Hi

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

f(x)= -10x^2 + 5x + 6   |putting into the vertex form by completing the square

f(x)= -10[x^2 - 1/2x] + 6 

f)x)= -10[(x - 1/4)^2 - 1/16] + 6 

f)x)= -10(x - 1/4)^2 - 10/16 + 6 

f)x)= -10(x - 1/4)^2 + 10/16 + 6 

f)x)= -10(x - 1/4)^2 + 106/16   |vertex is Pt(1/4, 53/8)



 

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