What is the vertices (x-4)^2=8y+16 ?

The vertices (x-4)^2=8y+16 is Minimum (4,-2)

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vertices (x-4)^2=8y+16

vertices

(x - 4)^{2} = 8 y + 16

Minimum (4, - 2)

Solution steps

(x - 4)^{2} = 8 y + 16

Rewrite

(x - 4)^{2} = 8 y + 16

in the form

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

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

The parabola parameters are:

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

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

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

Simplify

- \frac{- 1}{2 ( \frac{1}{8} )} : 4

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = - 2

Therefore the parabola vertex is

(4, - 2)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{8}

Minimum (4, - 2)

- 4 x^{2}

x^{2} - 4 x - 7

f (x) = - 3 x^{2} + 4 x - 1

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

f (x) = - 9.5 x^{2} - 47.5 x + 63

What is the vertices (x-4)^2=8y+16 ?
The vertices (x-4)^2=8y+16 is Minimum (4,-2)