Your math homework assignment is to find out which quadrants the graph of the function \begin{align*}f(x)= 4\ln(x + 3)\end{align*} falls in. On the way home, your best friend tells you, "This is the easiest homework assignment ever! All logarithmic functions fall in Quadrants I and IV." You're not so sure, so you go home and graph the function as instructed. Your graph falls in Quadrant I as your friend thought, but instead of Quadrant IV, it also falls in Quadrants II and III. Which one of you is correct?

### Graphing Logarithmic Functions

Now that we are more comfortable with using these functions as inverses, let’s use this idea to graph a logarithmic function. Recall that functions are inverses of each other when they are mirror images over the line \begin{align*}y=x\end{align*}. Therefore, if we reflect \begin{align*}y=b^x\end{align*} over \begin{align*}y=x\end{align*}, then we will get the graph of \begin{align*}y=\log_b x\end{align*}.

Recall that an exponential function has a horizontal asymptote. Because the logarithm is its inverse, it will have a *vertical* asymptote. The general form of a logarithmic function is \begin{align*}f(x)=\log_b(x-h)+k\end{align*} and the vertical asymptote is \begin{align*}x=h\end{align*}. The domain is \begin{align*}x>h\end{align*} and the range is all real numbers. Lastly, if \begin{align*}b>1\end{align*}, the graph moves *up* to the right. If \begin{align*}0 < b < 1\end{align*}, the graph moves *down* to the right.

Let's graph \begin{align*}y=\log_3(x-4)\end{align*} and state the domain and range.

To graph a logarithmic function without a calculator, start by drawing the vertical asymptote, at \begin{align*}x=4\end{align*}. We know the graph is going to have the general shape of the first function above. Plot a few “easy” points, such as (5, 0), (7, 1), and (13, 2) and connect.

The domain is \begin{align*}x>4\end{align*} and the range is all real numbers.

Now, let's determine if (16, 1) is on \begin{align*}y=\log (x-6)\end{align*}.

Plug in the point to the equation to see if it holds true.

\begin{align*}1 &= \log(16-6) \\ 1 &= \log 10 \\ 1 &= 1\end{align*}

Yes, this is true, so (16, 1) is on the graph.

Finally, let's graph \begin{align*}f(x)=2 \ln(x+1)\end{align*}.

To graph a natural log, we need to use a graphing calculator. Press \begin{align*}Y=\end{align*} and enter in the function, \begin{align*}Y=2 \ln(x+1)\end{align*}, **GRAPH**.

### Examples

#### Example 1

Earlier, you were asked to determine if your friend was correct.

The vertical asymptote of the function \begin{align*}f(x)= 4\ln(x + 3)\end{align*} is \begin{align*}x = -3\end{align*} since *x* will approach \begin{align*}-3\end{align*} but never quite reach it, *x* can assume some negative values. Hence, the function will fall in Quadrants II and III. Therefore, you are correct and your friend is wrong.

#### Example 2

Graph \begin{align*}y=\log_{\frac{1}{4}} x+2\end{align*} in an appropriate window.

First, there is a vertical asymptote at \begin{align*}x=0\end{align*}. Now, determine a few easy points, points where the log is easy to find; such as (1, 2), (4, 1), (8, 0.5), and (16, 0).

To graph a logarithmic function using a TI-83/84, enter the function into \begin{align*}Y=\end{align*} and use the Change of Base Formula: \begin{align*}log_a x = \frac{log_b x}{log_b a}\end{align*}. The keystrokes would be:

\begin{align*}Y= \frac{\log(x)}{\log \left ( \frac{1}{4} \right )}+2\end{align*}, **GRAPH**

To see a table of values, press \begin{align*}2^{nd} \rightarrow \end{align*} **GRAPH**.

#### Example 3

Graph \begin{align*}y=-\log x\end{align*} using a graphing calculator. Find the domain and range.

The keystrokes are \begin{align*}Y=-\log(x)\end{align*}, **GRAPH**.

The domain is \begin{align*}x>0\end{align*} and the range is all real numbers.

#### Example 4

Is (-2, 1) on the graph of \begin{align*}f(x)=\log_{\frac{1}{2}} (x+4)\end{align*}?

Plug (-2, 1) into \begin{align*}f(x)=\log_{\frac{1}{2}} (x+4)\end{align*} to see if the equation holds true.

\begin{align*}1 &= \log_{\frac{1}{2}} (-2+4) \\ 1 &= \log_{\frac{1}{2}} 2 \rightarrow \frac{1}{2}^x=2 \\ 1 & \ne -1 \end{align*}

Therefore, (-2, 1) is not on the graph. However, (-2, -1) is.

### Review

Graph the following logarithmic functions without using a calculator. State the equation of the asymptote, the domain and the range of each function.

- \begin{align*}y=\log_5 x\end{align*}
- \begin{align*}y=\log_2(x+1)\end{align*}
- \begin{align*}y=\log(x)-4\end{align*}
- \begin{align*}y=\log_{\frac{1}{3}}(x-1)+3\end{align*}
- \begin{align*}y=-\log_{\frac{1}{2}}(x+3)-5\end{align*}
- \begin{align*}y=\log_4(2-x)+2\end{align*}

Graph the following logarithmic functions using your graphing calculator.

- \begin{align*}y=\ln (x+6)-1\end{align*}
- \begin{align*}y=-\ln (x-1)+2\end{align*}
- \begin{align*}y=\log(1-x)+3\end{align*}
- \begin{align*}y=\log(x+2)-4\end{align*}
- How would you graph \begin{align*}y=\log_4 x\end{align*} on the graphing calculator? Graph the function.
- Graph \begin{align*}y=\log_{\frac{3}{4}}x\end{align*} on the graphing calculator.
- Is (3, 8) on the graph of \begin{align*}y=\log_3 (2x-3)+7\end{align*}?
- Is (9, -2) on the graph of \begin{align*}y=\log_{\frac{1}{4}} (x-5)\end{align*} ?
- Is (4, 5) on the graph of \begin{align*}y=5 \log_2 (8-x)\end{align*}?

### Answers for Review Problems

To see the Review answers, open this PDF file and look for section 8.7.