach this value from either the left or right, the function's values will increase or decrease without bound, heading towards infinity.

Vertical asymptotes occur in rational functions where the denominator goes to zero, yet the numerator doesn't, leading to infinitely large values. It's important to note this line of the asymptote helps shape the graph significantly, as it outlines the line that the function will never cross or touch.

Domain and Range

The domain and range of a function describe the set of possible inputs (domain) and outputs (range) for the function. For the function

• \( y = \frac{1}{(x-3)^2} \), the domain includes all real numbers except \( x = 3 \).
• Since division by zero is undefined, \( x = 3 \) creates a restriction, resulting in a domain of \((-\infty, 3) \cup (3, \infty)\).

As for the range, this function never produces zero or negative values because squares are always positive unless they're zero (which they're not here since zero is undefined). Therefore, as the function tends to positive infinity when approaching the asymptote on either side, the range becomes

• \((0, \infty)\), indicating all positive real numbers.

Domain and range are essential as they specify the "reachable" values of the function, shaping a complete understanding of its behavior across the real number line.