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

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

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vertices y=x^2+x

vertices

y = x^{2} + x

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

Solution steps

y = x^{2} + x

The parabola parameters are:

a = 1, b = 1, c = 0

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{1}{4}

Therefore the parabola vertex is

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

y = 9 x^{2}

y = 8 x^{2} + 22 x - 13

y = x^{2} + 8 x + 18

x^{2} - 5 x - 6

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