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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.

 Oct 10, 2018
 #1
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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

 Oct 10, 2018

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