What is the vertices x^2+4x-24y-92=0 ?

The vertices x^2+4x-24y-92=0 is Minimum (-2,-4)

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vertices x^2+4x-24y-92=0

vertices

x^{2} + 4 x - 24 y - 92 = 0

Minimum (- 2, - 4)

Solution steps

x^{2} + 4 x - 24 y - 92 = 0

Rewrite

x^{2} + 4 x - 24 y - 92 = 0

in the form

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

y = \frac{1 \cdot x ^{2}}{24} + \frac{1 \cdot x}{6} - \frac{23}{6}

The parabola parameters are:

a = \frac{1}{24}, b = \frac{1}{6}, c = - \frac{23}{6}

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

x_{v} = - \frac{( \frac{1}{6} )}{2 ( \frac{1}{24} )}

Simplify

- \frac{\frac{1}{6}}{2 ( \frac{1}{24} )} : - 2

x_{v} = - 2

Plug in

x_{v} = - 2

to find the

y_{v}

value

y_{v} = - 4

Therefore the parabola vertex is

(- 2, - 4)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{24}

Minimum (- 2, - 4)

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

f (x) = - x^{2} + 7 x + 18

y = (2 x - 1) (5 x - 6)

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

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

What is the vertices x^2+4x-24y-92=0 ?
The vertices x^2+4x-24y-92=0 is Minimum (-2,-4)