at what point is the tangent to f(x)=x^2+4x-1 horizontal?
That would be when the derivative, df/dx = 2x + 4, equals zero.
df/dx at any point of a curve f(x) is the slope of the tangent line.
To find the point at which the tangent to the function f(x) = x^2 + 4x - 1 is horizontal, we need to look for the values of x where the derivative of the function, denoted as df/dx or f'(x), equals zero.
The derivative of f(x) = x^2 + 4x - 1 can be found by applying the power rule of differentiation. In this case, since the function consists only of monomials, each term is differentiated independently. The derivative of x^2 is 2x and the derivative of 4x is 4. As a result, the derivative of f(x) is:
f'(x) = 2x + 4
To find the x-values where the tangent is horizontal, we set the derivative equal to zero and solve for x:
2x + 4 = 0
Subtracting 4 from both sides gives:
2x = -4
Dividing both sides by 2 yields:
x = -2
Therefore, the tangent to the function f(x) = x^2 + 4x - 1 is horizontal at the point (-2, f(-2)).