What is the vertices x^2+8y+2x-23=0 ?

The vertices x^2+8y+2x-23=0 is Maximum (-1,3)

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vertices x^2+8y+2x-23=0

vertices

x^{2} + 8 y + 2 x - 23 = 0

Maximum (- 1, 3)

Solution steps

x^{2} + 8 y + 2 x - 23 = 0

Rewrite

x^{2} + 8 y + 2 x - 23 = 0

in the form

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

y = - \frac{x ^{2}}{8} - \frac{x}{4} + \frac{23}{8}

The parabola parameters are:

a = - \frac{1}{8}, b = - \frac{1}{4}, c = \frac{23}{8}

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

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

Simplify

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

x_{v} = - 1

Plug in

x_{v} = - 1

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(- 1, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{8}

Maximum (- 1, 3)

y = x^{2} - 10 x + 24

2 x^{2} - 5 x + 5

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

y^{2} = 20 x

y = 3 x^{2}

What is the vertices x^2+8y+2x-23=0 ?
The vertices x^2+8y+2x-23=0 is Maximum (-1,3)