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

Sep 27, 2016

Vertex form is ${\left(x - 1\right)}^{2} = y + \frac{45}{4}$.
The vertex or this parabola is $V \left(1 , - \frac{45}{4}\right)$

Explanation:

The equation ${\left(x - \alpha\right)}^{2} = 4 a \left(y - \beta\right)$ represents the parabola with

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

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

Here, the given equation can be standardized as

${\left(x - 1\right)}^{2} = y + \frac{45}{4}$. giving $a = 1 ' 4 , \alpha = 1 \mathmr{and} \beta = - \frac{45}{4}$.

Vertex is $V \left(1 , - \frac{45}{4}\right)$

Axis is x=1.

Focus is S(1, -11).

Directrix is $y = - \frac{49}{4}$