Find the Horizontal Tangent of f(x)=x^2 + 4x - 1

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To find the horizontal tangent of the function f(x) = x^2 + 4x - 1, one must determine where the derivative equals zero. The derivative of the function is f'(x) = 2x + 4. Setting this equal to zero reveals that the horizontal tangent occurs at x = -2. Substituting x = -2 back into the original function yields the point (-2, 3). Thus, the horizontal tangent to the function is at the point (-2, 3).
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At what point is the tangent to f(x)=x^2 + 4x - 1 horizontal?

How do you do that? I think it's asking me at what point is x = 0, or undefined, right??
 
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Hint:

What is the unique feature about the equation of a horizontal line?
How does that relate to tangent lines?
 
Last edited:
Horizontal tangent is where derivative=0, that is x=-2.
 

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