What is the vertices y=+8-9x^2 ?

The vertices y=+8-9x^2 is Maximum (0,8)

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vertices y=+8-9x^2

vertices

y = + 8 - 9 x^{2}

Maximum (0, 8)

Solution steps

y = 8 - 9 x^{2}

Rewrite

y = + 8 - 9 x^{2}

in the form

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

y = - 9 x^{2} + 8

The parabola parameters are:

a = - 9, b = 0, c = 8

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

x_{v} = - \frac{0}{2 ( - 9 )}

Simplify

x_{v} = 0

Plug in

x_{v} = 0

to find the

y_{v}

value

y_{v} = 8

Therefore the parabola vertex is

(0, 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 9

Maximum (0, 8)

2 π r^{2}

f (x) = - 3 x^{2} + 6 x

y = 14 (x - 4) (x + 10)

x^{2} + 8 x - 80

f (x) = x^{2} - 6 x - 16