How do you write $y= - 0.05x^2 + 12.8$ into vertex form?

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Answer 1 · 2015-06-17T21:24:06

It is nearly in vertex form already, but you could write:

$y = -0.05(x-0)^2+12.8$

The vertex is at $(0, 12.8)$

Vertex form is:

$y = a(x-x_0)^2 + y_0$

where $a$ is some non-zero constant (positive or negative) and $(x_0, y_0)$ is the vertex.

How do you write y= - 0.05x^2 + 12.8 y= - 0.05x^2 + 12.8 into vertex form?