What is the vertices f(x)=-5x^2+10x-6 ?

The vertices f(x)=-5x^2+10x-6 is Maximum (1,-1)

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

vertices

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

Maximum (1, - 1)

Solution steps

y = - 5 x^{2} + 10 x - 6

The parabola parameters are:

a = - 5, b = 10, c = - 6

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

x_{v} = - \frac{10}{2 ( - 5 )}

Simplify

- \frac{10}{2 ( - 5 )} : 1

x_{v} = 1

Plug in

x_{v} = 1

to find the

y_{v}

value

y_{v} = - 1

Therefore the parabola vertex is

(1, - 1)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = - 5

Maximum (1, - 1)

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

(x - 4)^{2} + 2

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

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

y = 2 x^{2} + 15 x

What is the vertices f(x)=-5x^2+10x-6 ?
The vertices f(x)=-5x^2+10x-6 is Maximum (1,-1)