Sunday, December 13, 2009

6.3 Transfomations on Lagarithmic Functions

In class on Friday we learned how to graph and apply transformations to a logarithmic function. These transformations are very similar to the ones that we learned earlier in the year when we were dealing with the sin and cos graphs. The set up of these functions is the same as the set up for the log functions, y=alog[k(x-d)]+c, the variables of these functions affect the same parts of the graph as well. The a value will vertically stretch and compress it, the k value will horizontally stretch or compress it, the h will translate the graph horizontally and lastly the c value will shift it vertically.

*REMEMBER: the d value will shift the vertical asymptote and the c value will shift the horizontal asymptote, as well as the graph.

When applying transformations to a function follow these 5 steps…
Step 1: Ensure that the function is in y=alog[k(x-y)]+c form
Step 2: Apply vertical stretches, compressions and reflections
Step 3: Apply horizontal stretches, compressions and reflections
Step 4: Apply horizontal translations
Step 5: Apply vertical translations

From these 5 steps you should be able to create a mapping rule in which you can use to graph the function. (x,y) = (1/k x-d,ay+c)

You can now use the mapping rule to create a table of values so that you can graph the translated function.

*REMEMBER: a logarithmic graph is the opposite of an exponential graph so you must exchange the x and y values in the table of values.

x y = 10^x
-2 1/100
-1 1/10
0 1
1 10
2 100

x y= log10x
1/100 -2
1/10 -1
1 0
10 1
100 2
(then take the values from this table and apply them in your mapping rule to find the points so that you can graph your function)

The last things that we covered on Friday were the key features of a graph. Well, domain, range, and the asymptotes, can be found easily by looking at the graph, but how do we find an intercept when the point is not on the table of values? Well, just like in previous chapter what we do is solve for the variable using the log function. For example, if you were looking for the x-intercept you would make y=0 and then you could solve for x, we know this because at the intercept y=0. The opposite can be done for the y-intercept, x=0.

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