Question 189058
To find the vertex and examine if there is a maximum or minimum value you have to differentiate that function and put it equal to zero as following, I'll take the second problem in consideration, f(x) = - x^2 +6x +20

d/dx{f(x)} = d/dx(-x^2)+ d/dx (6x) + d/dx(20)
=-2x + 6 + 0
=-2x+6

now set, d/dx {f(x)} = 0
or, -2x+6=0
or, x= 3  put this in f(x), so y=f(3)=-3^2+6*3+20=29

so the vertex is (3,29)=(x,y)



Explanation: Differentiating with respect to (w.r.t) x means the rate of change of the value of the function w.r.t small change in value of x. so that point will be the highest (or lowest) where the change is zero. So we set it equal to zero.



Maximum or Minimum value: now at this point, the question is whether at this point the function has its highest value(maximum) or lowest value(minimum). If we again differentiate f(x) then if the differentiated value is positive (that means the value of the rate of change functions is increasing so it will be a minima) but if it is negative it means that the rate is decreasing indicating the value is maximum.

so d/dx^2 {f(x)}
= d/dx {d/dx {f(x)}
= d/dx (-2x) + d/dx(6) 
= -2+0 
= -2

so that at this point the function has a maximum value

and the value is, f(3) = 29