What is the vertices f(x)=5x^2-2x-1 ?

The vertices f(x)=5x^2-2x-1 is Minimum (1/5 ,-6/5)

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vertices f(x)=5x^2-2x-1

vertices

f (x) = 5 x^{2} - 2 x - 1

Minimum (\frac{1}{5}, - \frac{6}{5})

Solution steps

y = 5 x^{2} - 2 x - 1

The parabola parameters are:

a = 5, b = - 2, c = - 1

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

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

Simplify

- \frac{- 2}{2 \cdot 5} : \frac{1}{5}

x_{v} = \frac{1}{5}

Plug in

x_{v} = \frac{1}{5}

to find the

y_{v}

value

y_{v} = - \frac{6}{5}

Therefore the parabola vertex is

(\frac{1}{5}, - \frac{6}{5})

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 5

Minimum (\frac{1}{5}, - \frac{6}{5})

- 4 x^{2} + 6 x - 1

f (x) = x^{2} - 8 x + 18

y^{2} - 4 x - 4 = 0

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

(8 x - 23)^{2} - 10

What is the vertices f(x)=5x^2-2x-1 ?
The vertices f(x)=5x^2-2x-1 is Minimum (1/5 ,-6/5)