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Topic: [ap-calculus] larson pg 137 exercise 69
Replies: 3   Last Post: Oct 21, 2012 5:51 AM

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Kevin Sharitz

Posts: 8
Registered: 9/15/12
RE: [ap-calculus] larson pg 137 exercise 69
Posted: Oct 18, 2012 8:53 AM
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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)=2x-k. This has to equal the 5 from the line so 2x-k=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: [ap-calculus] larson pg 137 exercise 69

NOTE:
This ap-calculus 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/getting-started
and post messages there.
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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.
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