# How do you find the points where the graph of the function #y=x^4 - 8x^2 +2# has horizontal tangents and what is the equation?

Horizontal tangents are at points

graph{x^4-8x^2+2 [-5, 5, -20, 20]}

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To find the points where the graph of the function y=x^4 - 8x^2 +2 has horizontal tangents, we need to find the values of x where the derivative of the function is equal to zero.

First, we find the derivative of the function: y' = 4x^3 - 16x.

Next, we set the derivative equal to zero and solve for x: 4x^3 - 16x = 0.

Factoring out 4x, we get: 4x(x^2 - 4) = 0.

Setting each factor equal to zero, we have two possibilities: 4x = 0 or x^2 - 4 = 0.

Solving for x, we find x = 0 or x = ±2.

Therefore, the points where the graph of the function has horizontal tangents are (0, 2) and (0, -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.

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