What is the vertices 1/2 x^2+8x+5 ?

The vertices 1/2 x^2+8x+5 is Minimum (-8,-27)

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vertices 1/2 x^2+8x+5

vertices

\frac{1}{2} x^{2} + 8 x + 5

Minimum (- 8, - 27)

Solution steps

y = \frac{1}{2} x^{2} + 8 x + 5

Rewrite

y = \frac{1}{2} x^{2} + 8 x + 5

in the form

y = a x^{2} + b x + c

y = \frac{x ^{2}}{2} + 8 x + 5

The parabola parameters are:

a = \frac{1}{2}, b = 8, c = 5

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

x_{v} = - \frac{8}{2 ( \frac{1}{2} )}

Simplify

- \frac{8}{2 ( \frac{1}{2} )} : - 8

x_{v} = - 8

Plug in

x_{v} = - 8

to find the

y_{v}

value

y_{v} = - 27

Therefore the parabola vertex is

(- 8, - 27)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = \frac{1}{2}

Minimum (- 8, - 27)

y = 2 x^{2} - 12 x + 3

f (x) = x^{2} - 8 x + 15

f (x) = - x^{2} + 8 x

+ (y - 2) = 3 (x - 5)^{2}

y = (x + 1) (x + 2)

What is the vertices 1/2 x^2+8x+5 ?
The vertices 1/2 x^2+8x+5 is Minimum (-8,-27)