Lnx graph confusion

Hi, for the equation y=ln(4-x) why doesnt the graph cross the negative x axis since +4 shifts the graph to the left. I first reflected the graph over the y axis. I found that you have to changethe equation to this : y=ln-(x-4) ... Why ?thanks

Reply 1

RichE

Original post by coconut64

It's going to cross the x-axis at the solution of ln(4-x)=0.

Also the transformations that take y=lnx to y=ln(4-x) are

reflection in the y-axis
translation to the right by 4

Reply 2

coconut64

Original post by RichE

It's going to cross the x-axis at the solution of ln(4-x)=0.

Also the transformations that take y=lnx to y=ln(4-x) are

reflection in the y-axis
translation to the right by 4

My question is why is the translation to the right not to the left as its +4 thanks

Reply 3

NotNotBatman

Original post by coconut64

I asked this question last year,

Original post by notnek

Let's try that.

Start with

f(x) = \ln x

"Reflect the graph in the y axis" - this is a transformation

f(x)\rightarrow f(-x)

f(-x) = \ln (-x)

Call this new function

g(x)

then "shift the graph 4 units in the negative x direction"

is a transformation

g(x)\rightarrow g(x+4)

g(x) = \ln (-x)

g(x+4) = \ln -(x+4) = \ln (-x -4)

Can you see your mistake?

Reply 4

coconut64

Original post by NotNotBatman

I asked this question last year,

The last part is the bit i dont get. How does it go from g(x+4) to g-(x+4) thanks

Reply 5

RichE

Original post by coconut64

My question is why is the translation to the right not to the left as its +4 thanks

reflection in the y-axis takes it to y=ln(-x)

Translation to the right by 4 takes it to y = ln(-(x-4)) = ln(4-x)

Reply 6

NotNotBatman

Original post by coconut64

The last part is the bit i dont get. How does it go from g(x+4) to g-(x+4) thanks

It's g[-(x+4)] = g(-x-4)

When you multiply by -1, you have to do it to the whole quantity in the bracket.

Reply 7

coconut64

Original post by NotNotBatman

It's g[-(x+4)] = g(-x-4)

When you multiply by -1, you have to do it to the whole quantity in the bracket.

Why dovyou multiply it by -1?

Reply 8

NotNotBatman

Original post by coconut64

Why dovyou multiply it by -1?

After the reflection you shifted the graph 4 units to the left, so replace any x with an x+4, so f(-x) becomes f(-(x+4)) which is wrong.

Reply 9

B_9710

You can do the transformations in which ever order you're most comfortable with but depending on which one you do first the second transformation may be different in each case, you can either see it as a translation by vector (-4 0) followed by a reflection in the y axis OR you can say it is a reflection in the y axis followed by a translation by vector (4 0). Notice that the translations in each case are not the same.
Either of these are correct.

Reply 10

Zacken

Original post by coconut64

My question is why is the translation to the right not to the left as its +4 thanks

You start with this:

Let's start with

f(x) = \ln x

. If you want to do

f(x+ 4)

then you get

f(x+4) = \ln (4 + x)

which shifts it to the left. But

\ln (4+x) \neq \ln (4-x)

(unsurprisingly).

So to get

f(x) = \ln (4-x)

you need to do two things:

1.

f(-x) = \ln (-x)

which reflects the graph in the y-axis. This gets you this:

This is your new graph. Let's call it

g(x) = \ln (-x)

. The graph of it is as above. Now you are going to do

g(x - 4) = \ln (-(x-4)) = \ln (-x + 4) = \ln (4-x)

. This will shift the graph to the right by 4 units.

The reason why

g(x -4) = \ln (4-x)

is because you need to replace the the

x

g(x)

x-4

.

That is, pretend you have

g(x) = \ln (-(x))

. Then you need to replace

(x)

with

(x-4)

.

This gets you

f(x-4) = \ln (-(x-4)) = \ln (-x + 4)

.

If you had

\ln x

being transformed to

\ln 2x

then a shift, say

f(x) = \ln 2x

and you wanted to shift it 3 units to the left you'd do

\ln (2(x + 3)) = \ln (2x + 6)

. Not

\ln (2x + 3)

. Capiche?

Reply 11

coconut64

Original post by Zacken

You start with this:

Let's start with

f(x) = \ln x

. If you want to do

f(x+ 4)

then you get

f(x+4) = \ln (4 + x)

which shifts it to the left. But

\ln (4+x) \neq \ln (4-x)

(unsurprisingly).

So to get

f(x) = \ln (4-x)

you need to do two things:

1.

f(-x) = \ln (-x)

which reflects the graph in the y-axis. This gets you this:

This is your new graph. Let's call it

g(x) = \ln (-x)

. The graph of it is as above. Now you are going to do

g(x - 4) = \ln (-(x-4)) = \ln (-x + 4) = \ln (4-x)

. This will shift the graph to the right by 4 units.

The reason why

g(x -4) = \ln (4-x)

is because you need to replace the the

x

g(x)

x-4

.

That is, pretend you have

g(x) = \ln (-(x))

. Then you need to replace

(x)

with

(x-4)

.

This gets you

f(x-4) = \ln (-(x-4)) = \ln (-x + 4)

.

If you had

\ln x

being transformed to

\ln 2x

then a shift, say

f(x) = \ln 2x

and you wanted to shift it 3 units to the left you'd do

\ln (2(x + 3)) = \ln (2x + 6)

. Not

\ln (2x + 3)

. Capiche?

Ah, okay that does ring a bell. I think I get it now, I will just do more practice to make it stick in my head. Thanks for your very detailed answer. Also, thanks for helping me with mechanics previously, I managed to get 93 ums!

Reply 12

coconut64

Thanks guys I get it now, after a long time.

Reply 13

Zacken

Original post by coconut64

Basically, the best thing to do is to always stick a bracket around (x) in the function when you see it, or at least do so whilst learning about it - you'll get used to it after .

Like shifting

\sqrt{2x}

by 5 units. Write it as

\sqrt{2(x)}

then do the shift as

\sqrt{2(x +5)}

.

(And congratulations on your mechanics! 93 UMS is fantastic. Glad I helped)

Reply 14

coconut64

Original post by Zacken

Basically, the best thing to do is to always stick a bracket around (x) in the function when you see it, or at least do so whilst learning about it - you'll get used to it after .

Like shifting