What is the vertices x^2+5x+6 ?

The vertices x^2+5x+6 is Minimum (-5/2 ,-1/4)

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vertices x^2+5x+6

vertices

x^{2} + 5 x + 6

Minimum (- \frac{5}{2}, - \frac{1}{4})

Solution steps

y = x^{2} + 5 x + 6

The parabola parameters are:

a = 1, b = 5, c = 6

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

x_{v} = - \frac{5}{2 \cdot 1}

Simplify

x_{v} = - \frac{5}{2}

Plug in

x_{v} = - \frac{5}{2}

to find the

y_{v}

value

y_{v} = - \frac{1}{4}

Therefore the parabola vertex is

(- \frac{5}{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{5}{2}, - \frac{1}{4})

f (x) = 2 x^{2} + 6 x - 1

y^{2} = 12 x

x^{2} + 6 x + 5

y = 3 x^{2} - 5 x 7

y^{2} = - 8 x