Problem 55 - College Algebra Essentials

Question

Begin by graphing \(f(x)=\log _{2} x .\) Then use transformations of this graph to graph the given function. What is the vertical asymptote? Use the graphs to determine each function's domain and range. $$h(x)=1+\log _{2} x$$

Step-by-Step Solution

Answer

The vertical asymptote for both functions is at \(x=0\). The domain of both functions is \(x > 0\) and the range of \(f(x) = \log_2x\) is all real numbers. However, the range of \(h(x) = 1 + \log_2x\) is \(y > 1\).

Step 1 - Graphing the Function \(f(x) = \log_2x\)

Start with the basic function \(f(x) = \log_2x\). It is important to recall that the basic graph of a logarithmic function \(f(x) = \log_bx\) has a vertical asymptote at \(x=0\), passes through the point \((1, 0)\), and increases without bound as \(x\) increases. Draw the graph appropriately on the 2-dimensional plane according to these characteristics.

Step 2 - Graphing \(h(x) = 1 + \log_2x\)

The function \(h(x) = 1 + \log_2x\) can be obtained from \(f(x) = \log_2x\) by shifting the graph of \(f(x)\) upward by one unit. This is because adding a constant to a function causes a vertical shift in the graph of the function. Thus, the graph of \(h(x)\) will be the same as the graph of \(f(x)\), but moved upward by one unit.

Step 3 - Determining the Vertical Asymptote

The vertical asymptote for both functions \(f(x) = \log_2x\) and \(h(x) = 1 + \log_2x\) is at \(x=0\). This is true even after the transformation because vertical transformations do not affect the positioning of the vertical asymptote, they only shift the graph up or down.

Step 4 - Determining the Domain and Range

The domain of both functions is \(x > 0\) since the logarithmic functions are only defined for positive values of \(x\). The range of \(f(x) = \log_2x\) is all real numbers because the logarithm of any positive number can take any real value. However, the range of \(h(x) = 1 + \log_2x\) is translated to \(y > 1\), because of the transformation. The graph of \(h(x)\) is shifted up so all the \(y\) values are greater than 1.