How do you find the horizontal and vertical tangents of #4x^2 + y^2 -8x +4y +4=0#?
Horizontal tangents at:
Vertical tangents at:
We have:
Differentiating implicitly:
We can confirm this by completing the square:
graph{4x^2 + y^2 -8x +4y +4=0 [-1, 3, -6, 2]}
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To find the horizontal and vertical tangents of the equation 4x^2 + y^2 -8x +4y +4=0, we can differentiate the equation with respect to x and y separately.
Differentiating with respect to x, we get: 8x - 8 + 0 + 0 = 0 Simplifying, we find: x = 1
Differentiating with respect to y, we get: 0 + 2y + 0 + 4 = 0 Simplifying, we find: y = -2
Therefore, the horizontal tangent occurs at the point (1, -2) and the vertical tangent occurs at the same point (1, -2).
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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.
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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