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Topic: [ap-calculus] larson pg 137 exercise 69
Replies: 1   Last Post: Oct 17, 2012 3:47 PM

 Dosky, Douglas Posts: 12 Registered: 2/23/09
RE: [ap-calculus] larson pg 137 exercise 69
Posted: Oct 17, 2012 3:47 PM

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Dwayne,

Here is how I solved the problem that you asked.

step 1: we need to find the point that the tangent line and the graph intersect thus we need to set the function equal to the tangent line
5x-4 = x^2 -kx

Step 2: We need to find the derivative so we get f'(x) = 2x - k

Step 3: We know the value of the derivative at the intersection point is 5 so 5 = 2x - k or k = 2x - 5

Step 4: Now lets plug in for k into step 1 so we get
5x - 4= x^2 -x(2x-5)
5x -4 = x^2 -2x^2 + 5x
-4 = -x^2
x = +/- 2

Step 5: Now lets plug in our two x values into step 3 to find our k values
k = 2(-2) - 5 = -9
k = 2(2) - 5 = -1

Take care

Douglas A. Dosky
(c) 614-260-4699
email: ddosky@worthington.k12.oh.us
Thomas Worthington High School
AP Calculus BC and AB Teacher
________________________________________
From: Dwayne Wellington [mrdw27@gmail.com]
Sent: Wednesday, October 17, 2012 9:29 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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