Transformations with exponential and logarithmic functions

Question

F(x)=2^x+2

Answer 1

Hi Jazlyn,

 

Did the function need to be graphed on a set of axes?  If so, maybe I can walk you through it.  The answer from Shaung will give you the inverse function since it solved for x but didn't address the topic of transformation.

 

First off - it is a transformation of the graph f(x)=y=2^x (I will use y instead of f(x) which just says that the value of y is a function of the independent variable x).

 

Do you know the shape of an exponential function? Hard to explain, but it is a curve passing through (0,1) and moving toward y=>infinity when x=>infinity.  On the negative side of x, y=>0 as x=>-infinity but never touches the y-axis.  The entire graph is on the positive y side.

 

Since the base is 2, y=2^x will go through the points (-1,1/2), (0,1), (1,2), and (2,4) for starters.  since you are interested in y=2^x+2, think about the x values you need to get the y values above...

 

When x=-3 (x+2=-1), y=1/2.  When x=-2 (x+2=0), y=0.  When x=-1 (x+2=1), y=2.  When x=0 (x+2=2), y=4.  I hope you see that what is happening in this transformation is that the entire graph is moving 2 units to the left.  Another way of thinking of this is that the x axis is shifted 2 to the right, so that the "zero" value is now at x=-2.

 

You transformed the x starting location from 0 to -2.

 

Likewise if given y=2^x+2+3, you would need to add 3 to the original y values and this will shift the graph UP 3 units.

 

In general:  if you have y=b^x-h+k, this is a transformation of y=b^x with a (+h) shift in x and a (+k) shift in y.

 

Logs are similar:  if you want y=log_b(x-h)+k, this is a transformation of y=log_b(x) with a (+h) shift in x and a (+k) shift in y.

 

As a matter of fact - all linear transformations work this way and using h and k are common.

Answer 2

log f(X) = log 2^(x+2)

           = (x+2) log 2

so log f(X) = xlog2 + 2log2

x = (log f(x) - 2log2) / log2

 

hope that helps!