What is the vertices x^2+4x+4y-4=0 ?

The vertices x^2+4x+4y-4=0 is Maximum (-2,2)

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vertices x^2+4x+4y-4=0

vertices

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

Maximum (- 2, 2)

Solution steps

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

Rewrite

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

in the form

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

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

The parabola parameters are:

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

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

x_{v} = - \frac{( - 1 )}{2 ( - \frac{1}{4} )}

Simplify

- \frac{- 1}{2 ( - \frac{1}{4} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 2

Therefore the parabola vertex is

(- 2, 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{4}

Maximum (- 2, 2)

y^{2} = - 24 x

f (x) = 2 x^{2} - 12 x + 23

x^{2} + 6 x + 7

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

y = 2 x^{2} + 10 x + 12

What is the vertices x^2+4x+4y-4=0 ?
The vertices x^2+4x+4y-4=0 is Maximum (-2,2)