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RE: [apcalculus] larson pg 137 exercise 69
Posted:
Oct 18, 2012 8:53 AM


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.  I think assume you mean k= 1 and k =9.
For line to be tangent to curve they must share a point and the derivative of the curve at that point must match the slope of the line. So it turns into a little systems problem.
Take the derivative of f and get f'(x)=2xk. This has to equal the 5 from the line so 2xk=5. Now solve for x. x=(5+k)/2 can now be substituted into the line's equation to get y=(17+5k)/2. Take both of these and substitute into the function f(x) (since they share a point the x and y values from line can be substituted into the f(x)). From there simplify to get k^2 +10k +9=0. You can factor and solve from there.
________________________________________ From: Dwayne Wellington [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  To search the list archives for previous posts go to http://lyris.collegeboard.com/read/?forum=apcalculus



