What is the vertices f(x)=2x^2-80x+900 ?

The vertices f(x)=2x^2-80x+900 is Minimum (20,100)

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vertices f(x)=2x^2-80x+900

vertices

f (x) = 2 x^{2} - 80 x + 900

Minimum (20, 100)

Solution steps

y = 2 x^{2} - 80 x + 900

The parabola parameters are:

a = 2, b = - 80, c = 900

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

x_{v} = - \frac{( - 80 )}{2 \cdot 2}

Simplify

x_{v} = 20

Plug in

x_{v} = 20

to find the

y_{v}

value

y_{v} = 100

Therefore the parabola vertex is

(20, 100)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 2

Minimum (20, 100)

y = 5 x^{2}

f (x) = - (x + 3)^{2} + 5

4 t^{2}

x^{2} - 6 x - 4 y + 5 = 0

(x + 1)^{2} + 1

What is the vertices f(x)=2x^2-80x+900 ?
The vertices f(x)=2x^2-80x+900 is Minimum (20,100)