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

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Answer 1 · 2016-09-27T12:09:16

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

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

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

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

Here, the given equation can be standardized as

${(x - 1)}^{2} = y + \frac{45}{4}$ $(x-1)^2=y+45/4$ . 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=x2−x−11y= x^2 -x - 11 ?