What is the vertices y=2x^2+10x+12 ?

The vertices y=2x^2+10x+12 is Minimum (-5/2 ,-1/2)

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vertices y=2x^2+10x+12

vertices

y = 2 x^{2} + 10 x + 12

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

Solution steps

y = 2 x^{2} + 10 x + 12

The parabola parameters are:

a = 2, b = 10, c = 12

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

x_{v} = - \frac{10}{2 \cdot 2}

Simplify

- \frac{10}{2 \cdot 2} : - \frac{5}{2}

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

Plug in

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

to find the

y_{v}

value

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

Therefore the parabola vertex is

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

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

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

x^{2} + 4 x - 21

x^{2} + 4 x - 12

- x^{2} + 2 x + 3

2 x^{2} + 4 x - 1

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

What is the vertices y=2x^2+10x+12 ?
The vertices y=2x^2+10x+12 is Minimum (-5/2 ,-1/2)