What is the vertices x^2+x+1 ?

The vertices x^2+x+1 is Minimum (-1/2 , 3/4)

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vertices x^2+x+1

vertices

x^{2} + x + 1

Minimum (- \frac{1}{2}, \frac{3}{4})

Solution steps

y = x^{2} + x + 1

The parabola parameters are:

a = 1, b = 1, c = 1

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{3}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, \frac{3}{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{3}{4})

y = 2 x^{2} + 5 x - 12

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

- 2 x^{2} + 8 x - 1

y = x^{2} + 3 x

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