+0  
 
0
63
1
avatar

For the curve C with equation y = x 4 – 8x 2 + 3, (a) find x y d d , (2) The point A, on the curve C, has x-coordinate 1. (b) Find an equation for the normal to C at A, giving your answer in the form ax + by + c = 0, where a, b and c are integers.

Guest Oct 10, 2018
 #1
avatar+92723 
+1

y = x^4  - 8x^2  +  3

 

dy / dx  =  4x^3  - 16x

 

When the x coordinate is 1, the related y coordinate  is  (1)^4 - 8(1)^2  + 3  =  -4

 

The slope at x =  1  =  4(1)^3  - 16(1)  =  -12

 

A normal line at this point will have a negative reciprocal slope of   1 / 12

 

And using the point (1, -4)...the equation of the normal line   is

 

y  =(1/12)(x -1)   - 4

 

y = (1/12)x -1/12  - 4

 

y = (1/12)x -1/12 - 48/12

 

y = (1/12)x - 49/12      multiply through  by 12

 

12y =  1x  - 49      rearrange as

 

1x - 12y  -  49   =  0

 

Here's the graph : https://www.desmos.com/calculator/gwc3nmrm5i

 

 

cool cool cool

CPhill  Oct 10, 2018

28 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.