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

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Answer 1 · 2016-01-10T23:00:55

Vertex (8, -29)

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)

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