Graphing the Logarithmic Function

The logarithmic function has a vertical asymptote and its domain is restricted. The curve is located on the left or right side of the asymptote. The side that the curve is on can be determined by analyzing parameter |b.|

• If parameter |b| is positive, the curve lies on the right side of the asymptote.
• If parameter |b| is negative, the curve lies on the left side of the asymptote.

1. Graph the vertical asymptote at |x=h.|
2. Determine which side of the asymptote the curve lies on by analyzing parameter |b.|
3. Calculate the coordinates of a few points belonging to the curve.
4. Plot the points found in Step 3 on the Cartesian plane and draw the curve.
    

When graphing the logarithmic function, the curve must get closer and closer to the asymptote without ever touching it.

Graph the logarithmic function with the rule |f(x)=-4\log_{3}\left(\dfrac{1}{2}(x+10)\right)+5.|

3. Construct a table of values.
   Since the curve lies to the right of the asymptote, we choose |x|-values that are greater than |-10.|
   Let’s use |x=-8.| ||\begin{align}f(\boldsymbol{\color{#560FA5}x}) &= -4\log_{3}\left(\dfrac{1}{2}(\boldsymbol{\color{#560FA5}x}+10)\right)+5\\ f(\boldsymbol{\color{#560FA5}{-8}}) &= -4\log_{3}\left(\dfrac{1}{2}(\boldsymbol{\color{#560FA5}{-8}}+10)\right)+5\\ f(-8) &= -4\log_{3}\left(\frac{1}{2}\times2\right)+5\\ f(-8) &= -4\log_{3}(1)+5\end{align}||The logarithm of 1 is |0.| ||\begin{align}f(-8) &= -4(0)+5\\&= 5\end{align}||So, a point on the curve is |(-8,5).| Next, we repeat and find other coordinates with different |x|-values.
   Let’s use |x=-4.| ||\begin{align}f(\boldsymbol{\color{#560FA5}x}) &= -4\log_{3}\left(\dfrac{1}{2}(\boldsymbol{\color{#560FA5}x}+10)\right)+5\\ f(\boldsymbol{\color{#560FA5}{-4}}) &= -4\log_{3}\left(\dfrac{1}{2}(\boldsymbol{\color{#560FA5}{-4}}+10)\right)+5\\ f(-4) &= -4\log_{3}\left(\frac{1}{2}\times6\right)+5\\ f(-4) &= -4\log_{3}(3)+5\\ f(-4) &= -4(1)+5\\ f(-4) &= -4+5\\ f(-4) &= 1\end{align}||So, another point on the curve is |(-4,1).| 
   Let’s calculate the |y|-intercept by taking |x=0.| ||\begin{align}f(\boldsymbol{\color{#560FA5}x}) &= -4\log_{3}\left(\dfrac{1}{2}(\boldsymbol{\color{#560FA5}x}+10)\right)+5\\ f(\boldsymbol{\color{#560FA5}0}) &= -4\log_{3}\left(\dfrac{1}{2}(\boldsymbol{\color{#560FA5}0}+10)\right)+5\\ f(0) &= -4\log_{3}\left(\frac{1}{2}\times10\right)+5\\ f(0) &= -4\log_{3}(5)+5\end{align}||Applying the change of base law for logarithms, we get the following:
   ||\begin{align}f(0) &= -4\left(\dfrac{\log(5)}{\log(3)}\right)+5\\ &\approx -4(1.46)+5\\&\approx -5.86+5\\&\approx -0.86\end{align}||So, the |y|-intercept is |-0.86.|
   Continuing in the same way, we get the following table of values:

|\boldsymbol{x}|	|-8|	|-4|	|0|	|8|	|44|
|\boldsymbol{y}|	|5|	|1|	|-0.86|	|-3|	|-7|

4. Plot the points found in Step 3 on the Cartesian plane and draw the curve.

To make logarithm calculations easier and find integer coordinates, we select |x|-values so the argument of the logarithm is a power of the base.

In the previous example, the rule was |f(x)=-4\log_{3}\left(\dfrac{1}{2}(x+10)\right)+5.| The base is |\boldsymbol{\color{#FF55C3}3}| and the argument is |\boldsymbol{\color{#333FB1}{\dfrac{1}{2}(x+10)}}.|

We determine the powers of |\boldsymbol{\color{#FF55C3}3}.| ||\begin{align}&\boldsymbol{\color{#FF55C3}3^\color{#3A9A38}0}&&\boldsymbol{\color{#FF55C3}3^\color{#3A9A38}1}&&\boldsymbol{\color{#FF55C3}3^\color{#3A9A38}2}&&\boldsymbol{\color{#FF55C3}3^\color{#3A9A38}3}&&\boldsymbol{\color{#FF55C3}3^\color{#3A9A38}4}\\&\boldsymbol{\color{#333FB1}1}&&\boldsymbol{\color{#333FB1}3}&&\boldsymbol{\color{#333FB1}9}&&\boldsymbol{\color{#333FB1}{27}}&&\boldsymbol{\color{#333FB1}{81}}\end{align}||Then, we find the |x|-values that have these powers in the argument.

||\begin{align}\boldsymbol{\color{#333FB1}{\dfrac{1}{2}(x+10)}} &= \boldsymbol{\color{#333FB1}1}\\ x+10 &= 2\\ x &= -8\end{align}||When |x=-8,| the argument is |\boldsymbol{\color{#333FB1}1}.| ||\log_{\boldsymbol{\color{#FF55C3}3}}(\boldsymbol{\color{#333FB1}1})=\boldsymbol{\color{#3A9A38}0}||

||\begin{align}\boldsymbol{\color{#333FB1}{\dfrac{1}{2}(x+10)}} &= \boldsymbol{\color{#333FB1}3}\\ x+10 &= 6\\ x &= -4\end{align}||When |x=-4,| the argument is |\boldsymbol{\color{#333FB1}3}.| ||\log_{\boldsymbol{\color{#FF55C3}3}}(\boldsymbol{\color{#333FB1}3})=\boldsymbol{\color{#3A9A38}1}||

||\begin{align}\boldsymbol{\color{#333FB1}{\dfrac{1}{2}(x+10)}} &= \boldsymbol{\color{#333FB1}9}\\ x+10 &= 18\\ x &= 8\end{align}||When |x=8,| the argument is |\boldsymbol{\color{#333FB1}9}.| ||\log_{\boldsymbol{\color{#FF55C3}3}}(\boldsymbol{\color{#333FB1}9})=\boldsymbol{\color{#3A9A38}2}||

So, by choosing |x=-8,| |x=-4,| and |x=8,| there is no need to use the change of base law for logarithms to do the calculations. Also, we get coordinates that are integers.