My Math Forum  

Go Back   My Math Forum > High School Math Forum > Pre-Calculus

Pre-Calculus Pre-Calculus Math Forum


Reply
 
LinkBack Thread Tools Display Modes
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
frazza999 is offline  
 
June 20th, 2014, 02:19 PM   #2
Senior Member
 
aurel5's Avatar
 
Joined: Apr 2014
From: Europa

Posts: 584
Thanks: 177

Quote:
Originally Posted by frazza999 View Post
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.
aurel5 is offline  
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
frazza999 is offline  
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).
v8archie is offline  
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
frazza999 is offline  
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 $y=f(x)$.
v8archie is offline  
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.
perfect_world is offline  
Reply

  My Math Forum > High School Math Forum > Pre-Calculus

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





Copyright © 2019 My Math Forum. All rights reserved.