What is the vertices x^2=8(y+4) ?

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

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vertices x^2=8(y+4)

vertices

x^{2} = 8 (y + 4)

Minimum (0, - 4)

Solution steps

x^{2} = 8 (y + 4)

Rewrite

x^{2} = 8 (y + 4)

in the form

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

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

The parabola parameters are:

a = \frac{1}{8}, b = 0, c = - 4

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

x_{v} = - \frac{0}{2 ( \frac{1}{8} )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(0, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{8}

Minimum (0, - 4)

y = 3 x^{2} - 12 x + 8

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

y = x^{2} - 4 x + 7

x = (y - 3)^{2}

y^{2} = x + 1