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What is the vertex of #y= 2(x - 4)^2 - x^2+3 #?

1 Answer

Jan 10, 2016

Vertex (8, -29)

Explanation:

Develop #y = 2(x - 4)^2 - x^2 + 3 = 2(x^2 - 8x + 16) - x^2 + 3 =#
#= 2x^2 - 16x + 32 - x^2 + 3 = x^2 - 16x + 35.#

x-coordinate of vertex:
#x = -b/(2a) = 16/2 = 8#
y-coordinate of vertex:
#y(8) = 64 - 16(8) + 35 = -64 + 35 = -29#

Vertex (8, -29)

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