# How do you find the points where the graph of the function #x^2+7y^2-4x-2=0# has horizontal tangents?

At

graph{x^2+7y^2-4x-2=0 [-2, 5, -2, 2]}

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To find the points where the graph of the function x^2+7y^2-4x-2=0 has horizontal tangents, we need to find the values of x and y that satisfy the condition for horizontal tangents.

First, we differentiate the given function with respect to x to find its derivative.

d/dx (x^2+7y^2-4x-2) = 2x - 4

Next, we set the derivative equal to zero to find the critical points where the tangents are horizontal.

2x - 4 = 0

Solving this equation, we find x = 2.

Substituting this value of x back into the original equation, we can solve for y.

(2)^2 + 7y^2 - 4(2) - 2 = 0

4 + 7y^2 - 8 - 2 = 0

7y^2 - 6 = 0

Simplifying further, we have 7y^2 = 6.

Dividing both sides by 7, we find y^2 = 6/7.

Taking the square root of both sides, we get y = ±√(6/7).

Therefore, the points where the graph of the function has horizontal tangents are (2, √(6/7)) and (2, -√(6/7)).

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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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