y =-2f(2x+4) + 0
y =-2f[2(x+2)] + 0
So for the coordinates (x, y) x is x/2 -2
My Q. is, how did the x become x/2 -2 ???
[sarcasm]Here is an example of a clearly worded question.[/sarcasm]So for the coordinates (x, y) x is x/2 -2
You probably mean that one considers some point (x, y) lying on the graph of y = -2f(x). The question is, for what x' the point (x', y) lies on the graph of y = -2f(2x + 4)? This happens when 2x' + 4 = x, i.e., when x' = x/2 - 2.
[arrogance] + "[sarcasm]Here is an example of a clearly worded question.[/sarcasm]" [/arrogance]
Here is an example of sarcasm + arrogance. First of all, the latex compiler, is not working. At least not for me at the moment. Therefore, explaining Qs is bound to be less clear. Second of all, no it is not what I mean. I was wondering about the process of "isolating" x. Thirdly, I don't like sarcasm. Sarcasm is for kids. So if u can't comment without it, then don't answer Qs.
Your question still isn't clear. You can't solve for x if you don't know the form of the function. I find the comment about x being replaced by x/2 - 2 to make things even more unclear, since you have said emakarov (sarcasm aside) was wrong.
What exactly is your question asking? Can you give us another example of a solved problem, perhaps?
-Dan
Wanted to keep things short as possible to avoid confusion. Here is the solved equation:
y =ac^x
y =-2(3)^2x+4
y =-2f[2x +4] + 0 where f(x) =3^x
y =-2f[2(x+2)] + 0
The y value for coordinate on a graph (x, -2) is easy to see. It is -2
Now the x value is going to be inside the [2(x+2)]. I'm a little confused on the process of breaking it down to get x. I can tell you that the (x+2) part becomes x-2 when plotted on graph, but that is about it.
The typical way to solve this kind of equation is to, say, take the natural log of both sides, but this runs into some troubles because of that negative sign in front of the 2. So I'll do it this way:
and I'll let you do the rest from there. Notice that you cannot have a non-negative y value.
Does that answer your question, or do you need something else from this?
-Dan