What is the vertex of $y = x^{2} - 4 x + 18$ $y=x^2-4x+18$ ?

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Answer 1 · 2016-03-25T12:31:09

Vertex (2, 14)

x-coordinate of vertex:
$x = - \frac{b}{2 a} = \frac{4}{2} = 2$ $x = -b/(2a) = 4/2 = 2$
y-coordinate of vertex:
y(2) = 4 - 8 + 18 = 14
Vertex (2, 14)

What is the vertex of y=x2−4x+18y=x^2-4x+18?