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 August 27th, 2009, 04:09 PM #1 Senior Member   Joined: Apr 2009 Posts: 201 Thanks: 0 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
 August 27th, 2009, 04:56 PM #2 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 Re: clarification help... Hint: What does the distributive axiom say about the expression $2a(x\,-\,a)\mbox{?}$
August 27th, 2009, 05:04 PM   #3
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Re: clarification help...

Quote:
 Originally Posted by Scott Hint: What does the distributive axiom say about the expression $2a(x\,-\,a)\mbox{?}$
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

 August 28th, 2009, 01:56 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,104 Thanks: 1907 The author probably had in mind the point–slope form of the equation of a line rather than the slope-intercept form.
 August 28th, 2009, 05:46 AM #5 Senior Member   Joined: Dec 2008 Posts: 251 Thanks: 0 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 $a,\,x,\,y$, $a(x\,+\,y)\,=\,ax\,+\,ay$. For example, to get $5$ of $x\,+\,y$ we can just get $5$ of $x$ and $5$ of $y$. Applying the distributive axiom to $g(x)\,=\,2a(x\,-\,a)\,+\,a^2$, we obtain $g(x)\,=\,2a(x\,-\,a)\,+\,a^2\,=\,2a\cdot x\,-\,2a\cdot a\,+\,a^2.$
 August 29th, 2009, 11:40 AM #6 Senior Member   Joined: Apr 2009 Posts: 201 Thanks: 0 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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