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  In x = a, tangent equation is: $\displaystyle \color{blue} {yf(a) = f'(a)(xa)}$ 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? $\displaystyle f\left( x \right) =3{ x }^{ 2 }+2x1\\ \\ f\left( 1 \right) =3{ \left( 1 \right) }^{ 2 }+2\left( 1 \right) 1=4$ 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 }+2x1\\ \\ 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: $\displaystyle { x }_{ 1 }=1,\quad { y }_{ 1 }=4,\quad m=8\\ \\ So:\\ \\ y4=8\left( x1 \right) \\ \\ y4=8x8\\ \\ y=8x8+4\\ \\ y=8x4$ 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 
Search tags for this page 
Click on a term to search for related topics.

Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
equation of a tangent line  unwisetome3  Calculus  2  October 28th, 2012 06:52 PM 
Equation of tangent line  unwisetome3  Calculus  4  October 20th, 2012 07:38 AM 
Tangent line equation  kevpb  Calculus  3  May 25th, 2012 10:32 PM 
Tangent Line Equation  RMG46  Calculus  28  September 28th, 2011 09:21 AM 
equation of the tangent line at the given value  ArmiAldi  Calculus  1  April 16th, 2008 05:51 AM 