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RE: [apcalculus] larson pg 137 exercise 69
Posted:
Oct 17, 2012 1:39 PM


NOTE: This apcalculus EDG will be closing in the next few weeks. Please sign up for the new AP Calculus Teacher Community Forum at https://apcommunity.collegeboard.org/gettingstarted and post messages there.  Dwayne,
Start with (I) x^2  kx = 5x  4 (the yvalues must be the same at the point of tangency)
and
(II) 2x  k = 5 (the derivatives must also be the same at the points of tangency).
Then solve for k in (II). (IIa) k = 2x  5
Substitute into (I) (III) x^2  (2x5)*x = 5x  4
Simplify x^2  2x^2 + 5x = 5x  4 x^2  4 = 0 Factor (x  2)(x + 2) = 0 x = 2, 2
We are not done! Substitute back into (IIa) to solve for k. k = 2(2)  5 = 1 k = 2(2)  5 = 9
It seems that you just forgot to solve for k after finding x. Let me know if you have any questions!
Sincerely, Mat
Original Message From: Dwayne Wellington [mailto:mrdw27@gmail.com] Sent: Wednesday, October 17, 2012 8:29 AM To: AP Calculus Subject: [apcalculus] larson pg 137 exercise 69
NOTE: This apcalculus EDG will be closing in the next few weeks. Please sign up for the new AP Calculus Teacher Community Forum at https://apcommunity.collegeboard.org/gettingstarted and post messages there.  Good morning, I was working on exercise 69 from Larson early Transcendentals and am missing the key to answering the question.
Find the value of k such that the line is tangent to the graph of the function.
(function) f(x) = x^2  kx (line) y = 5x 4
the answers are x= 1, 9 but I can't seem to get that when I work out the problem. Thank you for the assistance in advance.  To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus
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 To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus



