My Math Forum Equation of a tangent line

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 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
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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: 7,671 Thanks: 2651 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: 7,671 Thanks: 2651 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.

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### equation for the line through the points (0,20) and (2,398)

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