Author Topic: cherylim23's methods question thread :) (Read 15665 times) Tweet Share

m@tty · « **Reply #105 on:** January 17, 2011, 04:31:53 pm »

Quote from: cherylim23 on January 17, 2011, 02:58:20 pm

/Edit: Also, comparing these two graphs: $f(x) = \frac{1}{(x-4)^2}$ and $g(x) = \frac{1}{(x-4)^4}$
What's the transformation from f(x) to g(x)?

Thanks for reading

$g(x)=\left[f(x)\right]^2$

|ll|lll| · « **Reply #106 on:** January 17, 2011, 05:06:51 pm »

^ Is that a dilation or some sort?
Because the graphs: $y = \log_{10}(x-3)^2$ and $y = 2\log_{10}(x-3)$ are technically speaking, equivalent of each other, but the latter is dilated by a factor of 2 from the x-axis...

nacho · « **Reply #107 on:** January 17, 2011, 05:44:35 pm »

Are these recent questions from MM or essentials? And what chapter?

vea · « **Reply #108 on:** January 17, 2011, 07:51:48 pm »

Quote from: cherylim23 on January 17, 2011, 05:06:51 pm

^ Is that a dilation or some sort?
Because the graphs: $y = \log_{10}(x-3)^2$ and $y = 2\log_{10}(x-3)$ are technically speaking, equivalent of each other, but the latter is dilated by a factor of 2 from the x-axis...

They are algebraically the same but graphically different... Not so sure how to explain this.

brightsky · « **Reply #109 on:** January 17, 2011, 07:55:18 pm »

They are both the same graph.

evaever · « **Reply #110 on:** January 17, 2011, 08:04:50 pm »

Quote from: cherylim23 on January 17, 2011, 05:06:51 pm

Because the graphs: $y = \log_{10}(x-3)^2$ and $y = 2\log_{10}(x-3)$ are technically speaking, equivalent of each other, .....

They are different, the square one has two branches

dude · « **Reply #111 on:** January 17, 2011, 08:06:54 pm »

I think the first one is defined for x < 3, but in the domain (3,infinity) they are the same

Andiio · « **Reply #112 on:** January 17, 2011, 08:07:33 pm »

IMO It's an issue of the domains, but they are the same graph

|ll|lll| · « **Reply #113 on:** January 17, 2011, 08:42:42 pm »

That wasn't what I meant.

Okay, maybe that example wasn't good enough.

But you would say that $y=\log_{10}(x-3)^{-1}$ and $y=-\log_{10}(x-3)$ are identical to each other algebraically and graphically.
Since the latter is reflected in the x-axis, would you say that the former is reflected in the x-axis too (although it doesn't directly imply so)?

dude · « **Reply #114 on:** January 17, 2011, 08:45:24 pm »

Is reflected in the x-axis the same as across the x-axis?

schnappy · « **Reply #115 on:** January 17, 2011, 08:57:49 pm »

$y = \log_{10}(x-3)^2$

$y = 2\log_{10}(x-3)$

As said they are the same, however the first is defied for x<3 because you sqaure a -ve you get a +ve, so you can take the log of that number. You should get the same graph, but also the same graph vertically reflected along the asymptote.

evaever · « **Reply #116 on:** January 17, 2011, 09:14:19 pm »

y=-logx is the reflection of y=logx in the x-axis, true

y=logx^-1 is the same function as y=-logx, so it is the reflection of y=logx in the x-axis, also true

|ll|lll| · « **Reply #117 on:** January 17, 2011, 09:28:27 pm »

^Thanks evaever

Quote from: schnappy on January 17, 2011, 08:57:49 pm

$y = \log_{10}(x-3)^2$

$y = 2\log_{10}(x-3)$

As said they are the same, however the first is defied for x<3 because you sqaure a -ve you get a +ve, so you can take the log of that number. You should get the same graph, but also the same graph vertically reflected along the asymptote.

Does this imply that $y = \log_{10}(x-3)^2$ is dilated by a factor of 2 in the x-axis and also reflected in the y-axis or both or neither?

brightsky · « **Reply #118 on:** January 17, 2011, 09:33:39 pm »

Both. (Not y-axis but yeah. :p)

evaever · « **Reply #119 on:** January 17, 2011, 11:30:43 pm »

Does this imply that $y = \log_{10}(x-3)^2$ is dilated by a factor of 2 in the x-axis and also reflected in the y-axis or both or neither?
[/quote]

No, because $y = \log_{10}(x-3)^2$ is not the same function as $y = 2\log_{10}(x-3)$

Keep in mind you are comparing with $y = \log_{10}(x-3)$

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