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At what point is the tangent to f(x) = x^2 + 4x - 1 horizontal?
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take the derivative, set it equal to zero, and solve. or else just find the vertex
vertex is \[-\frac{b}{2a}=-\frac{4}{2}=-2\] so the point is \[(-2,-5)\]
The tangent line will be horizontal at the vertex. You can either graph the equation to find it or you can complete the square.
oh didnt see ur answer satellite,
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