C3: How do you reflect equations in lines other than the axis?

Hi,

I know how to reflect graphs in either the x or y axis, but I'm not sure what to do when it's not the x or y axis. For example, If reflecting

y=2^x

in x=5 do I first need to translate the graph to the right by 5?, then make the x negative?

Reply 1

gavinlee

OP

Original post by reubenkinara

Shift 5 units to the left, so the line of reflection becomes the y axis, then flip, and finally remember to shift 5 units back to the right to put the center line back where it belongs.

Hi - thanks for your help.

Shift 5 units to the left:

y=2^{x+5}

flip it:

y=2^{-x+5}

shift 5 units to the right:

y=2^{-x}

Is this right? I don't think I've understood you correctly because I don't think I've done the right thing :frown:

Reply 2

F1Addict

12

Original post by gavinlee

shift 5 units to the right:

y=2^{-x}

This bit is wrong. If

f(x)=2^{-x+5}

then to shift that 5 units to the right, you want

f(x-5)

. Note that

f(x-5)\not=2^{-x}

Spoiler

(edited 11 years ago)

Reply 3

gavinlee

OP

Original post by F1Addict

This bit is wrong. If

f(x)=2^{-x+5}

then to shift that 5 units to the right, you want

f(x-5)

. Note that

f(x-5)\not=2^{-x}

Spoiler

F1Addict, thanks for posting the general rule, it's really helpful. You'd think the text book would also think to print it!

Do you know the general rule for reflections in the y=a line? Again...not in my book.

Reply 4

atsruser

11

Original post by gavinlee

Hi,

I know how to reflect graphs in either the x or y axis, but I'm not sure what to do when it's not the x or y axis. For example, If reflecting

y=2^x

in x=5 do I first need to translate the graph to the right by 5?, then make the x negative?

To figure this out systematically, consider that if the line

x=5

was where the y axis was located, then everything would be easy: we'd simply do an

x \rightarrow -x

transform.

We can arrange this by creating a new coordinate system with a variable

u

that is 0 when

x=5

. We do this by the substitution

u = x-5

. We now have our original coord system,

(x,y)

and our new one

(u,y)

where the y measurements are the same in both systems.

So now, our problem is to reflect in the new

u

axis (which corresponds to the old

x=5

line), and we do that with the substitution

u \rightarrow -u

. But first, we need to transform our

y=x^2

curve to

u

coords:

u=x-5 \Rightarrow x=u+5

so

y=x^2 \rightarrow y =(u+5)^2

Now we reflect in the

u

axis:

y=(u+5)^2 \rightarrow y = (-u+5)^2

But we really want the equation in terms of

x

, so we'd better transform back to

x

land:

y = (-u+5)^2 \rightarrow y = (-(x-5)+5)^2 = (-x+5+5)^2 = (-x+10)^2

In general, if we reflect in

x=a

, we create the

u

axis on the line

x=a

:

u = x-a

Then we transform our function to

u

coords:

y=f(x) \rightarrow y = f(u+a)

Then we reflect in

x=a

/

u=0

via

u \rightarrow -u

:

y=f(u+a) \rightarrow y=f(-u+a)

Then we return to

x

coords, which is where we really want to be:

y=f(-u+a) \rightarrow y =f(-(x-a)+a) = f(-x+a+a) = f(-x+2a)

or:

y=f(-(x-2a))

as an earlier post expressed it.

(edited 11 years ago)

Reply 5

F1Addict

12

Original post by gavinlee

F1Addict, thanks for posting the general rule, it's really helpful. You'd think the text book would also think to print it!

Do you know the general rule for reflections in the y=a line? Again...not in my book.