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September 1, 2014

September 1, 2014

Posted by **anonymous** on Monday, January 28, 2008 at 6:38pm.

find all points on the curve where x=2. Show there is a horizontal tangent to the curve at one of the points.

- calculus -
**Reiny**, Monday, January 28, 2008 at 9:02pmsub x = 2 into your equation to get

4 - 4y = y^2 - 1

0 = y^2 + 4y - 5

(y+5)(y-1) = 0

so when x=2, y =-5 or when x=2, y = 1

so the points are (2,-5) and (2,1)

check them by subbing them back into the original equation, they work

- calculus -
**Reiny**, Monday, January 28, 2008 at 9:38pmI found the derivative of your equation implicitly and got

dy/dx = (2x+2xy)/(2y = x^2)

for a horizontal tangent this has to be zero, which means the numerator should be zero

subbing in (2,-5) makes the top zero, so there is a horizontal tangent at the point (2,_5)

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