What is the vertices (x-3)^2=8(y+5) ?

The vertices (x-3)^2=8(y+5) is Minimum (3,-5)

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vertices (x-3)^2=8(y+5)

vertices

(x - 3)^{2} = 8 (y + 5)

Minimum (3, - 5)

Solution steps

(x - 3)^{2} = 8 (y + 5)

Rewrite

(x - 3)^{2} = 8 (y + 5)

in the form

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

y = \frac{1 \cdot x ^{2}}{8} - \frac{3 x}{4} + \frac{9}{8} - 5

The parabola parameters are:

a = \frac{1}{8}, b = - \frac{3}{4}, c = \frac{9}{8} - 5

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

x_{v} = - \frac{( - \frac{3}{4} )}{2 ( \frac{1}{8} )}

Simplify

- \frac{- \frac{3}{4}}{2 ( \frac{1}{8} )} : 3

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 5

Therefore the parabola vertex is

(3, - 5)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{8}

Minimum (3, - 5)

f (x) = - 3 x^{2} + 6 x - 4

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

f (x) = - 2 x^{2} - 12 x - 20

x^{2} + 6 x - 8 y + 1 = 0

y^{2} - 6 y - 8 x + 17 = 0

What is the vertices (x-3)^2=8(y+5) ?
The vertices (x-3)^2=8(y+5) is Minimum (3,-5)