What is the vertices (x-1)^2=-2y-11 ?

The vertices (x-1)^2=-2y-11 is Maximum (1,-11/2)

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vertices (x-1)^2=-2y-11

vertices

(x - 1)^{2} = - 2 y - 11

Maximum (1, - \frac{11}{2})

Solution steps

(x - 1)^{2} = - 2 y - 11

Rewrite

(x - 1)^{2} = - 2 y - 11

in the form

y = a x^{2} + b x + c

y = - \frac{x ^{2}}{2} + x - 6

The parabola parameters are:

a = - \frac{1}{2}, b = 1, c = - 6

x_{v} = - \frac{b}{2 a}

x_{v} = - \frac{1}{2 ( - \frac{1}{2} )}

Simplify

- \frac{1}{2 ( - \frac{1}{2} )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - \frac{11}{2}

Therefore the parabola vertex is

(1, - \frac{11}{2})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{2}

Maximum (1, - \frac{11}{2})

f (x) = x^{2} - 6 x + 10

- x^{2} + 4 x

f (x) = - 6 x^{2} + 17 x - 12

f (x) = 4 x^{2} - 16 x + 10

2 x^{2} - 6 x + 4

What is the vertices (x-1)^2=-2y-11 ?
The vertices (x-1)^2=-2y-11 is Maximum (1,-11/2)