Problem 50 - College Algebra

Question

Begin by graphing \(f(x)=\log _{2} x .\) Then use transformations of this graph to graph the given function. What is the graph's \(x\) -intercept? What is the vertical asymptote? $$g(x)=\log _{2}(x+2)$$

Step-by-Step Solution

Answer

The x-intercept of the graph of \(g(x)=\log _{2}(x+2)\) is at \(x=-1\), and the vertical asymptote is at \(x=-2\).

Step 1: Understanding the basic logarithmic function

First, recall that the graph of \(f(x)=\log _{2} x\) has an x-intercept at \(x=1\) and a vertical asymptote at \(x=0\).

Step 2: Applying the transformation

Now, to graph \(g(x)=\log _{2}(x+2)\), the transformation applied to \(f(x)=\log _{2} x\) is a horizontal shift of the graph 2 units to the left. This means that for every point \((x, y)\) on the graph of \(f(x)\), the corresponding point on \(g(x)\) is \((x-2, y)\).

Step 3: Identifying the x-intercept

The x-intercept of the graph of \(f(x)=\log _{2} x\) is \(x=1\). By the transformation, the x-intercept of \(g(x)=\log _{2}(x+2)\) is shifted 2 units to the left, meaning it is now at \(x=-1\). This is because when \(x=-1\), \(g(-1)=\log _{2}((-1)+2)=\log _{2}(1)=0\).

Step 4: Identifying the vertical asymptote

The vertical asymptote of the graph of \(f(x)=\log _{2} x\) is \(x=0\). By the transformation, the vertical asymptote of \(g(x)=\log _{2}(x+2)\) is shifted 2 units to the left, giving \(x=-2\). This is because as \(x\) approaches -2 from the right, \(g(x)\) approaches negative infinity.

Key Concepts

Logarithm Transformations

Understanding how to graph logarithmic functions often begins with the basic function, like f(x) = \(\log_2 x\). Transformations of this parent graph can help you graph related functions such as g(x) = \(\log_2(x+2)\). A transformation involves shifting, stretching, compressing, or reflecting the basic graph.

For example, adding a constant inside the logarithm function, as in g(x) = \(\log_2(x+2)\), results in a horizontal shift. Specifically, adding 2 inside the function shifts the graph 2 units to the left. To visualize this, imagine sliding the entire graph of f(x) leftward by 2 units on the x-axis. Each original point (x, y) on the parent graph will correspond to a new point (x-2, y) on the transformed graph. So, the graph's basic shape remains the same, but its position on the coordinate plane has changed.

X-Intercept of Logarithmic Functions

The x-intercept of a log function is a key feature that helps in graphing and understanding its behavior. For the basic logarithmic function f(x) = \(\log_2 x\), the x-intercept occurs at the point where the y-value is zero.

For f(x), this happens at x=1, since \f(1) = 0. Keeping in mind the horizontal shift we discussed in logarithm transformations, the x-intercept for g(x) = \(\log_2(x+2)\) is shifted left by 2 units, landing it at x=-1. It's important to note, this doesn't change the 'height' of the x-intercept, which remains at y=0, it only changes the position along the x-axis.

Vertical Asymptote of Logarithmic Functions

The vertical asymptote of a logarithmic function is a line that the graph approaches but never touches or crosses. It represents values that are not in the domain of the function. For the parent function f(x) = \(\log_2 x\), the vertical asymptote is at x=0, because the log of zero is undefined.

When a transformation like g(x) = \(\log_2(x+2)\) is applied, the vertical asymptote also shifts 2 units to the left, giving us a new asymptote at x=-2. As x gets closer to -2 from the right, g(x) heads towards negative infinity. This means that for g(x), you can get closer and closer to x=-2, but you can never reach it, and the function will never produce an output for x≤-2.