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Topic: [ap-calculus] larson pg 137 exercise 69
Replies: 1   Last Post: Oct 17, 2012 11:18 AM

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Steven Becker

Posts: 54
Registered: 7/14/08
RE: [ap-calculus] larson pg 137 exercise 69
Posted: Oct 17, 2012 11:18 AM
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Hello,

Since the line is tangent to the graph, you know two things: 1. the graphs must be equal at some x-value and 2. their derivatives must also be equal at that same x-value.
So, x^2-kx=5x-4 and 2x-k=5. Thus, k=2x-5 from the second equation. Sub that into the first equation for k to get x^2-(2x-5)x=5x-4. Simplify to get x^2=4, so x=+/-2. Since you now have x-values where it is possible, you can find the corresponding k values by using the k=2x-5 equation. HTH.

Steve Becker
Math Teacher
(503) 785-7948
"What lies behind us and what lies before us are tiny matters compared to what lies within us." - Oliver Wendell Holmes

________________________________________
From: Dwayne Wellington [mrdw27@gmail.com]
Sent: Wednesday, October 17, 2012 6:29 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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