What is the vertex form of $y= x^2 -x - 11$ ?

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Answer 1 · 2016-09-27T12:09:16

Vertex form is $(x-1)^2=y+45/4$ .
The vertex or this parabola is $V(1, -45/4)$

The equation $(x-alpha)^2=4a(y-beta)$ represents the parabola with

vertex at $V( alpha, beta)$ , axis VS along $x = alpha$ , focus at

$S(alpha, beta+a)$ and directrix as $y=beta-a$

Here, the given equation can be standardized as

$(x-1)^2=y+45/4$ . giving $a = 1'4, alpha = 1 and beta =-45/4$ .

Vertex is $V(1, -45/4)$

Axis is x=1.

Focus is S(1, -11).

Directrix is $y=-49/4$

What is the vertex form of y= x^2 -x - 11 y= x^2 -x - 11 ?