What is the vertices y=2x^2+12x+17 ?

The vertices y=2x^2+12x+17 is Minimum (-3,-1)

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vertices y=2x^2+12x+17

vertices

y = 2 x^{2} + 12 x + 17

Minimum (- 3, - 1)

Solution steps

y = 2 x^{2} + 12 x + 17

The parabola parameters are:

a = 2, b = 12, c = 17

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

x_{v} = - \frac{12}{2 \cdot 2}

Simplify

x_{v} = - 3

Plug in

x_{v} = - 3

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(- 3, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (- 3, - 1)

4 x^{2} + 8 x + 4

y = (x + 2)^{2} + 1

3 x^{2} + 6 x + 2

2 x^{2} - 3 x

f (x) = 4 (x - 2)^{2} + 11

What is the vertices y=2x^2+12x+17 ?
The vertices y=2x^2+12x+17 is Minimum (-3,-1)