What is the vertices+(-1)/2 x^2+18 ?

The vertices+(-1)/2 x^2+18 is Maximum (0,18)

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vertices+(-1)/2 x^2+18

vertices

+ \frac{- 1}{2} x^{2} + 18

Maximum (0, 18)

Solution steps

y = \frac{- 1}{2} x^{2} + 18

y = \frac{- 1}{2} x^{2} + 18

The parabola parameters are:

a = - \frac{1}{2}, m = 6, n = - 6

x_{v} = \frac{m + n}{2}

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

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 18

Therefore the parabola vertex is

(0, 18)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - \frac{1}{2}

Maximum (0, 18)

x^{2} - 4 x + 12

y = x^{2} - x

f (x) = 6 x^{2} - 1

f (x) = - 2 x^{2} + 16 x - 31

f (x) = - x^{2} - 6 x - 10

What is the vertices+(-1)/2 x^2+18 ?
The vertices+(-1)/2 x^2+18 is Maximum (0,18)