I am an idiot. I can't even do this properly

[flyingpig]

Homework Statement

If the point (6, -12) is on the graph of y = f(x), which point must be on the graph of y = f(-x/3 + 6)?

M/C

a) (-36,-12)
b) (-24, -12)
c) (0, -12)
d) (16,-12)

The Attempt at a Solution

First of all, I am a little confused why doesn't the y value change? Why is it staying as y = -12? Doesn't f(x) change?

Also I tried doing f(-6/3 + 6) = f(-2 + 6) = f(4) = ?

Yeah...

Okay okay okay, let's try this algorithm, and then someone explain to me why this works

-x/3 + 6 = 6

-x/3 = 0

x = 0

 

[SammyS]

You only know one y value: namely, you know that f(6) = -12.

In the case of f((-x/3 + 6) , what must x be in order for you to have f(6) ?

 

[flyingpig]

0...but why does y change?

 

[SammyS]

0...but why does y change?

y doesn't change.

You still have y = f(6) , which is -12.

 

[flyingpig]

But isn't f(-x/3 + 6) a transformation ?

 

[mathgeek4]

But isn't f(-x/3 + 6) a transformation ?

The answer is (0,-12). All you really know here is that (6,-12) lies on y=f(x), but you do not know what is f(-x/3+6)=?. So you can only use the given info. And I think the explanation given by SammyS is the best way to do this.

 

[I like Serena]

But isn't f(-x/3 + 6) a transformation ?

Yes, it is a transformation, but only in the x direction (scale and translate horizontally).

To transform in the y direction you would need for instance:
y = f(x) / 3 + 6.

Or to do both:
y = f(-x/3 + 6) / 3 + 6.

 

[flyingpig]

No, let y = f(x) = -2x

Clearly f(6) = -12, which gives us (-6,12)

Now for a transformation f(-x/3 + 6)

f(-x/3 + 6) = -x/3 + 6, this clearly changes the y-valuse.

 

[Mark44]

No, let y = f(x) = -2x

Clearly f(6) = -12, which gives us (-6,12)

You probably meant (6, -12), although (-6, 12) is also on the graph of this line.

Now for a transformation f(-x/3 + 6)

f(-x/3 + 6) = -x/3 + 6, this clearly changes the y-valuse.

?
Assuming that f(x) = -2x, then f(-x/3 + 6) = -2(-x/3 + 6) = 2x/3 - 12

 

[I like Serena]

No, let y = f(x) = -2x

Clearly f(6) = -12, which gives us (-6,12)

Now for a transformation f(-x/3 + 6)

f(-x/3 + 6) = -x/3 + 6, this clearly changes the y-valuse.

You've chosen an example where you can't really distinguish between the two.

This is a line with a slope, so any horizontal transformation can also be expressed as a vertical transformation, which is basically what you've shown.

 

[flyingpig]

You probably meant (6, -12), although (-6, 12) is also on the graph of this line.

?
Assuming that f(x) = -2x, then f(-x/3 + 6) = -2(-x/3 + 6) = 2x/3 - 12

Yeah so if you do f(-6/3 + 6) = -8

 

[flyingpig]

You've chosen an example where you can't really distinguish between the two.

This is a line with a slope, so any horizontal transformation can also be expressed as a vertical transformation, which is basically what you've shown.

But it's still generalizing that f(-x/3 + 6) has no effect on the vertical component.

 

[I like Serena]

But it's still generalizing that f(-x/3 + 6) has no effect on the vertical component.

I don't quite understand what you're saying here...

But I'll put it like this:

Let y = f(x) = -2x
Clearly f(6) = -12, which gives us (6,-12)

So f(6) = f(-0/3 + 6) = -12

The horizontal transformation is that x=0 is sent to f(-0/3 + 6)=f(6)=-12, which gives us (0, -12).

No change in y coordinate, but the x coordinate of 6 has been transformed to x=0.