How do you find the vertex of a parabola $y=x^2 + .6x - 11$ ?

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Answer 1 · 2015-07-05T16:43:41

Vertex (0.3, -11.09)

Write the equation in vertex form by completing the square of terms containing x, as follows:

y= $x^2 +.6x +.09 -.09-11$

= $(x-0.3)^2 -11.09$

The vertex is therefor (0.3, -11.09)

How do you find the vertex of a parabola y=x^2 + .6x - 11y=x^2 + .6x - 11?