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*}

### 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*}

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*}*up* to the right. If \begin{align*}0 < b < 1\end{align*}*down* to the right.

Let's graph \begin{align*}y=\log_3(x-4)\end{align*}

To graph a logarithmic function without a calculator, start by drawing the vertical asymptote, at \begin{align*}x=4\end{align*}

The domain is \begin{align*}x>4\end{align*}

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*}**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*}*x* will approach \begin{align*}-3\end{align*}*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*}

First, there is a vertical asymptote at \begin{align*}x=0\end{align*}

To graph a logarithmic function using a TI-83/84, enter the function into \begin{align*}Y=\end{align*}

\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*}

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

The domain is \begin{align*}x>0\end{align*}

#### 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*}

\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*}
y=log5x - \begin{align*}y=\log_2(x+1)\end{align*}
y=log2(x+1) - \begin{align*}y=\log(x)-4\end{align*}
y=log(x)−4 - \begin{align*}y=\log_{\frac{1}{3}}(x-1)+3\end{align*}
y=log13(x−1)+3 - \begin{align*}y=-\log_{\frac{1}{2}}(x+3)-5\end{align*}
y=−log12(x+3)−5 - \begin{align*}y=\log_4(2-x)+2\end{align*}
y=log4(2−x)+2

Graph the following logarithmic functions using your graphing calculator.

- \begin{align*}y=\ln (x+6)-1\end{align*}
y=ln(x+6)−1 - \begin{align*}y=-\ln (x-1)+2\end{align*}
y=−ln(x−1)+2 - \begin{align*}y=\log(1-x)+3\end{align*}
y=log(1−x)+3 - \begin{align*}y=\log(x+2)-4\end{align*}
y=log(x+2)−4 - How would you graph \begin{align*}y=\log_4 x\end{align*}
y=log4x on the graphing calculator? Graph the function. - Graph \begin{align*}y=\log_{\frac{3}{4}}x\end{align*}
y=log34x on the graphing calculator. - Is (3, 8) on the graph of \begin{align*}y=\log_3 (2x-3)+7\end{align*}
y=log3(2x−3)+7 ? - 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.