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 September 27th, 2011, 09:23 PM #1 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Tangent Line Equation Alright, I've been able to find the slope of a tangent line at the given points, and find the derivative by limit process, but I don't know how to "find an equation of the tangent line to the graph of 'f' at the given points." f(x) = x�+3x+4, (-2,2) September 27th, 2011, 09:29 PM #2 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Tangent Line Equation Differentiating f wish respect to x, we find: Thus: So, we have the slope and the point, thus we may use the point-slope formula: September 27th, 2011, 09:44 PM #3 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Tangent Line Equation So what exactly is the Tangent Line equation? y = -x ? For example: f(x) = x�+3, (1,4) I know the derivative is f?(x) = 2x, but the answer I'm looking for is an Tangent Line equation at the points (1,4). The answer in the back of the book shows: y = 2x + 2. How did they arrive to that equation? And how do I correlate that with my original problem? September 27th, 2011, 09:51 PM   #4
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Re: Tangent Line Equation

Quote:
 Originally Posted by RMG46 So what exactly is the Tangent Line equation? y = -x ?...
Yes, it passes through the given point (-2, 2) and is tangent to the quadratic at that point.

Quote:
 Originally Posted by RMG46 ... For example: f(x) = x�+3, (1,4) I know the derivative is f?(x) = 2x, but the answer I'm looking for is an Tangent Line equation at the points (1,4). The answer in the back of the book shows: y = 2x + 2. How did they arrive to that equation? And how do I correlate that with my original problem?
Okay, you have correctly differentiated f(x). Now evaluate the derivative at x = 1 (the x-coordinate of the given point).

f'(1) = 2

So, we know the slope of the tangent line is 2 and it must pass through the given point (1,4). Using the point-slope formula, we get: September 27th, 2011, 10:01 PM #5 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Tangent Line Equation Ah, that makes sense. Was forgetting what the point-slope formula was at first... Thanks!  September 27th, 2011, 10:03 PM #6 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Tangent Line Equation Glad to assist!  September 27th, 2011, 10:21 PM #7 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Tangent Line Equation Okay, got a derivative question now. h(x) = (2x� - 3x + 1) / (x) Since it's dividing x, how would I rewrite that into an equation I can just use the power rule? Or am I going to have to use some weird formula? :/ September 27th, 2011, 10:25 PM #8 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 522 Math Focus: Calculus/ODEs Re: Tangent Line Equation No, (2x� - 3x + 1)^(-1) = 1/(2x� - 3x + 1). What we are given is: September 27th, 2011, 10:30 PM #9 Member   Joined: Jun 2010 From: USA, Texas Posts: 85 Thanks: 0 Re: Tangent Line Equation Oh, I see... Tricky! So I'm getting to: h?(x) = 2 + -1?� So: 2 + (-1/1�) => 2 + (-1) = 1? September 27th, 2011, 10:36 PM #10 Member   Joined: Aug 2009 Posts: 69 Thanks: 0 Re: Tangent Line Equation (MarkFL might ninja post me but oh well) You forgot an important thing. so, if n=-1, , NOT -1. Also, just a tip on the tangents: If you're not sure if something is a tangent line, and want to verify your work, type the graphs into wolfram alpha. http://www.wolframalpha.com/input/?i...2B3x%2B4%2C+-x Tags equation, line, tangent Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post math999 Calculus 4 February 25th, 2013 08:36 PM unwisetome3 Calculus 2 October 28th, 2012 07:52 PM unwisetome3 Calculus 4 October 20th, 2012 08:38 AM kevpb Calculus 3 May 25th, 2012 11:32 PM arron1990 Calculus 5 February 9th, 2012 02:29 AM

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