Calculus Calculus Math Forum

 October 12th, 2014, 09:25 AM #1 Newbie   Joined: Oct 2014 From: macau Posts: 4 Thanks: 0 About diff. problem I need your solution. ~please!! The question is: find all tangent lines through the origin to the graph of y=1+(x-1)^2 . Last edited by skipjack; October 13th, 2014 at 10:04 AM. October 12th, 2014, 10:17 AM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 3,094 Thanks: 1677 a line tangent to the curve $\displaystyle y = 1+(x-1)^2$ will pass thru the points $\displaystyle (0,0)$ and $\displaystyle \left(x,1+(x-1)^2\right)$ ... slope of any such line is $\displaystyle \frac{\Delta y}{\Delta x} = \frac{1+(x-1)^2 - 0}{x-0} = \frac{1+(x-1)^2}{x}$ now ... what do you know about the slope of a tangent line from a calculus perspective? Thanks from topsquark and superzhong October 12th, 2014, 03:31 PM #3 Global Moderator   Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,974 Thanks: 1156 Math Focus: Elementary mathematics and beyond $\displaystyle y=kx$ $\displaystyle y=1+(x-1)^2$ $\displaystyle k=2x-2$ $\displaystyle (2x-2)x=1+(x-1)^2\Rightarrow x=\pm\sqrt2$ $\displaystyle y=2(\sqrt2-1)x$ $\displaystyle y=-2(\sqrt2+1)x$ Thanks from topsquark and superzhong October 13th, 2014, 08:38 AM   #4
Newbie

Joined: Oct 2014
From: macau

Posts: 4
Thanks: 0

Quote:
 Originally Posted by greg1313 $\displaystyle y=kx$ $\displaystyle y=1+(x-1)^2$ $\displaystyle k=2x-2$ $\displaystyle (2x-2)x=1+(x-1)^2\Rightarrow x=\pm\sqrt2$ $\displaystyle y=2(\sqrt2-1)x$ $\displaystyle y=-2(\sqrt2+1)x$
thanks sensei !!! that is very helpful~~

Quote:
 Originally Posted by skeeter a line tangent to the curve $\displaystyle y = 1+(x-1)^2$ will pass thru the points $\displaystyle (0,0)$ and $\displaystyle \left(x,1+(x-1)^2\right)$ ... slope of any such line is $\displaystyle \frac{\Delta y}{\Delta x} = \frac{1+(x-1)^2 - 0}{x-0} = \frac{1+(x-1)^2}{x}$ now ... what do you know about the slope of a tangent line from a calculus perspective?
Thanks, I almost get it. ~XD

Last edited by skipjack; October 13th, 2014 at 10:22 AM. October 13th, 2014, 10:21 AM #5 Global Moderator   Joined: Dec 2006 Posts: 21,110 Thanks: 2326 Each tangent line through the origin has equation y = kx. If y = kx = 1 + (x - 1)² = x² - 2x + 2, x² - (k + 2)x + 2 = 0. For the required tangents, that quadratic's discriminant must be 0, so k = -2 + 2√2 or -2 - 2√2. Thanks from topsquark Tags diff, problem Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Niko Bellic Calculus 2 July 8th, 2013 11:01 AM zacmz Calculus 1 February 23rd, 2012 06:08 AM David_Lete Calculus 0 November 1st, 2009 10:11 AM thummel1 Calculus 2 September 1st, 2009 06:58 PM kien Geometry 1 April 12th, 2009 10:19 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      