# Find the point(s) (if any) of horizontal tangent lines for the equation #x^2+xy+y^2=6#. If none exist, why?

graph{(x^2+xy+y^2-6)((x-sqrt2)^2+(y+2sqrt2)^2-0.01)((x+sqrt2)^2+(y-2sqrt2)^2-0.01)=0 [-8.89, 8.89, -4.444, 4.445]}

So

Now intersect with the equation

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To find the point(s) of horizontal tangent lines for the equation x^2+xy+y^2=6, we need to find the values of x and y where the derivative of the equation with respect to y equals zero.

Taking the derivative of the equation with respect to y, we get: 2x + x(dy/dy) + 2y(dy/dy) = 0

Simplifying, we have: 2x + x + 2y(dy/dy) = 0

Combining like terms, we get: 3x + 2y(dy/dy) = 0

Since we are looking for horizontal tangent lines, the derivative dy/dy is zero. Therefore, the equation becomes: 3x = 0

Solving for x, we find that x = 0.

Substituting x = 0 back into the original equation, we get: 0^2 + 0y + y^2 = 6

Simplifying, we have: y^2 = 6

Taking the square root of both sides, we find that y = ±√6.

Therefore, the points of horizontal tangent lines for the equation x^2+xy+y^2=6 are (0, √6) and (0, -√6).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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