Personally, I've never heard of the vertex of a denominator? ...
If you have to find the vertex of a function (in this case a rational one) then the first derivative can be a good option.
Say you have a function f(x)=1/(x^2+6x+11). Would you have to complete the square to figure out the vertex of the parabola in the denominator(then sub in the x-value in the function to figure out the vertex in the rational function), or is there an easier way to draw the graph?
See the graph. The vertex is at ( ).
This could also have been calculated using differentiation.
Since, the parabola is of the form , the slope at the vertex must be equal to zero.
Therefore, <=> <=> <=>