A "horizontal tangent" simply means that the slope (#dy/dx#) is #0#.
Therefore, we find #f'(x)#, set it equal to #0#, and solve:
#f(x)=x^2+6x#
#f'(x)=2x+6-># using the power rule
#0=2x+6#
#2x=-6->x=-3#
Now we just need to find the corresponding #f(x)# value, and we'll be all set:
#f(x)=x^2+6x#
#f(-3)=(-3)^2+6(-3)#
#f(-3)=9-18=-9#
Thus the point of horizontal tangency is #(-3,-9)#.
Extra Info for Math Geeks
If you remember from your Algebra days, the #x#-coordinate of the vertex of a parabola is given by #-b/(2a)#. You may notice that our function, #x^2+6x#, is a parabola, so its vertex can be found using that expression. As long as you realize that a parabola has a horizontal tangent at its vertex, this problem becomes a simple matter of algebra:
#x#-coordinate of the vertex = #-b/(2a)#
#x#-coordinate of the vertex = #-(6)/(2(1))=-3#
This is what we found earlier, but we used calculus instead. Note that you still have to plug #-3# into #f(x)#, so you can find the whole point - which is what the problem is asking for.
