What is the vertices+(x^2+x)/2 ?

The vertices+(x^2+x)/2 is Minimum (-1/2 ,-1/8)

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vertices+(x^2+x)/2

vertices

+ \frac{x ^{2} + x}{2}

Minimum (- \frac{1}{2}, - \frac{1}{8})

Solution steps

y = \frac{x ^{2} + x}{2}

Rewrite

y = + \frac{x ^{2} + x}{2}

in the form

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

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

The parabola parameters are:

a = \frac{1}{2}, b = \frac{1}{2}, c = 0

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

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

Simplify

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

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

Plug in

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

to find the

y_{v}

value

y_{v} = - \frac{1}{8}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{1}{8})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{2}

Minimum (- \frac{1}{2}, - \frac{1}{8})