Transformations of Graphs of Logarithmic Functions

Vertical Shifts of the Parent Function $y=log_b(x)$

For any constant $d$, the function $f(x)=log_b(x)+d$

• shifts the parent function $y=log_b(x)$ up $d$ units if $d > 0$.
• shifts the parent function $y=log_b(x)$ down $d$ units if $d < 0$.
• has the vertical asymptote $x=0$.
• has domain $(0, \infty)$.
• has range $(−\infty, \infty)$.

HOW TO

Given a logarithmic function with the form $f(x)=log_b(x)+d$, graph the translation.

1. Identify the vertical shift:
   • If $d > 0$, shift the graph of $f(x)=log_b(x)$ up $d$ units.
   • If $d < 0$, shift the graph of $f(x)=log_b(x)$ down $d$ units.
2. Draw the vertical asymptote $x=0$.
3. Identify three key points from the parent function. Find new coordinates for the shifted functions by adding $d$ to the $y$ coordinate.
4. Label the three points.
5. The domain is $(0, \infty)$, the range is $(−\infty, \infty)$, and the vertical asymptote is $x=0$.

Example 5

Graphing a Vertical Shift of the Parent Function $y = log_b(x)$

Sketch a graph of $f(x)=log_3(x)−2$ alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.

Solution

Since the function is $f(x)=log_3(x)−2$, we will notice $d=–2$. Thus $d < 0$.

This means we will shift the function $f(x)=log_3(x)$ down 2 units.

The vertical asymptote is $x=0$.

Consider the three key points from the parent function, $(\frac{1}{3},−1), (1,0)$, and $(3,1)$.

The new coordinates are found by subtracting 2 from the $y$ coordinates.

Label the points $(\frac{1}{3},−3), (1,−2)$, and $(3,−1)$.

The domain is $(0, \infty)$, the range is $(−\infty, \infty)$, and the vertical asymptote is $x=0$.

Figure 9

The domain is $(0, \infty)$, the range is $(−\infty, \infty)$, and the vertical asymptote is $x=0$.

Try It #5

Sketch a graph of  $f(x)=log_2(x)+2$  alongside its parent function. Include the key points and asymptote on the graph. State the domain, range, and asymptote.