My Math Forum Tangent Line Equation

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 September 27th, 2011, 08:23 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) = x²+3x+4, (-2,2)
 September 27th, 2011, 08:29 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Tangent Line Equation Differentiating f wish respect to x, we find: $f'(x)=2x+3$ Thus: $f'(-2)=2(-2)+3=-1$ So, we have the slope and the point, thus we may use the point-slope formula: $y-2=-(x+2)$ $y=-x$
 September 27th, 2011, 08:44 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) = x²+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, 08:51 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) = x²+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:

$y-4=2(x-1)$

$y=2x+2$

 September 27th, 2011, 09:01 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, 09:03 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Tangent Line Equation Glad to assist!
 September 27th, 2011, 09:21 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) = (2x² - 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, 09:25 PM #8 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Tangent Line Equation No, (2x² - 3x + 1)^(-1) = 1/(2x² - 3x + 1). What we are given is: $h(x)=\frac{2x^2-3x+1}{x}=\frac{2x^2}{x}-\frac{3x}{x}+\frac{1}{x}=2x-3+x^{\small{-1}}$
 September 27th, 2011, 09:30 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?² So: 2 + (-1/1²) => 2 + (-1) = 1?
 September 27th, 2011, 09:36 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. $f(x)=x^n, f#39;(x)=n\cdot x^{n-1}$ so, if n=-1, $f'(x)=-1\cdot x^{-2}=\frac{-1}{x^2}$, 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

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