transformation problem

[khorven]

Hi. First of all the problem at hand:

Identify h and k, then write the equation of the transformed function in the form y - k = f(x - h)

f(x) = 1/x, translated 5 units to the left and 4 units up.

When I check the answers it seems 1/x changes to just x. Why is that? Also, how does the f part of the equation translate? From f(x) = x whatever to y = f(x - h) + k?

 

[stapel]

Identify h and k, then write the equation of the transformed function in the form y - k = f(x - h)

f(x) = 1/x, translated 5 units to the left and 4 units up.

When I check the answers it seems 1/x changes to just x. Why is that?

I have no idea. Please reply with the full and exact "answer" for this exercise, so we can try to figure out what they're trying to say (or if maybe there's a typo in the solutions manual). Thank you!

Also, how does the f part of the equation translate? From f(x) = x whatever to y = f(x - h) + k?

I'm sorry, but I don't understand what you're saying here...? The "f part of the equation" is the name of the particular function, f(x). You were given that f(x) = 1/x, not that f(x) = x. And, by "translate", they mean "move around, according to the rules" that they covered in the book and in class. (here) Try applying those rules.

 

[Ishuda]

Hi. First of all the problem at hand:

Identify h and k, then write the equation of the transformed function in the form y - k = f(x - h)

f(x) = 1/x, translated 5 units to the left and 4 units up.

When I check the answers it seems 1/x changes to just x. Why is that? Also, how does the f part of the equation translate? From f(x) = x whatever to y = f(x - h) + k?

Hints:
When you have a curve given by y= f(x) a translation of that curve up or down just means you have the same curve shape but moved up or down respectively. So, assuming k is positive, if the curve is moved up, y is larger or y= f(x) + k. If the curve is moved down, y is smaller or y=f(x)-k (again assuming k is positive). Think of holding on to a point of the curve and moving it up, then looking at the y value and seeing it is larger than before (or moving it down and ...).

When you have a curve given by y= f(x) a translation of that curve to the left or to the right just means you have the same curve shape but moved left or right respectively. So, assuming h is positive, if the curve is moved to the left, the argument of f gets smaller [EDIT: so you have to increase it to keep the same argument for the point and x is replaced by x+h]. If the curve is moved the right, the argument of f gets larger [EDIT: so you have to decrease it to keep the same argument for the point and x is replaced by x-h] (again assuming h is positive). Think of holding on to a point of the curve and moving it to the left, then looking at the x value and seeing it is smaller than before (or moving it to the right and ...).

 

[khorven]

Here is the answer from the solutions manual:
h = -5, k = 4; y - 4 = f(x + 5)

And my second question is if y = x, and f(x) = x, then how does the f move to the other side of the y when you go y = f(x), or y = f(x - h) + k?

 

[stapel]

Here is the answer from the solutions manual:
h = -5, k = 4; y - 4 = f(x + 5)

What rule did you apply to the instructions to "move left/right" and "move up/down" by the given amounts?

And my second question is if y = x, and f(x) = x, then how does the f move to the other side of the y when you go y = f(x), or y = f(x - h) + k?

I'm sorry, but I don't know what this means...? Is this for a different exercise (distinct from what you'd originally posted, where y = 1/x)?

Note: For ALL y and f(x), y = f(x), because "f(x)" is just another name for "y". This is not restricted to the one particular function "y = x".

 

[khorven]

What rule did you apply to the instructions to "move left/right" and "move up/down" by the given amounts?

I said y - 4 = f(1/x + 5). But, first of all, that doesn't look right on my calculator. It doesn't translate to the left. Also my teacher came up with the same answer as the solutions manual but wasn't able to explain it to me. I'm in correspondence so the teacher is not as primed on the material as usual.

I'm sorry, but I don't know what this means...? Is this for a different exercise (distinct from what you'd originally posted, where y = 1/x)?

Note: For ALL y and f(x), y = f(x), because "f(x)" is just another name for "y". This is not restricted to the one particular function "y = x".

Okay I think I see.

 

[khorven]

Mind you, I might be using the calculator incorrectly and putting two vertical translations in. Does anyone know how to do that sort of thing?

 

[ksdhart]

I'm working a little bit blindly here, but from what I can tell, the difficulty is in translating the function left and right. The equation you're trying to graph is \(\displaystyle y-4=f\left(x\right)+5\) where \(\displaystyle f\left(x\right)=\frac{1}{x}\). Now, instead of passing in just x as the argument to the function f, you're passing in (x+5). It looks like you've attempted that and gotten this as a result:

\(\displaystyle y-4=\frac{1}{x}+5\) or \(\displaystyle y-9=\frac{1}{x}\).

Can you see why that's not what you want? Think about what it means to pass in a different argument to a function. If you wanted to evaluate f(a) what would the function look like? If you wanted to evaluate f(q-1)? Then apply the same logic when evaluating f(x+5) and you're set.

 

[khorven]

I got it! Haha. Thanks, bro.