# Translations math problem (gr 12 level)

1. Dec 26, 2008

### Senjai

1. The problem statement, all variables and given/known data
If the function y=f(x) is transformed to -2y-2=f(0.5x-3), and the point (-3, -2) is on f(x), which of the following will be on the new function -2y-2=f(0.5x-3)?

2. Relevant equations
$$y = {af}\left[b(x-h)\right] + k$$

3. The attempt at a solution
When i attempted this question, i got (-9, -1) which is wrong, i don't know the answer still. I am still unsure really how to show my work so i pretty much did the horizontal translations and scale factors to x, and vertical translations and scale factors to y.

after rewriting the function, i have $y = -\frac{1}{2}(0.5x-3) + 2$

in an attempt to show work i showed the transformations.

$x \rightarrow 0.5x$
$x \rightarrow x - 3$
$y \rightarrow 2y$
$y \rightarrow -y{}\textit{(reflection on x axis)}$
$y \rightarrow y - 2$

I then simply tried to run those transformations on the x and y values seperatly.
(-3, -2)

x = -3(2) -3 = -9
y = -((-1/2)(-2)) + 2 = 1?? im lost,

i dont really know how to show my owkrk for this question, nor how to do it properly.
any help would be appreciated.

2. Dec 26, 2008

### Senjai

This was also a multiple choice question. Possible answers were
(-9, -1)
(1, 0)
(0, 0)
(-12, 2)

3. Dec 26, 2008

### HallsofIvy

Staff Emeritus
Saying that "(-3, -2) is on f(x)" means that f(-3)= -2. In order that we be able to use that, without any other knowledge of f, we need to be able to apply f to -3 so would have to have 0.5x- 3= -3 so 0.5x= 0 or x= 0. In that case, we have -2y- 2= f(-3)= -2 so -2y= 0 and y= 0. The point is (0, 0).

Last edited: Dec 26, 2008
4. Dec 26, 2008

### Senjai

wow, i feel really really dumb, makes sense putting it that way.. looked at it on a graph, was able to do it that way, algebraically i wasn't able to see it like that.. so its easy, transformation applied to x = previous x value? i dont understand why you cant sub -3 in for x and use that to transform it though, sorry for being all dumb about it...

~Senjai

5. Dec 28, 2008

### yeongil

That's not right. What happened to the f? If the function $$y = f(x)$$ is transformed to $${-}2y{-}2 = f(0.5x{-}3)$$, then
$$y = {-}\frac{1}{2}[f(0.5x{-}3)] {-} 1$$.

But you don't need to do this at all; HallsofIvy's solution is the way to go.

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