What is the vertices x^2+4x-8y+12=0 ?

The vertices x^2+4x-8y+12=0 is Minimum (-2,1)

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vertices x^2+4x-8y+12=0

vertices

x^{2} + 4 x - 8 y + 12 = 0

Minimum (- 2, 1)

Solution steps

x^{2} + 4 x - 8 y + 12 = 0

Rewrite

x^{2} + 4 x - 8 y + 12 = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{8} + \frac{1 \cdot x}{2} + \frac{3}{2}

The parabola parameters are:

a = \frac{1}{8}, b = \frac{1}{2}, c = \frac{3}{2}

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

x_{v} = - \frac{( \frac{1}{2} )}{2 ( \frac{1}{8} )}

Simplify

- \frac{\frac{1}{2}}{2 ( \frac{1}{8} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(- 2, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{8}

Minimum (- 2, 1)

y = - \frac{1}{4} x^{2} + 8 x - 4

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

- \frac{x ^{2}}{2} - 2 x

y = - x^{2} + 5

f (x) = x^{2} - x - 2

What is the vertices x^2+4x-8y+12=0 ?
The vertices x^2+4x-8y+12=0 is Minimum (-2,1)