Problem 41 - A Graphical Approach to Precalculus with Limits

Question

In each of the following. (a) explain how the graph of the given finction can be obtained from the graph of \(g(x)=\log _{2} x,\) and (b) graph the function. $$f(x)=3 \log _{2} x+1$$

Step-by-Step Solution

Answer

The graph of \(f(x) = 3\log_{2} x + 1\) is a vertically stretched and shifted version of \(g(x) = \log_{2} x\).

Step 1: Identify the Base Function

The base function is given as \(g(x) = \log_{2} x\). This is a logarithmic function with a base of 2, which represents the power to which the base, 2, must be raised to produce a given number, \(x\). Its graph passes through the point (2,1) and has a vertical asymptote at \(x=0\).

Step 2: Understand the Transformation by 3

The function \(f(x) = 3\log_{2} x\) involves multiplying the base function by 3. Multiplying by 3 stretches the graph vertically, increasing the output value for any input \(x\). The graph becomes steeper compared to \(g(x)\).

Step 3: Apply the Vertical Shift

The function \(f(x) = 3\log_{2} x + 1\) adds 1 to every value of the function \(3\log_{2} x\). This represents a vertical shift upward by 1 unit, shifting the entire graph of \(3\log_{2} x\) up by 1. Thus, the y-intercept becomes 1 unit higher than it was before the shift.

Step 4: Sketch the Graph

Start with the graph of \(g(x) = \log_{2} x\). Then apply the vertical stretch from Step 2, and finally apply the vertical shift from Step 3. The final graph, \(f(x) = 3\log_{2} x + 1\), has a steeper incline and is shifted up by 1 unit. It retains the vertical asymptote of the original function at \(x=0\).

Key Concepts

Function Transformation

Transformations help us modify the original graph of a function, turning it into a new one. For the function \( g(x) = \log_{2} x \), we start with a basic logarithmic function. This function shows us how exponentiation works, with the graph passing through key points like \((2,1)\), having a vertical asymptote at \(x = 0\).

When we talk about transforming this base function into \( f(x) = 3\log_{2} x + 1 \), we perform operations that change the appearance of the original graph.

• Vertical Stretch: By multiplying by a constant (in this case, 3), we modify its steepness.
• Vertical Shift: Adding or subtracting numbers from the function to move it up or down.

These transformations can be layered, so the sequence of transformation matters for getting the final graph right.

Vertical Stretch

A vertical stretch alters the height of a graph by multiplying the function's values. For example, in the transition from \( g(x) = \log_{2} x \) to \( 3\log_{2} x \), the multiplier 3 stretches the graph vertically.

Imagine stretching a rubber band. Similarly, the function becomes taller, with larger function values for the same \(x\) inputs:

• Points that were previously at \( (2, 1) \) under the base function \( g(x) = \log_{2} x \) would now be at \( (2, 3) \) due to the multiplier 3.

This makes the graph steeper, enhancing its vertical aspect, possibly making it harder to distinguish small values, but enlarging the magnitude of larger values.

Vertical Shift

A vertical shift occurs when a constant is added or subtracted to a function. Following the vertical stretch, adding 1 unit in the equation \( f(x) = 3\log_{2} x + 1 \) signifies a vertical shift.

This means we move every point on the stretched graph of \( 3\log_{2} x \) up by 1 unit:

• The point that was shifted to \((2, 3)\) after the stretch is now at \((2, 4)\).

The whole graph, therefore, rises, making the y-intercept 1 unit higher. Importantly, this transformation doesn't change the shape or inclination of the function—only its position relative to the vertical direction changes. The graph still maintains the vertical asymptote from the original function at \(x=0\), just displaced up.