# Need calculus help, again (important)... **URL:** \ **Category:** Miscellaneous and Personal Stuff I Must Share\ **Created:** [March 12, 2002, 11:32am UTC](https://boards.straightdope.com/t/need-calculus-help-again-important/97821 "2002-03-12T11:32:55Z")\ **Posts on this page:** 3\ **Page:** 2
**Author:** ![RTFirefly](https://avatars.discourse-cdn.com/v4/letter/r/c77e96/32.png) [@RTFirefly](https://boards.straightdope.com/u/RTFirefly)\ **Post date:** [March 13, 2002, 1:06am UTC](https://boards.straightdope.com/t/need-calculus-help-again-important/97821/21 "2002-03-13T01:06:30Z")
The short version is that it involves a minuscule bit of calculus, and a shitload of algebra. I kept on getting tripped up on the part that **BabaBooey** had done correctly as shown in the OP, which was why I didn’t realize exactly where he was in the problem. Made my advice kinda lacking, for which I apologize. Turned out he’d already done the heavy lifting, and just needed the last easy step. The part where he says: > [@](#): > > I get the slope as being (-2c^2 +12c -14)/(2c-3) is in fact exactly what you need on one side of the equation. I got there by letting f(x)=x[sup]2[/sup], g(x)=-x[sup]2[/sup]+6x-5, with (c,g©) being the tangent point to g(x), and (c-k,f(c-k)) being the tangent point to f(x). That allows you to tie your functions together. f’(x)=2x and g’(x)=-2x+6, of course, and let’s let m be the slope of the tangent line. So m=g’© and m=f’(c-k) also. Which means m= **-2c+6** and m=2(c-k), which means 2c-2k=-2c+6 or _k=2c-3_. So we’ve got m,c,k all in terms of each other. Which **Baba** got all the pieces of, and used them together to get his slope formula - and all he needed from there was to set the bolded slope formula equal to the quoted slope formula, solve for c, and he was home free. OK, so the quoted part: m = (y[sub]2[/sub]-y[sub]1[/sub])/(x[sub]2[/sub]-x[sub]1[/sub]) = (g©-f(c-k))/(c-(c-k)) = (g©-f(3-c))/(2c-3) (using k=2c-3) = (-c[sup]2[/sup]+6c-5-(3-c)[sup]2[/sup])/(2c-3) = (-2c[sup]2[/sup]+12c-14)/(2c-3). And I couldn’t see that **Baba** was almost there because I was doing this problem in between stuff at work, and kept on screwing up the algebra. Anyhow, the rest is easy: set the mess on the previous line equal to -2c+6, multiply through by 2c-3 (noting that c can’t be 3/2, although that doesn’t come into play), getting (2c-3)(-2c+6) = (-2c[sup]2[/sup]+12c-14) which eventually gets you c=1, c=2 as solutions. That’s the x-coordinate of each of the solutions for the tangent line of g(x). We’ve got k and m defined in terms of c (c-k is the x-coord of the tangent line where it’s tangent to f(x)), so the rest is easy. Can’t believe I got so tripped up on algebra. :o ---
**Author:** ![c\_goat](https://avatars.discourse-cdn.com/v4/letter/c/f4b2a3/32.png) [@c\_goat](https://boards.straightdope.com/u/c_goat)\ **Post date:** [March 13, 2002, 5:01am UTC](https://boards.straightdope.com/t/need-calculus-help-again-important/97821/22 "2002-03-13T05:01:22Z")
> [@](#): > > \*Originally posted by RTFirefly \* > \*\*Can’t believe I got so tripped up on algebra. :o \*\* Well at least you answered the question correctly, and didn’t send the poor guy to class with the wrong answer. 😃 ---
**Author:** ![BabaBooey](https://avatars.discourse-cdn.com/v4/letter/b/dfb087/32.png) [@BabaBooey](https://boards.straightdope.com/u/BabaBooey)\ **Post date:** [March 13, 2002, 8:24am UTC](https://boards.straightdope.com/t/need-calculus-help-again-important/97821/23 "2002-03-13T08:24:59Z")
I ended up just turning in what I had so far. I knew that I had done the work correctly, so since I had done everything right up to the slope’s equation, I should get a little credit. Thanks a lot for the help you guys. [Previous page](https://boards.straightdope.com/t/need-calculus-help-again-important/97821.md?page=1)