Post by: peterpan101 on November 19, 2012, 12:03:49 pm
Post by: nisha on November 19, 2012, 12:36:03 pm
Post by: Lasercookie on November 19, 2012, 12:46:18 pm
Anyway it doesn't matter if it is or isn't, it's pretty much the same process you'd take for any graph. To sketch this, just note what the base graph would have been, and then look at each of the transformations applied to it to get the graph we want.
In this case the base graph would be something like
So it's going to be some kind of exponential graph, with a few transformations
We can see that the transformations on the graph are
- Translated down 3 units parallel to the y-axis
- Dilated along the y-axis by a factor of 2
- Reflection in the x-axis (so it's going to be flipped upside down)
- Dilated along the x-axis by a factor of 1/2
So that gives us a pretty good idea of what the graph will look like. (an upside down exponential graph that's been moved down by three units)
Before drawing those in, let's find the intercepts and asymptotes.
What do we know about
Due to that asympote there won't be any x-intercepts (it's an upside down exponential graph). We can find the y-intercept by subbing in x = 0
So we can mark that down, and just sketch in the general shape. For showing the dilations just have the graph a bit more squished or stretched than it would have been otherwise.
Post by: charmanderp on November 19, 2012, 12:48:02 pm
Post by: peterpan101 on November 19, 2012, 12:50:44 pm
sketch 2log10(x-3)
Post by: Hancock on November 19, 2012, 12:50:54 pm
It's just an exponential curve.
So we know that the horizontal asymptote will occur when y = -3.
It's been reflected in the x-axis due to the -2. N
Now all we need are the x and y intercepts.
y-int occurs when x = 0. Therefore y(0) = -2 - 3 = -5
x - int occurs when y = 0 Therefore, once you solve this equation, you will find there is no x-intercept. (you will get log_10 (-2/3), which doesn't exist in the real number line).
Post by: Lasercookie on November 19, 2012, 12:52:58 pm
thanks for your help, I have another oneI'm going to test you here :P can you give us the outline of some working on how you think you'd start off and from there we can figure out at what step you're having trouble
sketch 2log10(x-3)
Post by: peterpan101 on November 19, 2012, 01:09:44 pm
thanks for your help, I have another oneI'm going to test you here :P can you give us the outline of some working on how you think you'd start off and from there we can figure out at what step you're having trouble
sketch 2log10(x-3)
ok so you would get the inverse
x=log10(y-3)^2
converte to index so
10^x= (y-3)
now im stuck !
Post by: Jenny_2108 on November 19, 2012, 01:15:37 pm
thanks for your help, I have another oneI'm going to test you here :P can you give us the outline of some working on how you think you'd start off and from there we can figure out at what step you're having trouble
sketch 2log10(x-3)
ok so you would get the inverse
x=log10(y-3)^2
converte to index so
10^x= (y-3)
now im stuck !
The shape of log10 graph is just similar to loge, so you just need to find vertical asymptote and x-intercept