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

The vertices y=x^2+x+8 is Minimum (-1/2 , 31/4)

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vertices y=x^2+x+8

vertices

y = x^{2} + x + 8

Minimum (- \frac{1}{2}, \frac{31}{4})

Solution steps

y = x^{2} + x + 8

The parabola parameters are:

a = 1, b = 1, c = 8

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

x_{v} = - \frac{1}{2 \cdot 1}

Simplify

x_{v} = - \frac{1}{2}

Plug in

x_{v} = - \frac{1}{2}

to find the

y_{v}

value

y_{v} = \frac{31}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{31}{4})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (- \frac{1}{2}, \frac{31}{4})

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

2 x^{2} - 4 x + 7

x^{2} - 5 x - 84

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

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

What is the vertices y=x^2+x+8 ?
The vertices y=x^2+x+8 is Minimum (-1/2 , 31/4)