Question

Question #b423f

Answer 1

There are no points on the curve that have a horizontal tangent line. Please see the explanation.

Explanation:

Implicitly differentiate the equation:

`-1+2y+2xdy/dx+2ydy/dx=0`

`2(x+y)dy/dx = 1-2y`

`dy/dx = (1-2y)/(2(x+y)), y !=-x`

Because a horizontal tangent line implies that the first derivative is 0, we set the first derivative equal to 0:

`0 =(1-2y)/(2(x+y)), y !=-x`

`0 =1-2y, y !=-x`

`y = 1/2`

Substitute `y = 1/2` into the equation `-x+2xy+y^2=0`:

`-x+2x(1/2)+(1/2)^2=0`

`-x+x+1/4=0`

`1/4=0`

There is no point on the curve where `y = 1/2`, therefore, there is no point that has a horizontal tangent line.

graph{-x+2xy+y^2=0 [-10, 10, -5, 5]}

Please observe that `y = 1/2` is an asymptote