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

The vertices f(x)=x^2-6x+2 is Minimum (3,-7)

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

vertices

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

Minimum (3, - 7)

Solution steps

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

The parabola parameters are:

a = 1, b = - 6, c = 2

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} = - 7

Therefore the parabola vertex is

(3, - 7)

a < 0,

then the vertex is a maximum value

a > 0,

then the vertex is a minimum value

a = 1

Minimum (3, - 7)

\frac{y ^{2}}{25} - \frac{x ^{2}}{16} = 1

\frac{( x + 3 ) ^{2}}{16} + \frac{( y - 5 ) ^{2}}{25} = 1

9 x^{2} + y^{2} = 9

y \geq x^{2}

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

What is the vertices f(x)=x^2-6x+2 ?
The vertices f(x)=x^2-6x+2 is Minimum (3,-7)