What is the vertices f(x)=-2x^2+8x-5 ?

The vertices f(x)=-2x^2+8x-5 is Maximum (2,3)

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vertices f(x)=-2x^2+8x-5

vertices

f (x) = - 2 x^{2} + 8 x - 5

Maximum (2, 3)

Solution steps

y = - 2 x^{2} + 8 x - 5

The parabola parameters are:

a = - 2, b = 8, c = - 5

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} = 3

Therefore the parabola vertex is

(2, 3)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (2, 3)

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

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

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

(x - 4)^{2} + 2

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

What is the vertices f(x)=-2x^2+8x-5 ?
The vertices f(x)=-2x^2+8x-5 is Maximum (2,3)