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September 27th, 2011, 09:23 PM   #1
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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)
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September 27th, 2011, 09:29 PM   #2
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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:



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September 27th, 2011, 09:44 PM   #3
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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?
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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:



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September 27th, 2011, 10:01 PM   #5
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Re: Tangent Line Equation

Ah, that makes sense.

Was forgetting what the point-slope formula was at first...

Thanks!
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September 27th, 2011, 10:03 PM   #6
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Re: Tangent Line Equation

Glad to assist!
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September 27th, 2011, 10:21 PM   #7
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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? :/
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September 27th, 2011, 10:25 PM   #8
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Re: Tangent Line Equation

No, (2x - 3x + 1)^(-1) = 1/(2x - 3x + 1). What we are given is:

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September 27th, 2011, 10:30 PM   #9
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Re: Tangent Line Equation

Oh, I see... Tricky!

So I'm getting to: h?(x) = 2 + -1?

So: 2 + (-1/1) => 2 + (-1) = 1?
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September 27th, 2011, 10:36 PM   #10
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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
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