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Old 02-07-07, 05:16   #1 (permalink)
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Help with Calculus?

Can anyone solve this for me in y=mx+b form?

- Find the equations of the tangent lines at the points (1,5) and (2,3).

How you got the answer would be greatly appreciated.


P.S. I feel pretty silly asking a detailing forum about this but I know alot of you guys are smart so its worth a shot

function is y = 3 + 4x^2 - 2x^3
 
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Old 02-07-07, 05:29   #2 (permalink)
Not too old to learn.
 
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Re: Help with Calculus?

I feel for you, honestly. I'm one of those that could not comprehend Jr. High algebra......with a brilliant tutor. Sorry I can't be of help.
 
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Old 02-07-07, 05:50   #3 (permalink)
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Re: Help with Calculus?

Giving you the answer will be cheating. :-)

Go to this link for an explanation of the y=mx+b form (which is the slope intercept form).
Straight-Line Equations: Slope-Intercept Form

Walk through the example that is given. If you understand that example, you can derive your answer as posed above.
 
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Old 02-07-07, 05:52   #4 (permalink)
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Re: Help with Calculus?

Alright well I asked my bro and he explained it to me. Thanks anyway guys!
 
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Old 02-07-07, 07:35   #5 (permalink)
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Re: Help with Calculus?

Just in case you're still unclear or anyone else needs to know...here are the steps:

1. Differentiate the function. The derivative will be the slope ("m") of the y=mx+b form, because derivatives are, by definition, the slope of the tangent line.

2. Assuming you're doing this in point-slope form, take the coordinates (in this case first (1,5)) and plug it into the form so that the result is (y-5)=m(x-1), which then you should be able to simplify to get a resulting y=mx+b equation.
 
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Old 02-07-07, 08:30   #6 (permalink)
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Re: Help with Calculus?

y=mx+b is the equation for finding the slope of a line (as mentioned).

That's Algebra not Calculus.

MorBiD
 
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Old 02-07-07, 09:31   #7 (permalink)
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Re: Help with Calculus?

Yep, it's algebra... linear equation by plotting the y-intercept, and plotting another point by using the slope of the line. It could be a course that leads you into Calculus with a little warm-up from Algebra.

Wait a minute... this is a detailing forum. I'm here to de-stress. If anything, we should be discussing chemistry. Like what is the cross-linking molecular structure of polymers vs. simple carnuba wax?
 
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Old 02-07-07, 09:59   #8 (permalink)
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Re: Help with Calculus?

You know, there are threads here that discuss that very topic. But at university my majors required lots of math and no chemistry so I stay out of it
 
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Old 02-08-07, 07:44   #9 (permalink)
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Re: Help with Calculus?

I could tell you the answer but I'd have to kill you afterwards.
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Old 02-08-07, 04:53   #10 (permalink)
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Re: Help with Calculus?

If you still can't get it let me know... I will be checking this post for an hour, if you need it I will give you my cell phone number and I will talk you through the equation/calculations.

Math is one of those things that when you see the path you never forget it. If you don't see the path, then no map in the world can help.
 
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Old 02-08-07, 06:15   #11 (permalink)
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Re: Help with Calculus?

Quote:
Originally Posted by tyymm
If you still can't get it let me know... I will be checking this post for an hour, if you need it I will give you my cell phone number and I will talk you through the equation/calculations.

Math is one of those things that when you see the path you never forget it. If you don't see the path, then no map in the world can help.
Thanks for the generous offer, but I got the answer (turned out to be pretty simple). I really appreciate you wanting to help!

BTW - to the previous posters, this is Calculus, you need limits to solve the problem.
 
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Old 02-08-07, 08:21   #12 (permalink)
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Re: Help with Calculus?

limits? Hum, have you all learned differentiation yet? It'll get a lot easier after that. You won't have to worry about that long formula with limits and all.
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