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August 27th, 2009, 03:09 PM   #1
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clarification help...

Hey, there's something I'm confused about from my textbook, I was hoping that you guys could help me:

It's about finding the union points between a tangent line and some function..

If the function is x^2,
dy/dx f(a) = 2a.

g(x) = 2a(x-a)+a^2
= 2ax-a^2

x^2 = 2ax - a^2
(x^2 - 2ax + a^2 ) = 0
(x - a ) ^2 = 0 , x = a

I don't understand the 2a(x-a)+a^2 part.. if you wanted to have the equation of the tangent line, wouldn't y= a^2 when x = a? but here, the equation has a^2 as it's "b" in mx+b and "x" is substituted with (x-a).. I'm confused..

The equation should look like like 2a(x) - a^2 from the beginning, if x=a, y=a^2

thanks :S
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August 27th, 2009, 03:56 PM   #2
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Re: clarification help...

Hint: What does the distributive axiom say about the expression

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August 27th, 2009, 04:04 PM   #3
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Re: clarification help...

Quote:
Originally Posted by Scott
Hint: What does the distributive axiom say about the expression

I'm not sure what you're getting at, but I know that when you expand everything you'll end up with 2ax - a^2.. but why does the author put the (x-a) and +a^2 into the equation to begin with?

Sorry, I can be slow sometimes

but thanks for replying
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August 28th, 2009, 12:56 AM   #4
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The author probably had in mind the point–slope form of the equation of a line rather than the slope-intercept form.
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August 28th, 2009, 04:46 AM   #5
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Re: clarification help...

The author might have wanted readers to practice using the distributive axiom in finding the solution.

Here's another hint. The distributive axiom states that for any ,

.

For example, to get of we can just get of and of .

Applying the distributive axiom to , we obtain

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August 29th, 2009, 10:40 AM   #6
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Re: clarification help...

Ah, I've got the idea. Thanks guys. I just never seen the slope point form before, but it makes sense: the change in y is the change in x times the slope
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