What is the vertices (x+1)^2=-6(y-4) ?

The vertices (x+1)^2=-6(y-4) is Maximum (-1,4)

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vertices (x+1)^2=-6(y-4)

vertices

(x + 1)^{2} = - 6 (y - 4)

Maximum (- 1, 4)

Solution steps

(x + 1)^{2} = - 6 (y - 4)

Rewrite

(x + 1)^{2} = - 6 (y - 4)

in the form

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

y = - \frac{x ^{2}}{6} - \frac{x}{3} - \frac{1}{6} + 4

The parabola parameters are:

a = - \frac{1}{6}, b = - \frac{1}{3}, c = - \frac{1}{6} + 4

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

x_{v} = - \frac{( - \frac{1}{3} )}{2 ( - \frac{1}{6} )}

Simplify

- \frac{- \frac{1}{3}}{2 ( - \frac{1}{6} )} : - 1

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 4

Therefore the parabola vertex is

(- 1, 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{6}

Maximum (- 1, 4)

x^{2} - 2 x - 4 y + 13 = 0

5 x^{2} - 10

f (x) = 2 x^{2} - 4 x + 3

y = 18 x^{2} - 6 x + 4

f (x) = - x^{2} + 14 x - 48

What is the vertices (x+1)^2=-6(y-4) ?
The vertices (x+1)^2=-6(y-4) is Maximum (-1,4)