What is the vertices f(x)=-2x^2+16x-31 ?

The vertices f(x)=-2x^2+16x-31 is Maximum (4,1)

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vertices f(x)=-2x^2+16x-31

vertices

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

Maximum (4, 1)

Solution steps

y = - 2 x^{2} + 16 x - 31

The parabola parameters are:

a = - 2, b = 16, c = - 31

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

x_{v} = - \frac{16}{2 ( - 2 )}

Simplify

- \frac{16}{2 ( - 2 )} : 4

x_{v} = 4

Plug in

x_{v} = 4

to find the

y_{v}

value

y_{v} = 1

Therefore the parabola vertex is

(4, 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 2

Maximum (4, 1)

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

y = 2 x (x - 4) + 7

x = y^{2} + 2 y - 1 - (2 y^{2} - 3 y + 6)

f (x) = x^{2} + 6 x - 8

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

What is the vertices f(x)=-2x^2+16x-31 ?
The vertices f(x)=-2x^2+16x-31 is Maximum (4,1)