The Transformation y = f(bx) - Problem 2


Let’s graph another transformation. I want to graph y equals 4 minus 2 to the ½x plus 1. Now this is a transformation of the parent function; y equals 2 to the x. Now recall that that function looks like this. It’s an upward sweeping curve, it passes through the point 0 1 and 1 2. And it’s got this horizontal asymptote at y equals 0. That means that as we go up to the left the graph gets closer and closer to the line y equals 0. So I have to keep in mind as I’m graphing both as I’m graphing the transformation. But first can we make a table of data for this parent function, and I’ll use the variable u and 2 to the u.

I usually use nice easy values like -1 0 and 1. 2 to the -1 is ½. 2 to the 0 is 1, 2 to the 1 is 2. And I want to record that y equals 0 is the horizontal asymptote. Because I’m going to plot the transformation to the asymptote as well us to the points.

I have x and I’m going to rewrite this as -2 to the ½ x plus 1 plus 4. I’ll put the plus 4 at the end. First of all I, want to rename this part u. So if u equals 1/2x plus 1 what is x equal? I subtract 1 for both sides and I get 1/2x and then I multiply both sides by 2. And I get I’ll write down here x equals 2 u minus 2.

So this is how I’m going to get my x values I’ll take the u values, double them and subtract 2. So if I double this is get -2 and subtract 2 I get -4. If I take this and double it I get 0 subtracting 2 I get -2. If I take 1 and double it I get 2 minus 2 is 0.

Now what about the y values? Well I’ve got 2 to the u, I have to take the opposite of 2 to the u and add 4. So the opposite of a half is -1/2 plus 4 ,3.5. The opposite of 1 is -1 plus 4 is 3. The opposite of 2 is -2 plus 4 is 2.

And let’s not forget the horizontal asymptote. Since these are all y values I can plot the same transformation to this y value. So the opposite of 0 is 0 plus 4 is 4. So y equals 4 will be my new asymptote. And in fact that’s the first thing I want to graph.

I will graph my asymptote at y equals 4. Let me pick another color for that. So this is going to be my horizontal asymptote. And then I’ll fill in my points; -4, 3.5. So here is -4, this is 4 so 3.5 is half way between 3 and 4, right about there, -2, 3, -2 is here and 3 is here. So and 0 2 that’s here. Looks like I can use one more point. So why don’t I use let’s say 2. 2 to the 2 is 4.

Remember I take the u values, I double them and subtract 2. So 2 times 2 is 4, minus 2 is 2. And I take this y value, take its opposite. And add 4 so I get minus 4 plus 4, 0. So 2,0 is another point and that’s this point here. That will give me a pretty good graph. So let me just draw, there it is. Asymptote at all we have this horizontal asymptote y equals 4 and this downward curving graph. And it curves downward because of this minus sign.

That’s a reflection across the x axis. It's been shifted up 4 units. That’s where the asymptote is up at y equals 4 rather than at 0. And also there’s been some horizontal shifting and horizontal stretching as well. So you could see that in the effect of the graph being moved a little bit it’s been made a little bit wider. So this is our final graph.

exponential functions horizontal shift vertical shift horizontal asymptotes reflection