## Thomas' Calculus 13th Edition

$(-2,-5)$
Given that $f(x)=x^2+4x-1$, the function will have a horizontal tangent when the derivative is equal to 0, since the slope will indicate that there is no change in the y value of the tangent when x changes. $f'(x)$ is found using the power rules: $f'(x)=2x+4$ We let $f'(x)$ be equal to zero. $2x+4=0$ $x=-2$ $f(-2)=-5$ Thus, at the point $(-2,-5)$, there is maximum or a minimum where the graph will have a horizontal tangent.