What is the vertices x^2+10x-2y+23=0 ?

The vertices x^2+10x-2y+23=0 is Minimum (-5,-1)

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vertices x^2+10x-2y+23=0

vertices

x^{2} + 10 x - 2 y + 23 = 0

Minimum (- 5, - 1)

Solution steps

x^{2} + 10 x - 2 y + 23 = 0

Rewrite

x^{2} + 10 x - 2 y + 23 = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{2} + 5 x + \frac{23}{2}

The parabola parameters are:

a = \frac{1}{2}, b = 5, c = \frac{23}{2}

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

x_{v} = - \frac{5}{2 ( \frac{1}{2} )}

Simplify

- \frac{5}{2 ( \frac{1}{2} )} : - 5

x_{v} = - 5

Plug in

x_{v} = - 5

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 5, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{2}

Minimum (- 5, - 1)

+ x^{2} - 36

f (x) = 2 (x + 14)^{2} - 6

y = 3 x^{2} - 6 x + 5

x^{2} - 2

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

What is the vertices x^2+10x-2y+23=0 ?
The vertices x^2+10x-2y+23=0 is Minimum (-5,-1)