What is the vertices y=2x^2-4x+15 ?

The vertices y=2x^2-4x+15 is Minimum (1,13)

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vertices y=2x^2-4x+15

vertices

y = 2 x^{2} - 4 x + 15

Minimum (1, 13)

Solution steps

y = 2 x^{2} - 4 x + 15

The parabola parameters are:

a = 2, b = - 4, c = 15

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

x_{v} = - \frac{( - 4 )}{2 \cdot 2}

Simplify

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = 13

Therefore the parabola vertex is

(1, 13)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (1, 13)

4 (x + 6) + 3 (2 + x + 3 x^{2}) - 2 x^{2}

f (x) = - x^{2} + 6 x - 11

y^{2} = - 20 x

2 x^{2} - 6 x + 15

(x + 1)^{2} = - 6 (y - 4)