# Wednesday 8 November class

Topics:

• Graphs of logarithmic functions: transforming the basic log function graphs (see Session 5, section 5.2 to review the transformations)

• Finding the equation of the asymptote, the domain, the x-intercept, and the-y-intercept (if it exists) for the transformed log functions.

To find the equation of the vertical asymptote, set the argument (the input) of the function equal to 0. (You can also find the equation simply by looking at how the transformations have acted on the position of the vertical asymptote, and you can also find the position of the x-intercept this way some of the time.)

For example, $f(x) = \log_{2}(x+5)$

We added 5 to the input, so this moves the graph of $\log_{2}(x)$ 5 units to the right. So the asymptote $x=0$ moves to $x=-5$, and the x-intercept at (1,0) moves to (-4,0).

Alternatively, you could find the equation of the vertical asymptote by setting  the argument x+5=0 and solving: x=-5 is the equation of the vertical asymptote.

To find the domain, either consider how the transformations act on the graph, or else you can look for the x-values where the argument is greater than 0.

In the example, we can see from the fact that the graph was shifted 5 units to the left, that the domain is now x>-5, in other words the domain is the interval $(5, \infty)$

Alternatively, we can find the domain by setting x+5 >0 and solving: x>-5.

To find the x-intercept: sometimes it is easy to see how the x-intercept has been moved, but we can always find the x-intercept (as for ANY function) by setting the function =0 and solving for x.

In the example, we set $\log_{2}(x+5) = 0$ and solve by rewriting in exponential form:

$2^{0} = x+5$

$1 = x+5 \implies x=-4$, so the x-intercept is (-4,0).

To find the y-intercept, if there is one: there will only be a y-intercept if x=0 is in the domain of the function. In that case, set x=0 in the formula for the function to find the y-intercept (as you would do for ANY function).

In the example, since 0 is in the domain, we set x=0 to get $y=\log_{2}(0+5) = \log_{2}(5)$.

So the y-intercept is at $(0, \log_{2}(5))$. It is not possible to compute $latex \log_{2}(5)$ exactly, but we can estimate it using a calculator – we will do this later. It is possible to see that $latex \log_{2}(5)$ is between 2 and 3 by translating to exponential form – try it!

When you sketch a graph, make sure that you show all of these features clearly.

Homework:

• Review the examples worked in class: one of them is in the notes above. Make sure that you understand how to find the asymptote, intercept or intercepts, and domain for the transformed logarithmic functions.

• Do the WeBWorK except you can omit problem #12 and 13 for now. Please do all the rest and do not wait to the last minute!

• Also do the following problems from the textbook, for extra practice: Exercise 13.6(a-h), find the equation of the vertical asymptote, the domain, the x-intercept, and the y-intercept if there is one, and sketch the graph. I will post answers to these.

• Don’t forget that Test 3 is scheduled for next Wednesday. The review materials will be on a separate post.

Don’t forget, if you get stuck on a problem, you can post a question on Piazza. Make sure to give your question a good subject line and tell us the problem itself – we need this information in order to answer your question. And please only put one problem per posted question!

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