What is the vertices x^2+x-6 ?

The vertices x^2+x-6 is Minimum (-1/2 ,-25/4)

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vertices x^2+x-6

vertices

x^{2} + x - 6

Minimum (- \frac{1}{2}, - \frac{25}{4})

Solution steps

y = x^{2} + x - 6

The parabola parameters are:

a = 1, b = 1, c = - 6

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{25}{4}

Therefore the parabola vertex is

(- \frac{1}{2}, - \frac{25}{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{25}{4})

f (x) = 0.75 x^{2} - 5

x^{2} + 9 x + 4

5 x^{2} + 10 x + 6

+ x^{2} - 5

x^{2} + 8 x + 19