What is the vertex of #y=x^2+9x+8#?

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Answer 1 · 2017-04-19T06:37:52

Vertex is #(-9/2,-49/4)#.

For finding vertex of the equation, we should convert it in the form #(y-k)=(x-h)^2#, where #(h,k)# is the vertex.

As #y=x^2+9x+8#

= #x^2+2×9/2×x+(9/2)^2-(9/2)^2+8#

= #(x+9/2)^2-81/4+8#

= #(x+9/2)^3-49/4#

i.e. #y+49/4=(x+9/2)^2#

or #(y-(-49/4))=(x-(-9/2))^2#

Hence, vertex is #(-9/2,-49/4)#.

graph{x^2+9x+8 [-15.08, 4.92, -12.72, -2.72]}