What is the vertices f(x)=-2x^2+8x-9 ?

The vertices f(x)=-2x^2+8x-9 is Maximum (2,-1)

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vertices f(x)=-2x^2+8x-9

vertices

f (x) = - 2 x^{2} + 8 x - 9

Maximum (2, - 1)

Solution steps

y = - 2 x^{2} + 8 x - 9

The parabola parameters are:

a = - 2, b = 8, c = - 9

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

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

Simplify

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

x_{v} = 2

Plug in

x_{v} = 2

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(2, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (2, - 1)

f (x) = - 5 x^{2} + 10 x - 6

y = x^{2} - 4 x - 1

(x - 4)^{2} + 2

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

y = - x^{2} + 4 x + 5

What is the vertices f(x)=-2x^2+8x-9 ?
The vertices f(x)=-2x^2+8x-9 is Maximum (2,-1)