User Name Remember Me? Password

 Pre-Calculus Pre-Calculus Math Forum

 June 20th, 2014, 02:06 PM #1 Newbie   Joined: May 2014 From: Sydney Posts: 29 Thanks: 0 Equation of a tangent line The question: Here is how i have tried to do it. Where am i going wrong? I am Guessing Tangent line at points (1, 0). f(x) = 3x^2 + 2x - 1 f'(x) = 6x + 2 f'(1) = 6 * 1 + 2 m = 8 and y = 8x + c 0 = 8 + c c = -8 so: y = 8x - 8 Cheers June 20th, 2014, 02:19 PM   #2
Senior Member

Joined: Apr 2014
From: Europa

Posts: 584
Thanks: 177

Quote:
 Originally Posted by frazza999 The question: In x = a, tangent equation is:

$\displaystyle \color{blue} {y-f(a) = f'(a)(x-a)}$

Last edited by aurel5; June 20th, 2014 at 02:38 PM. June 20th, 2014, 02:25 PM #3 Newbie   Joined: May 2014 From: Sydney Posts: 29 Thanks: 0 I dont understand :/ What would the equation of the tangent line be at the end June 20th, 2014, 02:28 PM #4 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Your problem is that the function you gave doesn't go through (1,0). June 20th, 2014, 02:52 PM #5 Newbie   Joined: May 2014 From: Sydney Posts: 29 Thanks: 0 it goes through what then? How can i determine what the y coordinate is with out drawing it? cheers cheers June 20th, 2014, 03:35 PM #6 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,690 Thanks: 2669 Math Focus: Mainly analysis and algebra Set $x=1$ and see what value you get for $y=f(x)$. June 25th, 2014, 02:58 PM #7 Senior Member   Joined: Jul 2013 From: United Kingdom Posts: 471 Thanks: 40 Firstly, you have to look for the point that the tangent line goes through. What is y, when x=1? $\displaystyle f\left( x \right) =3{ x }^{ 2 }+2x-1\\ \\ f\left( 1 \right) =3{ \left( 1 \right) }^{ 2 }+2\left( 1 \right) -1=4$ So you now have the point (1,4). Now you have to find the derivative of the function of x, then see what the slope is at x=1. The slope can be thought of as m for gradient. $\displaystyle f\left( x \right) =3{ x }^{ 2 }+2x-1\\ \\ f'\left( x \right) =6x+2,\\ \\ f'\left( 1 \right) =6\left( 1 \right) +2=8\\ \\ \therefore \quad m=8\quad at\quad \left( 1,4 \right)$ You then use this formula to derive the equation of the tangent line to the function: $\displaystyle y-{ y }_{ 1 }=m\left( x-{ x }_{ 1 } \right)$ So: $\displaystyle { x }_{ 1 }=1,\quad { y }_{ 1 }=4,\quad m=8\\ \\ So:\\ \\ y-4=8\left( x-1 \right) \\ \\ y-4=8x-8\\ \\ y=8x-8+4\\ \\ y=8x-4$ Graph can be found at: https://www.desmos.com/calculator/f8am1o3lig Last edited by perfect_world; June 25th, 2014 at 03:02 PM. Tags equation, line, tangent equation for the line through the points (0,20) and (2,398)

Click on a term to search for related topics.
 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 unwisetome3 Calculus 2 October 28th, 2012 06:52 PM unwisetome3 Calculus 4 October 20th, 2012 07:38 AM kevpb Calculus 3 May 25th, 2012 10:32 PM RMG46 Calculus 28 September 28th, 2011 09:21 AM ArmiAldi Calculus 1 April 16th, 2008 05:51 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top      