Hi

f(x) = x^2 -10x-4

Finding vertex

Using the vertex form of a parabola,  where(h,k) is the vertex

f(x) = x^2 -10x-4  |completing square to put into vertex form

f(x) = 1*(x-5)^2 - 25 -4

f(x) = (x-5)^2 - 29 | Vertex is Pt(5,-29) Line of symmetry is x= 5

1 = a > 0, parabola opens upward f(5) = -29 is a minimum point for f(x) 





 

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