# Simple reflection and translation question

1. Mar 11, 2013

### nothingsus

1. The problem statement, all variables and given/known data

If the graph of 1/x is reflected in the y-axis, translated 3/4 units to the right and 2 units up, the resulting graph would have the equation...

A) 1/(x-2) + 0.75
B) 3/(4x) +2
C) 2 - 1/(3/4x)
D) -1/(4x-3) + 2
E) 1/(3-4x) + 2

2. The attempt at a solution

So the general formula of a hyperbola is y = a/(x-b) + c (in this context anyway, I know there's another form)

reflection in the y axis -> y=f(-x)
translated 3/4 right -> b=3/4
translated 2 up -> c=2

So if I put it all together I get
y = 1/(-x-b) + c
y = 1/(-x-0.75) + 2
= 1/-(x+0.75) + 2
= -1/(x+0.75) +2

2. Mar 11, 2013

### Curious3141

When you do the reflection about the y-axis, $y = f(x)$ becomes $y = f(-x) = g(x)$ (say).

When you do the translation to the RIGHT by 0.75, $y = g(x)$ becomes $y = g(x-0.75)$ (note the minus sign here).

Now $g(x - 0.75) = g(-(0.75 - x)) = f(0.75 - x)$ since $g(-x) = f(x)$.

When you finally do the vertical translation upward, you get $y = f(0.75 - x) + 2$.

So the final answer is $y = f(0.75 - x) + 2 = \frac{1}{0.75-x} + 2 = \frac{4}{3-4x} + 2$ after simplification.

If there wasn't a 4 in the numerator in any of the suggested answers, the question is likely wrong.

3. Mar 11, 2013

### ehild

D is not quite correct. It should be y=-1/(x-3/4)+2 or y=-4/(4x-3)+2.
There is a mistake in your derivation:
If you shift a function y=f(x) to the right by A the new function is y=f(x-A). (Think: the new function has its asymptote where it goes to infinity at x=3/4)

ehild