What is the vertices f(x)=x^2-6x+1 ?

The vertices f(x)=x^2-6x+1 is Minimum (3,-8)

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vertices f(x)=x^2-6x+1

vertices

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

Minimum (3, - 8)

Solution steps

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

The parabola parameters are:

a = 1, b = - 6, c = 1

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

x_{v} = - \frac{( - 6 )}{2 \cdot 1}

Simplify

x_{v} = 3

Plug in

x_{v} = 3

to find the

y_{v}

value

y_{v} = - 8

Therefore the parabola vertex is

(3, - 8)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 8)

x^{2} + \frac{y ^{2}}{64} = 1

x^{2} - y = 0

y^{2} = - 4 x

9 x^{2} - 4 y^{2} - 90 x - 32 y = - 305

f (x) = 3 x^{2} - 6 x + 4

What is the vertices f(x)=x^2-6x+1 ?
The vertices f(x)=x^2-6x+1 is Minimum (3,-8)