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May 27th, 2009, 07:36 AM   #1
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equation of a secant

Hello, I was wondering if I could ask for your help for this one questions. I have to determine the general equation of a secand from any point (a, f(a)) to the pont f (x) = 4^x where x = 2 and then i have to describe how i would use this to find the equation of the tangent at the point (2,f(2)). I'm really stuck and can't figure it out, I have m = (f(a) - f(16)/a - 2) (x - 2) + 16. But i know that this is way wrong, and i'm hoping one of you guys or girls could give me a nudge in the right direction. Thank you.
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May 27th, 2009, 08:01 AM   #2
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Re: equation of a secant

The tangent is the limiting secant as a point on the curve, (a,f(a)) approaches the general point (x,f(x)), or, specifically in this case, (2, f(2)) = (2,2^4), or (2,16).

The general equation of the secant would be found as (y-16)/(x-2) = (f(a)-16)/(a-2)
For tangency, a -->2 in the limit, so you want to find lim[a-->2] (f(a)-16)/(a-2), which is lim[a-->2] (a^4-16)/(a-2).

Try factoring. You should wind up with (y-16)/(x-2) = 64 if I'm right.
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May 27th, 2009, 02:09 PM   #3
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A couple of mistakes there methinks.
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May 27th, 2009, 03:46 PM   #4
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Re:

Quote:
Originally Posted by skipjack
A couple of mistakes there methinks.
e.g. using a^4 rather than 4^a?

Oh, well, a bad hair day all around as the lawn tractor also went belly up. Things happen. How about removing a factor of 16, then using L'Hopital?
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