Hi i was wondering if i could get help with finding the equation of these exponential functions. I am able to find a simple translation but i do not know how to find a equarion with more than one transformation at play.
Definitely a pretty tough question! We will need to make some approximations, and work through each logically. First, we can assume that the function will look something like
Where a and b are constants. The graphs could actually be any number of functions (eg. there could be a constant before x, or before e), but I believe that this is the most standard type of shift that will be assessed. We can quickly find b; that's going to be the value that the function was shifted UP by. Clearly, it's going to be 2, giving us the function.
Now, we can plot a point to find the value of a. When x=0, y=5 (from the graph). Therefore
This gives us a function
Let's test another point, to check that this function is correct. We know that at x=-1, y=3.
Now, the right hand side equals 4.1, not 3, therefore we have found the wrong equation. Looks like we need to do this entire question properly, which is a bit more difficult. I'll leave the above for reference.
Our new function is going to look like the following
Now, when x=0, y=5. So,
When x=-1, y=3. So,
When x=1, y=11. So,
Also, when x approaches negative infinity, y approaches 2. So,
This last equation gives us d=2 (as e to the negative infinity is zero!).
Now, we need to have a crack at the simultaneous equations. I want to get rid of a, and so I'll divide the first equation by the second. Note that I will bring d (2) to the left hand side before dividing through, so I can easily cancel out the 'a' term.
Great! We managed to grab a value for b; our equation now looks like this;
Now, I want to KEEP the c term, so dividing isn't going to work. The two equations I think we could use are
(sorry I can't figure out how the number the equations; hopefully it's clear-ish!).
The problem is, if we multiply the terms we keep both a and c, and if we divide we lose both a and c. Using wolfram alpha, turns out that a=3, c=0. Can't directly see why (actually, it has to do with indice laws; trying expanding out the indice, and simplifying the e^log(3) etc!).
This gives us the final equation,
That was an absolute trekk and a half. I hope that you could follow that working; I really don't think this could possibly be examined, so I'm not going to go into more detail. Give the second one a go using a similar method, and post your working if you still need help!