# How do you find the points where the graph of the function #f(x) = -x^2-3x+5# has horizontal tangents?

The given function:

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To find the points where the graph of the function f(x) = -x^2-3x+5 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 f(x) by applying the power rule and the sum rule of differentiation. The derivative of -x^2 is -2x, the derivative of -3x is -3, and the derivative of 5 is 0. Therefore, the derivative of f(x) is -2x - 3.

Next, we set the derivative equal to zero and solve for x: -2x - 3 = 0. Adding 3 to both sides gives -2x = 3, and dividing by -2 gives x = -3/2.

So, the graph of the function f(x) = -x^2-3x+5 has horizontal tangents at the point (-3/2, f(-3/2)).

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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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