Logarithms graphing problem - can't get help now :(

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I'm having a lot of trouble right now and I have a test tomorrow. The problem being. my notes' examples don't seem to add up and the book is just absolutely useless, so i thought the GTP community may be able to help

This is what i need to figure out. When your graphing an equation such as: y=log4(x-2)+4

-what is the original reference point? My teacher said something like the reference point being (1,0), but im not sure.

-What do each of the 4's do? I know they move the reference point, but im still confused.

-When your reference point is to the right of the asymptote, in what direction should the line be graphed?

I'm really getting tired of this teacher. I pass in all my homework, but can't seem to get more than between 60-75 on tests because her idea of a chapter test, is FIVE FREAKIN QUESTIONS and she gives us a quiz every Friday. Its like she purposely wants to completely kill my grade and GPA. This is really starting to piss me off.

If you have any more information that may be helpful, id be glad to hear it also. Thank you
 
TEAMDiablo3
I'm having a lot of trouble right now and I have a test tomorrow. The problem being. my notes' examples don't seem to add up and the book is just absolutely useless, so i thought the GTP community may be able to help
Click to expand...

This is likely coming too late for you, since the school day has probably started by now. Anyways....

This is what i need to figure out. When your graphing an equation such as: y=log4(x-2)+4
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Stop right there... I'm confused by your notation.
For starters, is this in log base 10, or 4? y = log10 4(x-2)+4 or y = log4 (x-2)+4
I'm assuming you don't mean a natural logarithm. y = loge 4(x-2)+4 = ln 4(x-2)+4

If it's in log base 10, then what are you actually taking the log of? I need to see another pair of brackets...
Is it y = [log10 4(x-2)]+4, or y = log10 [4(x-2)+4] = log10 [4(x-1)]

Each of the above considerations makes a significant difference in the appearance of the plot.


Anyways, I'm hoping it's y = [log4 (x-2)]+4 because that most closely represents the sort of thing I saw in highschool. I'm going to assume that through the rest of the reply.

-what is the original reference point? My teacher said something like the reference point being (1,0), but im not sure.

-What do each of the 4's do? I know they move the reference point, but im still confused.
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In y = loga (x) there is always a vertical asymptote at x=0, and a root at x=1. ie. your reference point is (1,0) as your teacher says.

The (x-2) shifts everything 2 units in the positive x direction. Your asymptote is relocated to x=2, and your new reference point is (3,0).

The +4 outside the brackets shift things in the positive y direction. The asymptote is unaffected, but your reference point now moves to (3,4).

If I've made the correct assumption that it is in log base 4, then the first 4 kind of affects the slope of the curve. I don't know a great way to explain it, so I'm hoping your textbook has a graph to visually compare functions with different log bases. Basically though, as you increase the base the curve flattens out and between your reference point and asymptote it will approach the asymptote more rapidly. If you consider what a logarithm is and how they're computed this is fairly intuitive.

If I've made the wrong assumption and the function is something like y = [log10 4(x-2)]+4, then I have no idea what to say.... that's a weird one and I don't know if there's a straight forward explanation for how it affects the shape and position of the graph. I'd head straight for MATLAB or my TI to plot it, or start plugging in x or y values manually into (10^(y-4))/4 = x-2.

-When your reference point is to the right of the asymptote, in what direction should the line be graphed?
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Depends on the placement of negative signs. See the bottom of the page in this link.
http://www.purplemath.com/modules/graphlog3.htm



edit: If you're ever in doubt when graphing functions, just plot a few points and connect the dots. If you aren't allowed a calculator it can become tricky with logarithmic functions (it's the worst with natural logs), but if you understand how logarithms work and you're dealing with a fairly simple function then you can make life easy by picking friendly x or y values.

For instance... in y = [log4 (x-2)]+4 , rearrange the equation into 4^(y-4) = (x-2). Then you can start plugging in values either for x-2, or for y-4. I'd probably go with the base exponent, (y-4), in this case because it looks likely to give me integer answers for x and y.

For y-4 = 0, y=4, then x-2=1, therefore x=3 -- ie. 4^0 = 1
For y-4 = 1, y=5, then x-2=4, therefore x=6 -- ie. 4^1 = 4
For y-4 = 2, y=6, then x-2=16, therefore x=18 -- ie. 4^2 = 16
For y-4 = 3, y=7, then x-2=64, therefore x=66 -- ie. 4^3 = 64
etc...
 
Thanks for the help, but i do get it a bit more now after talking with my friend and reading your post.

Thank god we had a substitute teacher today and the whole sophomore class had a pizza party (we scored 1st for average test scores on the Florida FCAT in our county) so my class didn't have to take a test.

I'm only in Algebra 2 so were not doing things such as using different points, and approach.

Finals are 9 school days away :grumpy: I should really get a head start on math because i have forgotten how to do a lot of things that i learned and i really need to get a high grade on the final
 
TEAMDiablo3
I'm having a lot of trouble right now and I have a test tomorrow. The problem being. my notes' examples don't seem to add up and the book is just absolutely useless, so i thought the GTP community may be able to help

This is what i need to figure out. When your graphing an equation such as: y=log4(x-2)+4

-what is the original reference point? My teacher said something like the reference point being (1,0), but im not sure.

-What do each of the 4's do? I know they move the reference point, but im still confused.

-When your reference point is to the right of the asymptote, in what direction should the line be graphed?

I'm really getting tired of this teacher. I pass in all my homework, but can't seem to get more than between 60-75 on tests because her idea of a chapter test, is FIVE FREAKIN QUESTIONS and she gives us a quiz every Friday. Its like she purposely wants to completely kill my grade and GPA. This is really starting to piss me off.

If you have any more information that may be helpful, id be glad to hear it also. Thank you
Click to expand...


I'm in pre-calc but can't give you a great explaination like Boundry did but I'll give you the best I can.

I'm assuming that in log base 4, so its the same graph as log base 4 moved 2 units to the right and 4 units up. It's 2 to the right becuase that -2 is inside those parenthenses and 4 up because the 4 is outside the parenthenses. If it was a +2 within the parenthenses it would be two to the left and if it was a -4 outside it would've been 4 down.

The regular graph of log base 4 passes through (1,0), (4,1), (16,2), (64,3) and so on. It has a vertical asymptote at x = 0.
 
hehe, ya. I know it was just a typo.

You described the function y = (x^3) - (81x^2) + (1104x) - 1024, based on those roots
 
Quite a few courses... I'm not a math major (because how much more of a nerd would I need to be to do that), but I'm only 4 courses away from obtaining my B.Sc. in Mechanical Engineering. I believe I've taken or 10 or 11 math courses at the university level: 6 in calculus, and the rest in statistics, numerical methods for solving equations (ie, iterative and/or matrix methods), and advanced methods for creating curves and surfaces patches (b-spline's and beziers, etc).

The last math course I took was on the Fourier Series and on methods for solving partial differential equations, but geared specifically towards situations encountered in engineering practice. I got one of my few A's in that course, but be damned if I can remember any of it now. ;)
 
Boundary Layer
Quite a few courses... I'm not a math major (because how much more of a nerd would I need to be to do that), but I'm only 4 courses away from obtaining my B.Sc. in Mechanical Engineering. I believe I've taken or 10 or 11 math courses at the university level: 6 in calculus, and the rest in statistics, numerical methods for solving equations (ie, iterative and/or matrix methods), and advanced methods for creating curves and surfaces patches (b-spline's and beziers, etc).

The last math course I took was on the Fourier Series and on methods for solving partial differential equations, but geared specifically towards situations encountered in engineering practice. I got one of my few A's in that course, but be damned if I can remember any of it now. ;)
Click to expand...

Awesome! I'm just beginning my college career and Mechanical engineering and Eletrical Engineering are my top two choices. Right now I'm just Engineering as my associates and then will decide which of the two I'm more inclined to.
 

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