What is the vertices y=+0.5x^2-6x+21 ?

The vertices y=+0.5x^2-6x+21 is Minimum (6,3)

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vertices y=+0.5x^2-6x+21

vertices

y = + 0.5 x^{2} - 6 x + 21

Minimum (6, 3)

Solution steps

y = 0.5 x^{2} - 6 x + 21

The parabola parameters are:

a = 0.5, b = - 6, c = 21

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

x_{v} = - \frac{( - 6 )}{2 \cdot 0.5}

Simplify

x_{v} = 6

Plug in

x_{v} = 6

to find the

y_{v}

value

y_{v} = 3

Therefore the parabola vertex is

(6, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 0.5

Minimum (6, 3)

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

f (x) = x^{2} + 2 x + 5

f (x) = x^{2} + 2 x + 2

2 (x - 1)^{2} + 1

f (x) = (x - 7)^{2} - 1

What is the vertices y=+0.5x^2-6x+21 ?
The vertices y=+0.5x^2-6x+21 is Minimum (6,3)