## University Calculus: Early Transcendentals (3rd Edition)

The graph of $f$ has a horizontal tangent at $(-2,-5)$.
$$f(x)=x^2+4x-1$$ 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 $(-2,-5)$.