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how do you find the vertex of the graph of the quadratic function of y=-2x^2-16x-20? can someone explain it to me?
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OK so you have the graph of a quadratic in the form of f(x)=ax^2+bx+c The vertex (x,y) is located at \[(\frac{-b}{2a},f(\frac{-b}{2a})\]
-b/2a
y=- 2x² -16x-20 = -2 ( x² + 8x + 10) Vertex ( h, k) h = -b/2a = -8/2 = -4 ->k = f ( -4) = -2 ( 16 -32 + 10 ) = 12 => vertex ( -4, 12)
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