What is the vertices f(x)=5x^2-10x+4 ?

The vertices f(x)=5x^2-10x+4 is Minimum (1,-1)

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vertices f(x)=5x^2-10x+4

vertices

f (x) = 5 x^{2} - 10 x + 4

Minimum (1, - 1)

Solution steps

y = 5 x^{2} - 10 x + 4

The parabola parameters are:

a = 5, b = - 10, c = 4

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

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

Simplify

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

Minimum (1, - 1)

(y - 9)^{2} = - 4 (x - 2)

(x + 2)^{2} = - 8 (y + 3)

y = x^{2} + 3 x + 4

y = 24 x - 3 x^{2}

(x - 5)^{2} = 4 (y - 3)

What is the vertices f(x)=5x^2-10x+4 ?
The vertices f(x)=5x^2-10x+4 is Minimum (1,-1)