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Topic: [ap-calculus] larson pg 137 exercise 69
Replies: 1   Last Post: Oct 17, 2012 1:39 PM

 Mathew Deram Posts: 34 Registered: 1/4/12
RE: [ap-calculus] larson pg 137 exercise 69
Posted: Oct 17, 2012 1:39 PM

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Dwayne,

(I) x^2 - kx = 5x - 4 (the y-values 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 - (2x-5)*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: [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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