Answer
The graph of $f$ has a horizontal tangent at $(-2,-5)$.
Work Step by Step
$$f(x)=x^2+4x-1$$
Having a horizontal tangent means that having the slope, and the derivative, there equal $0$.
In other words, we need to find $x$ such that $$\lim_{h\to0}\frac{f(x+h)-f(x)}{h}=0\hspace{1cm}(1)$$
First, we examine $f(x+h)$:
$f(x+h)=(x+h)^2+4(x+h)-1=x^2+2xh+h^2+4x+4h-1=(x^2+4x-1)+(2xh+h^2+4h)$
Therefore,
$$\frac{f(x+h)-f(x)}{h}=\frac{2xh+h^2+4h}{h}=2x+h+4$$
Replacing the result into $(1)$: $$\lim_{h\to0}(2x+h+4)=0$$
$$2x+0+4=0$$
$$2x+4=0$$
$$x=-2$$
We have $f(-2)=(-2)^2+4\times(-2)-1=4-8-1=-5$
Therefore, the graph of $f$ has a horizontal tangent at $(-2,-5)$.
