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December 21, 2014

December 21, 2014

Posted by **Alex** on Friday, November 30, 2012 at 6:39pm.

y/(x–9y)=x^6–2

at the point (1,-1/-8).

Can someone please help me?

I dont understand

- Math (Calculus) -
**Reiny**, Friday, November 30, 2012 at 7:00pmpuzzled, why would your write the point as

(1 , -1/-8) and not as simply (1 , 1/8) ?

first expand it

y = x^7 - 2x - 9x^6 y + 18y

0 = x^7 - 2x - 9x^6 y + 17y

0 = 7x^6 - 2 - 9x^6 dy/dx - 54x^5y + 17dy/dx

dy/dx(17 - 9x^6) = 54x^5y - 7x^6

dy/dx = (54x^5y - 7x^6))/(17 - 9x^6)

at x = 1

dy/dx = (54(1/8) -7)/(17 - 9)

= (-1/4) / 8

= -1/32

equation of tangent:

y + 1/8 = (-1/32)(x-1)

y = (-1/32)x + 1/32 - 1/8

y = (-1/32)x -3/32

check my arithmetic

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