How would I solve this problem?
Find the point where the tangent line is horizontal in the following function:
I computed the derivative: $\quad f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$.
But what would I do next?
I've included a graph of the function $\;f(x)=(x-2)(x^2-x-11)\;$(in blue), along with the two horizontal lines tangent to the function: $\;y = -5,\;\; y = 27$ (violet and "brown", respectively) to see in a "picture" what is happening here.
You are on your way there. Consider your first derivative $$f'(x)=(x-2)(2x-1)+(1)(x^2-x-11)$$
Let's simplify this:$$f'(x)=(2x^2-5x+2)+(x^2-x-11)$$
Now, for a horizontal line, $f'(x) = 0$. So let's solve $$3x^2-6x-9 = 0$$ $$x^2-2x-3 = 0$$
$$(x-3)(x+1) = 0$$ $x = 3 $ or $x = -1$